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#1
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Mains transformer question.
I have now got the first channel of my KT88
50W monobloc amp up and running, powered by a lab psu. The B+ rails is at 500V. The amp draws 320mA when driven at full power continuous (20V into 8 Ohms) with a sine wave I plan to use a SS full wave bridge in the psu. and rate the mains transformer at 1.61 times the DC current drawn by the amp. It is highly unlikely that the amplifier will ever run at full power, but classical music requires plenty of headroom. So, should I rate the transformer at 1.61 x 320mA, or add some headroom to this, or specify a lower rating for the winding? Iain |
#2
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"Iain M Churches" wrote in message ... I have now got the first channel of my KT88 50W monobloc amp up and running, powered by a lab psu. The B+ rails is at 500V. The amp draws 320mA when driven at full power continuous (20V into 8 Ohms) with a sine wave I plan to use a SS full wave bridge in the psu. and rate the mains transformer at 1.61 times the DC current drawn by the amp. It is highly unlikely that the amplifier will ever run at full power, but classical music requires plenty of headroom. So, should I rate the transformer at 1.61 x 320mA, or add some headroom to this, or specify a lower rating for the winding? Iain Wouldn't it be prudent to check the amp at the full spectrum of audio frequencies? Current draw can widely vary with different frequencies. Just my $.02. west |
#3
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"west" wrote in message om... Wouldn't it be prudent to check the amp at the full spectrum of audio frequencies? Current draw can widely vary with different frequencies. Just my $.02. west Driving white noise into a dummy load, I get the same current requirement - 320mA. Your $.02. much appreciated :-) Iain |
#4
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"Iain M Churches" I have now got the first channel of my KT88 50W monobloc amp up and running, powered by a lab psu. The B+ rails is at 500V. The amp draws 320mA when driven at full power continuous (20V into 8 Ohms) with a sine wave ** Err - so you have built a power stage that is only 31 % efficient at full power ?? ( 50 / (500 x 0.32) = 0.31 ) Even pure, bloody class A is better. I plan to use a SS full wave bridge in the psu. and rate the mains transformer at 1.61 times the DC current drawn by the amp. It is highly unlikely that the amplifier will ever run at full power, but classical music requires plenty of headroom. ** Err - you have not supplied the range of DC supply current. So, should I rate the transformer at 1.61 x 320mA, or add some headroom to this, or specify a lower rating for the winding? ** The AC tranny needs to be of 250 + VA rating or more to run that electron sucking POS. ( 500 x 0.32 x 1.6 = 256 VA , plus heater current ) ................. Phil |
#5
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Iain M Churches wrote: I have now got the first channel of my KT88 50W monobloc amp up and running, powered by a lab psu. The B+ rails is at 500V. The amp draws 320mA when driven at full power continuous (20V into 8 Ohms) with a sine wave So it must be operating in mainly class B to make the 50 watts. A KT88 would have Ia = 60 mA at idle, and Ea = 500v, if you have fixed bias, which you should. So two tubes in class A would draw 120mA if the poer was pure class A, but for class AB, the input power rises, and the Ia could double, but you have more again. Its probably because the load the tubes are seeing at 50 watts is about 3k, rather low imho. I like to have the maximum power at clipping from a pair of KT88 at 54 watts at 4 ohms, and RL a-a is around 5k. Then at 8 ohms the power falls to 34 watts, but with a huge class A %, and the sound is good and it measures very well. I plan to use a SS full wave bridge in the psu. and rate the mains transformer at 1.61 times the DC current drawn by the amp. Allowing for 3 times the idle current is OK. In practice, a good class AB amp rarely ever gets out of class A, and spends 99.9% of its life with the power draw being nearly = to the idle condition. It is highly unlikely that the amplifier will ever run at full power, but classical music requires plenty of headroom. Then use large caps to store a big charge. No need to make the PT rated too high. But it should have Bmax = 0.9 tesla or less, and have the core rated for 3 times the max VA, then copper rated for twice the VA, then the PT will run quiet, and cool. Use only GOSS, or have a toroidal especially wound, varnished and potted. Plitron make some nice gear. So, should I rate the transformer at 1.61 x 320mA, or add some headroom to this, or specify a lower rating for the winding? 320 mA of draw is enormous for a pair of KT88. OK, so the drive amp might use 20 mA, but that still leaves 300 mA, and that's 150 mA per tube, or more than twice what i would use at idle, but some folks run KT88 at 40 watts, or at 80 mA of plate and screen current. Its too much. The total power needs in total have to be examined. Heater power :- KT88, 2 x 6.3 x 1.8 = 22.68VA input tubes, allow 3 x 6.3 x 0.6 = 11.34VA idle power, B+, = 500v x 2 x 80 mA, but allow for 500v x 2 x 160 mA = 160 VA, Input stage power, 500v x 20 mA = 10VA bias supply, 100v x 10 mA = 1 VA, Total VA = 205VA. Don't muck about, get a 300VA tranny which has very adequate ratings. This will mean a power tranny with 2.7" stack of 1.5" tongue E&I lams, so round that up to a 3" stack. Use only GOSS lams, lest the tranny run too warm. The Bmax should be under 0.9 T for low losses, but most importantly for quite op with a rectifier. the best possible regulation with the tranny isn't needed, since the amp will operate mostly class A and its rare that you will ever seen the EA change more than a volt or 3 even with a high resistance tube rectifier. Have a CLC filter before the CT, with series 470 uF x 350v rated caps, so you have at least 235 uF, about 3H, 235 uF, and that should be OK. If dcr of the choke is less than 30 ohms, allow the loaded vrms of the B+ winding to be 0.74 tims the B+ proposed, so for +500v, the winding is 370v. I have often used a doubler, and a 190 volt winding with thicker wire. Have a few taps on the winding at about 20vrms postions to allow B+ = 528v, 500v, 472v, 444v and 416v, so that if other tubes are selected such as EL34 or 6L6, you are not stuck. Then if you were sensible, and wanted more class A % in the output power at the slight expense of maximum class AB power, then you would have B+ = 416v, or use cathode bias with say B+ = 444v, so Ea would be around 410v, and raise the Ia to 78 mA, and retain the load value you have now, and settle for the outcome with greater fidelity for the same amount of NFB. Seen this? http://www.turneraudio.com.au/htmlwe...0ulabinteg.htm I know the SE triode tribe hates this sort of amp but I don't care a hoot, I know it sounds fantabulistic. If you are doing mono amp chassis, with a separate PS, that PS can have 600VA tranny, which will not be much larger than a 300VA, and then run an umbilical cable to each amp chassis. For the cables, use no less than a decent heavy duty mobile crane 5 way cable with at least five 5 amp wires and very flexible. I generally hard wire the cable into the amp and have an octal socket to fit into a recessed socket hole on the PS, so the accidental drag or yank on the cable won't **** up the cable or plug. The 5 way wire allows for B+, B-, OV, and heaters. On the octal socket&plug, 4 of the 8 pins should be used for the high heater current, and that leaves 4 for the other things. If you get some 7 way cable with each wire rated at 3 amps, heaters and the rest can still be catered for without hot cable and losses. There are military grade plugs and sockets, but I have never used them, and they cost heaps. Just make sure the male plug is on a cable from the amp, not the PS, lest you have the PS turned on and exposed lethal B+ if you have forgotten to plug in a cable. Its good idea to araldite plates to the plug, and screw it to the PS chassis to prevent a child pulling out a cable and sucking it before the B+ has subsided. Patrick Turner. Iain |
#6
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Iain M Churches wrote: "west" wrote in message om... Wouldn't it be prudent to check the amp at the full spectrum of audio frequencies? Current draw can widely vary with different frequencies. Just my $.02. west Driving white noise into a dummy load, I get the same current requirement - 320mA. I doubt white noise is the right stuff for a test like this; certainly it is uneccessary. Pink noise I have used, and the amp will start to clip on peaks in the noise at a power output level well below what is gained with a sine wave at clipping. Pink noise is like a music signal, and again, busy rock and roll that has been compressed will give a power output about 1/2 the sine wave power. Full range pink noise will also stimulate the OPT to saturate with large LF signals, and cause a knocking noise, and current pulses in the OPVs, and the HF part of the noise will perhaps saturate the amp with HF, depending on the amp quality. A sine wave at 1 kHz is fine. Patrick Turner. Your $.02. much appreciated :-) Iain |
#7
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Phil Allison wrote: "Iain M Churches" I have now got the first channel of my KT88 50W monobloc amp up and running, powered by a lab psu. The B+ rails is at 500V. The amp draws 320mA when driven at full power continuous (20V into 8 Ohms) with a sine wave ** Err - so you have built a power stage that is only 31 % efficient at full power ?? ( 50 / (500 x 0.32) = 0.31 ) Even pure, bloody class A is better. Agreed. It's OK Phil. I have him thinking about the loading in another post. I think he has RLa-a rather too low. I also think that if there was some way to alter the load match on his OPTs, he ought to investigate. SS amps can be up to 78% efficient at full power, class aB. But tube amps rarely more than 66%, and class A pentode amps at about 40%. Class AB UL at full power should reach about 50%. I plan to use a SS full wave bridge in the psu. and rate the mains transformer at 1.61 times the DC current drawn by the amp. It is highly unlikely that the amplifier will ever run at full power, but classical music requires plenty of headroom. ** Err - you have not supplied the range of DC supply current. So, should I rate the transformer at 1.61 x 320mA, or add some headroom to this, or specify a lower rating for the winding? ** The AC tranny needs to be of 250 + VA rating or more to run that electron sucking POS. ( 500 x 0.32 x 1.6 = 256 VA , plus heater current ) I said 300VA trannies are appropriate... A 600VA tranny for both channels is better, since the larger tranny can have better regulation than a pair of trannies. Patrick Turner. ................ Phil |
#8
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"Patrick Turner" wrote in message ... Iain M Churches wrote: "west" wrote in message om... Wouldn't it be prudent to check the amp at the full spectrum of audio frequencies? Current draw can widely vary with different frequencies. Just my $.02. west Driving white noise into a dummy load, I get the same current requirement - 320mA. I doubt white noise is the right stuff for a test like this; certainly it is uneccessary. I am wondering if the fig of 320mA is correct. The Solartron psu has a single meter which can be switched between voltage and current. Perhaps I should check by measuring the voltage drop across a series 1 Ohm resistor. Iain |
#9
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"Patrick Turner" wrote in message ... I said 300VA trannies are appropriate... A 600VA tranny for both channels is better, since the larger tranny can have better regulation than a pair of trannies. This amp is the first of a pair of monoblocs, so each will have its own mains transformer. Iain |
#10
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I now have a little more info regarding the prototype 50W amp. The OPT has Ra-a of 4.5k and rated at 100W. The meter on the Solartron bench psu is difficult to read. The voltage scale is clear, as it is printed in black, but the ammeter scale in in red, and faded over the years. So, I put a 1R 5W resistor in series with the psu, with a DVM across it. The amp has fixed bias. I have set each of the KT88's to draw 60mA idle. The idle current is measured across a 33 Ohm resistor to give 2V. (2/33 = 0.060) Now here's the new bit. With a sine wave input and driving at 50W (20V into 8 Ohms) the total current drawn is 240mA (not 320mA as I thought earlier) So that's an efficiency of 50/(500 x 0.240) = 42% Better? So, to ask my question again, "At what current rating should I specify the HT winding on the transformer secondary" Thanks, Iain |
#11
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On Tue, 15 Mar 2005 10:49:41 +0200, "Iain M Churches" wrote:
I have now got the first channel of my KT88 50W monobloc amp up and running, powered by a lab psu. The B+ rails is at 500V. The amp draws 320mA when driven at full power continuous (20V into 8 Ohms) with a sine wave Are you sure about that 320ma??? Sounds awful high to me! The last 50 watt amp I made ran at closer to 120ma... Maybe your load is a bit off? Or your meter? I plan to use a SS full wave bridge in the psu. and rate the mains transformer at 1.61 times the DC current drawn by the amp. According to my Fairchild PS book, the transformer VA could be 2.16 PO with a capacitive loaded bridge, no CT. Not sure what the peak current would be... maybe you could calculate it. It is highly unlikely that the amplifier will ever run at full power, but classical music requires plenty of headroom. With music, you probably will fill the peak demands with energy from the filter caps, so you may want to make these quite large. The average current from the tranny will probably be low, I forget the percent that music requires. Of course, it could depend on the music as well! I build guitar amps, and they have a more stringent current requirement, since they run at 11 all the time! So, should I rate the transformer at 1.61 x 320mA, or add some headroom to this, or specify a lower rating for the winding? Iain You realize you're talking about a 500ma transformer in a 50 watt amp!?!? I have a huge Hammond here, the 475ma one, and I have it reserved for a 200 watt stereo... I'd suggest you run some more tests! Good luck! |
#12
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"Iain M Churches" I have set each of the KT88's to draw 60mA idle. The idle current is measured across a 33 Ohm resistor to give 2V. (2/33 = 0.060) Now here's the new bit. With a sine wave input and driving at 50W (20V into 8 Ohms) the total current drawn is 240mA (not 320mA as I thought earlier) ** Is IMC measuring the voltage drop with an average responding meter or a "true rms" one ?? And does he have the slightest idea of the difference ??? ............ Phil |
#13
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"Bob" wrote in message ... Are you sure about that 320ma??? No. See my second post. I checked the current drawn with two other meters (Fluke) and they give the total as 240mA. (0.240V dropped across a 1 Ohm resistor) Maybe your load is a bit off? Or your meter? The load is a 100W wirewound. The transformer Ra-a = 4.5k With the new reading of 240mA, that makes the efficiency about 42%, which is probably OK. Patrick has suggested that the transformer should be specificied at 3x idling current, which seems like a good idea. Iain |
#14
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Iain M Churches wrote: I now have a little more info regarding the prototype 50W amp. The OPT has Ra-a of 4.5k and rated at 100W. How do you work that out, bearing in mind that your other statements about your amp were a bit wonky to start with? The meter on the Solartron bench psu is difficult to read. The voltage scale is clear, as it is printed in black, but the ammeter scale in in red, and faded over the years. So, I put a 1R 5W resistor in series with the psu, with a DVM across it. The amp has fixed bias. I have set each of the KT88's to draw 60mA idle. The idle current is measured across a 33 Ohm resistor to give 2V. (2/33 = 0.060) Now here's the new bit. With a sine wave input and driving at 50W (20V into 8 Ohms) the total current drawn is 240mA (not 320mA as I thought earlier) Where did you measure that? with 33 ohms between each k and 0V? What allowance have you made for the signal voltage present at each k? What if you had a 10 ohm R in series with the CT, but before the cap to 0V, so that plate current has virtually no signal present? What of the screen dissipation? So that's an efficiency of 50/(500 x 0.240) = 42% Better? Not including the screen current, you'd get a better calculated plate Is the amp UL, if so, what part in the efficiency measure does the screen play? And BTW, at 50 watts, the B+ will have dropped, and power input to the plates is less than 500v at idle, so power input at 50 watts won't be 500 x 0.24 amps, but maybe 470 x 0.24. To know reality, be conscious of variations of the conditions where it is measured. IE, when conditions change always *expect* to have to make allowances in your measurements. So, to ask my question again, "At what current rating should I specify the HT winding on the transformer secondary" The rating can be for the maximum current draw condition for the load that produces the most power. Its not a strict design premise. If you observe the wave output wave form you will see when clipping just begins. That's where thd = approx 2%. That's where max PO is usually measured. Then plot a graph of power output vs load value using your dummy loads of 1,2,3,4,5,6,8,12,16, 24, and 32 ohms. This graph will show a steep eise of PO from o ohms to say a peak at 4 ohms, then a sag downwards towards zero PO as RL approaches very high value. Then where is the peak in the PO? Is it at 3 ohms, 5 ohms, 8 ohms? Is it where you want it? If you have speakers nominally 8 ohms, you should assume part of the Z will be at 6 ohms or even lower, while the average is maybe 8 ohms across the band. If I know a speaker is said to be 8, I still set the amp up to ideally suit a 5.6 ohm load, which is 1/2 way between 4 and 8 ohms. Then any speaker between 3 and 16 ohms is fine. There are almost no 16 ohm speakers around now but where there are they are old and also usually sensitive, so it doesn't matter if the amp is "set to a low Z match" for such speakers. Modern speakers are often nominally 6 to 8 ohms, often with a did to maybe 4 ohnms, so you want your amp to cope, and be able to give the extra current grunt without lots of distortion, so the choice of 5.6 ohms is not absurd. I usually set up my PP amps to make max PO into about 3 ohms, below which PO rolls off if we keep to the 1% thd rule. It means that the RLa-a has to be carefully chosen for that and the turn ratio, and hence impedance ratio depends on the B+, and class. There is some reading material about EH6550, which are identical to KT88 at http://www.turneraudio.com.au/htmlwe...50pographs.htm At the KT88 UL schematic at http://www.turneraudio.com.au/htmlwe...0ulabinteg.htm I state I get 53 watts at 4 ohms, 35 watts at 8 ohms. But there would be slightly more than 53 watts at 3 ohms. The UL% = 50%, and the PO is less than pure beam tetrode. But its OK, because AB tetrode has far higher thd and Rout, and I don't need 80 watts. In the UL amp of mine the RLa-a = approximately 9k when 8 ohms, 4.5k with 4 ohms, and so with 3 ohms, RLa-a = is used For pure class A operation, and Pd at 30 watts for each tube, Ia quiescent = 60 mA, Ea = 500v. When you plot the load lines for a single tube at 60 mA x 500v, and in tetrode or UL mode, theoretical maximum class A is where RL = 7k approx. The -ve and +ve V swings are around 450v either side of the 500v quiescent point which won't move when class A is used, because Pin stays the same for class A. Where you have a PP circuit, for pure class A, each of the tubes is working with a 7 k load, so the RLa-a is 14k, and voltage swing is 900 peak volts a-a, so we get 636 vrms, and that gives us 28.9 watts of pure class A. Pin = 2 x 500 x 0.06 = 60 watts, so class A plate efficiency ( neglecting any effects the screens may have, since they are minor, ) is 100% x ( 28.9 / 60 ) = 48% which is near the theroetcal maximum amount of class A ever available from any pure class A circuit. Losses in the OPT of 10% would reduce the secondary load output power to 26 watts, so efficiency is 43%. Plate efficiency is plate efficiency, and should be measured in the plate circuit, lest we measure plate efficiency wrongly by including the OPT. So if you don't know the winding R, you have to have 10 ohm plate current sensors, and so the measuremts carefully. Its easier to simply ascertain the general amp efficiency with OPT losses included, which saves us having to measure distorted wave forms in the plate currents when examining class AB operation. So, a surprisingly high load value is needed for pure class A if we sart with Ea = 500v and Ia = 60 mA for each tube. As soon as we reduce Ia at idle, the RLa-a for pure class A has to be increased, and if Ia is increased, RLa-a can be reduced and still have all class A. With all class A the circuit gain is highest, and applied NFB is the greatest, so thd will perhaps be 0.07% at 28 watts, not a bad result. With a 14k a-a load, even triode will give 25 watts, and we don't need so much global NFB and the Rout, thd will still be just as low. But nobody is much impressed with this low power hi-fi idea, they say they must have 50 watts. Or 100. The tetrode plate resistance line on the KT88/6550 data sheet for where Eg1 = 0V is a curve that starts at 0V and 0amps, and rises nearly straight to 0.35 amps at Ea = 80v, then swings down to near flat to pass through 0.4 amps at 500v. The "corner" of the grapgh is the called the knee, and the load lines for all proposed loads for AB1 should intersect the Ra line to the left of the knee. Let us suppose we have 5k a-a as a load for the UL amp with the KT88 and 500v and 60 mA. Now to determine the v swing, imagine we had the amp working in class B. The loadline condition can be drawn for each tube separately, since to draw composite load line for both tubes is utterly confusing to a beginner. So I never do that. Back to the single tube, biased for class B. For class B, the tube being turned off by a -ve going Eg1, its as if it was not in the amp. The one that is being turned on by the +ve Eg1 going voltage Is connected to only 1/2 the primary, and thus sees a load which is 1/4 of the class A a-a load. So if RL a-a was a nominal 5k, then the load for one tube in class B is 5,000 / 4 = 1,250 ohms. Since the class B idle cindition has Ia = 0amps, we can draw the loadline for the tube from the 0.0 amps axis and from the Ea = 500v spot. 500v / 1,250 = 0.4 amps on the verical Ia axis, and we can dtaw a line there. It will intersect the Ra tube line at about Ea = 80v and Ia = 0.35 amps. Regardless of whatever class A % of power produced, there cannot be any greater load voltage swing than indicated by the class B load line for 1 tube, in this case, 500v - 80v, = 420 peakv The same thing happens with two tubes in a PP circuit, and the max swing for the 5 ka-a load is 2 x 840 peakv = 594v. So the maximum power into 5k is ( 594 x 594 ) / 5,000 = 70.5 watts. This is further spelled out down the page for the 6550 at my website at the above url. The initial pure class A watts can be calculated by loadline analysis and plotting class A load lines on the same graph for the class B situation, which BTW is *never actually used* but unless one becomes quite used to imaginary conditions for tube operation, and models of perfect tubes, you won't learn anything fast. There is always some plate current, and class B is impossible to arrange without enormous amounts of unwanted X-over distortion. If the load is increased from 5k, the PO will decrease, and if the load is reduced to say 3k a-a, then the class B load line will become more vertical, and pass through Eg1 = 0V at a point to the right of the knee in the curve. This usually then brings the load line into the region where Pd for the tube is greater than the limit, for 6550 its 42 watts, and a line should be drawn on the plate curves indicating this, and where load lines should never be placed. If RLa-a = 3k, the class B load line = 750 ohms, and the class B loadline runs from 500v to 0.666 amps, and well into the "more than 42 watts Pd area". The load swing for one tube will be from 500v down to 200v, so 300peakv, so 600pv across 3k and that gives 60 watts. So the power peaks at some load value between 5k and 3k. At that peak, say 4.5k, is where i would match the amp to a 3 ohm load. With UL operation, i won't get the 70 watts for tetrode, and nor will you. For 100 watts from a pair of KT88/6550, even in tetrode mode, higher Ea is needed, and/or class AB2 operation needed. Then things get complex, the reliability suffers, and thd skyrockets. Far better to use 4 output tubes if over 50 watts is expected for hi-fi. So where we have and OPT to give 4.5k to 3 ohms, the ZR = 1,500:1, and to match to 5.6 gives 8.4k a-a, and for UL we get 48 watts, with the first 16 watts in class A. For 8 ohms, the RL a-a becomes 12.0 k, and PO = 37 watts, with the first 24 watts in pure class A. I have found all other ways to match loads to tubes to be wrong for me. If you use the above load matching technique and ideology, and you have about 16 db of global NFB when 8 ohms is connected, then expect Ro to be under 0.5 ohms, and thd under 0.2% at 37 watts, and at 2 watts, expect about 0.02% thd, and even slightly lower Rout. Operation with triodes is possible with the same load values as I like to use with UL. As for the PT rating, follow the ideas in the last detailed post I sent; I hate having to repeat myself. Always use a far bigger VA rating than the actual VA likely to be drawn. Bean counters pair everything down to save a penny here and there and everyfukkinwhere, but such folks could never find employment at this establishment. The last thing the diyer/hobbiest person needs is to allow penny pinching ideologies to stuff the righteousness of a design. So don't think like the guys in the office at Quad, Leak, Radford, Dynaco et all; Think BETTER. Spread the wings of your mind and fly!. Patrick Turner. Thanks, Iain |
#15
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On Tue, 15 Mar 2005 10:49:41 +0200, Iain M Churches wrote:
So, should I rate the transformer at 1.61 x 320mA, or add some headroom to this, or specify a lower rating for the winding? Generally speaking, power transformers are rated based on their temperature rise at full load over a long period of continuous use (like between 8 to 24 hours). For shorter periods or on peak demand they can normally supply somewhat more than the rated current, how much more depends on how conservatively they're rated, ie: what temperature rise is considered acceptable at rated load. 320 ma is 160 watts of plate input power, sounds like plenty of headroom for a 50 watt amp! -- Ned Carlson Triode Electronics Chicago,IL USA www.triodeelectronics.com |
#16
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In message , Phil Allison
writes "Iain M Churches" I have set each of the KT88's to draw 60mA idle. The idle current is measured across a 33 Ohm resistor to give 2V. (2/33 = 0.060) Now here's the new bit. With a sine wave input and driving at 50W (20V into 8 Ohms) the total current drawn is 240mA (not 320mA as I thought earlier) ** Is IMC measuring the voltage drop with an average responding meter or a "true rms" one ?? And does he have the slightest idea of the difference ??? ........... Phil He's using a sine wave, so if the meter is correctly calibrated it won't matter. Now with pink noise it's another thing altogether. -- Chris Morriss |
#17
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"Chris Morriss" wrote
** Is IMC measuring the voltage drop with an average responding meter or a "true rms" one ?? He's using a sine wave, so if the meter is correctly calibrated it won't matter. But he's measuring supply current. Quite *where* he's measuring is not very clear. Not much of all this is making sense to me. To optimise a mains transformer, the load on the transformer itself must be known. Do we know what kind of rectification is being used? I tend to miss stuff if there's too much crap in a post. Iain, have you tried the free power supply modelling program at http://www.duncanamps.com/software.html? cheers, Ian |
#18
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"Chris Morriss" Phil Allison "Iain M Churches" I have set each of the KT88's to draw 60mA idle. The idle current is measured across a 33 Ohm resistor to give 2V. (2/33 = 0.060) Now here's the new bit. With a sine wave input and driving at 50W (20V into 8 Ohms) the total current drawn is 240mA (not 320mA as I thought earlier) ** Is IMC measuring the voltage drop with an average responding meter or a "true rms" one ?? And does he have the slightest idea of the difference ??? He's using a sine wave, * The DC supply current to a class AB tube amplifier is **NOT** sine wave - dickhead. so if the meter is correctly calibrated it won't matter. ** Better go work out what that wave shape is - dickhead. Now with pink noise it's another thing altogether. ** **** off. ............... Phil |
#19
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Patrick Turner wrote:
Iain M Churches wrote: I now have a little more info regarding the prototype 50W amp. The OPT has Ra-a of 4.5k and rated at 100W. How do you work that out, bearing in mind that your other statements about your amp were a bit wonky to start with? The meter on the Solartron bench psu is difficult to read. The voltage scale is clear, as it is printed in black, but the ammeter scale in in red, and faded over the years. So, I put a 1R 5W resistor in series with the psu, with a DVM across it. The amp has fixed bias. I have set each of the KT88's to draw 60mA idle. The idle current is measured across a 33 Ohm resistor to give 2V. (2/33 = 0.060) Now here's the new bit. With a sine wave input and driving at 50W (20V into 8 Ohms) the total current drawn is 240mA (not 320mA as I thought earlier) Where did you measure that? with 33 ohms between each k and 0V? What allowance have you made for the signal voltage present at each k? What if you had a 10 ohm R in series with the CT, but before the cap to 0V, so that plate current has virtually no signal present? What of the screen dissipation? So that's an efficiency of 50/(500 x 0.240) = 42% Better? Not including the screen current, you'd get a better calculated plate Is the amp UL, if so, what part in the efficiency measure does the screen play? And BTW, at 50 watts, the B+ will have dropped, and power input to the plates is less than 500v at idle, so power input at 50 watts won't be 500 x 0.24 amps, but maybe 470 x 0.24. To know reality, be conscious of variations of the conditions where it is measured. IE, when conditions change always *expect* to have to make allowances in your measurements. So, to ask my question again, "At what current rating should I specify the HT winding on the transformer secondary" The rating can be for the maximum current draw condition for the load that produces the most power. Its not a strict design premise. If you observe the wave output wave form you will see when clipping just begins. That's where thd = approx 2%. That's where max PO is usually measured. Then plot a graph of power output vs load value using your dummy loads of 1,2,3,4,5,6,8,12,16, 24, and 32 ohms. This graph will show a steep eise of PO from o ohms to say a peak at 4 ohms, then a sag downwards towards zero PO as RL approaches very high value. Then where is the peak in the PO? Is it at 3 ohms, 5 ohms, 8 ohms? Is it where you want it? If you have speakers nominally 8 ohms, you should assume part of the Z will be at 6 ohms or even lower, while the average is maybe 8 ohms across the band. If I know a speaker is said to be 8, I still set the amp up to ideally suit a 5.6 ohm load, which is 1/2 way between 4 and 8 ohms. Then any speaker between 3 and 16 ohms is fine. There are almost no 16 ohm speakers around now but where there are they are old and also usually sensitive, so it doesn't matter if the amp is "set to a low Z match" for such speakers. Modern speakers are often nominally 6 to 8 ohms, often with a did to maybe 4 ohnms, so you want your amp to cope, and be able to give the extra current grunt without lots of distortion, so the choice of 5.6 ohms is not absurd. I usually set up my PP amps to make max PO into about 3 ohms, below which PO rolls off if we keep to the 1% thd rule. It means that the RLa-a has to be carefully chosen for that and the turn ratio, and hence impedance ratio depends on the B+, and class. There is some reading material about EH6550, which are identical to KT88 at http://www.turneraudio.com.au/htmlwe...50pographs.htm At the KT88 UL schematic at http://www.turneraudio.com.au/htmlwe...0ulabinteg.htm I state I get 53 watts at 4 ohms, 35 watts at 8 ohms. But there would be slightly more than 53 watts at 3 ohms. The UL% = 50%, and the PO is less than pure beam tetrode. But its OK, because AB tetrode has far higher thd and Rout, and I don't need 80 watts. In the UL amp of mine the RLa-a = approximately 9k when 8 ohms, 4.5k with 4 ohms, and so with 3 ohms, RLa-a = is used For pure class A operation, and Pd at 30 watts for each tube, Ia quiescent = 60 mA, Ea = 500v. When you plot the load lines for a single tube at 60 mA x 500v, and in tetrode or UL mode, theoretical maximum class A is where RL = 7k approx. The -ve and +ve V swings are around 450v either side of the 500v quiescent point which won't move when class A is used, because Pin stays the same for class A. Where you have a PP circuit, for pure class A, each of the tubes is working with a 7 k load, so the RLa-a is 14k, and voltage swing is 900 peak volts a-a, so we get 636 vrms, and that gives us 28.9 watts of pure class A. Pin = 2 x 500 x 0.06 = 60 watts, so class A plate efficiency ( neglecting any effects the screens may have, since they are minor, ) is 100% x ( 28.9 / 60 ) = 48% which is near the theroetcal maximum amount of class A ever available from any pure class A circuit. Losses in the OPT of 10% would reduce the secondary load output power to 26 watts, so efficiency is 43%. Plate efficiency is plate efficiency, and should be measured in the plate circuit, lest we measure plate efficiency wrongly by including the OPT. So if you don't know the winding R, you have to have 10 ohm plate current sensors, and so the measuremts carefully. Its easier to simply ascertain the general amp efficiency with OPT losses included, which saves us having to measure distorted wave forms in the plate currents when examining class AB operation. So, a surprisingly high load value is needed for pure class A if we sart with Ea = 500v and Ia = 60 mA for each tube. As soon as we reduce Ia at idle, the RLa-a for pure class A has to be increased, and if Ia is increased, RLa-a can be reduced and still have all class A. With all class A the circuit gain is highest, and applied NFB is the greatest, so thd will perhaps be 0.07% at 28 watts, not a bad result. With a 14k a-a load, even triode will give 25 watts, and we don't need so much global NFB and the Rout, thd will still be just as low. But nobody is much impressed with this low power hi-fi idea, they say they must have 50 watts. Or 100. The tetrode plate resistance line on the KT88/6550 data sheet for where Eg1 = 0V is a curve that starts at 0V and 0amps, and rises nearly straight to 0.35 amps at Ea = 80v, then swings down to near flat to pass through 0.4 amps at 500v. The "corner" of the grapgh is the called the knee, and the load lines for all proposed loads for AB1 should intersect the Ra line to the left of the knee. Let us suppose we have 5k a-a as a load for the UL amp with the KT88 and 500v and 60 mA. Now to determine the v swing, imagine we had the amp working in class B. The loadline condition can be drawn for each tube separately, since to draw composite load line for both tubes is utterly confusing to a beginner. So I never do that. Back to the single tube, biased for class B. For class B, the tube being turned off by a -ve going Eg1, its as if it was not in the amp. The one that is being turned on by the +ve Eg1 going voltage Is connected to only 1/2 the primary, and thus sees a load which is 1/4 of the class A a-a load. So if RL a-a was a nominal 5k, then the load for one tube in class B is 5,000 / 4 = 1,250 ohms. Since the class B idle cindition has Ia = 0amps, we can draw the loadline for the tube from the 0.0 amps axis and from the Ea = 500v spot. 500v / 1,250 = 0.4 amps on the verical Ia axis, and we can dtaw a line there. It will intersect the Ra tube line at about Ea = 80v and Ia = 0.35 amps. Regardless of whatever class A % of power produced, there cannot be any greater load voltage swing than indicated by the class B load line for 1 tube, in this case, 500v - 80v, = 420 peakv The same thing happens with two tubes in a PP circuit, and the max swing for the 5 ka-a load is 2 x 840 peakv = 594v. So the maximum power into 5k is ( 594 x 594 ) / 5,000 = 70.5 watts. This is further spelled out down the page for the 6550 at my website at the above url. The initial pure class A watts can be calculated by loadline analysis and plotting class A load lines on the same graph for the class B situation, which BTW is *never actually used* but unless one becomes quite used to imaginary conditions for tube operation, and models of perfect tubes, you won't learn anything fast. There is always some plate current, and class B is impossible to arrange without enormous amounts of unwanted X-over distortion. If the load is increased from 5k, the PO will decrease, and if the load is reduced to say 3k a-a, then the class B load line will become more vertical, and pass through Eg1 = 0V at a point to the right of the knee in the curve. This usually then brings the load line into the region where Pd for the tube is greater than the limit, for 6550 its 42 watts, and a line should be drawn on the plate curves indicating this, and where load lines should never be placed. If RLa-a = 3k, the class B load line = 750 ohms, and the class B loadline runs from 500v to 0.666 amps, and well into the "more than 42 watts Pd area". The load swing for one tube will be from 500v down to 200v, so 300peakv, so 600pv across 3k and that gives 60 watts. So the power peaks at some load value between 5k and 3k. At that peak, say 4.5k, is where i would match the amp to a 3 ohm load. With UL operation, i won't get the 70 watts for tetrode, and nor will you. For 100 watts from a pair of KT88/6550, even in tetrode mode, higher Ea is needed, and/or class AB2 operation needed. Then things get complex, the reliability suffers, and thd skyrockets. Far better to use 4 output tubes if over 50 watts is expected for hi-fi. So where we have and OPT to give 4.5k to 3 ohms, the ZR = 1,500:1, and to match to 5.6 gives 8.4k a-a, and for UL we get 48 watts, with the first 16 watts in class A. For 8 ohms, the RL a-a becomes 12.0 k, and PO = 37 watts, with the first 24 watts in pure class A. I have found all other ways to match loads to tubes to be wrong for me. If you use the above load matching technique and ideology, and you have about 16 db of global NFB when 8 ohms is connected, then expect Ro to be under 0.5 ohms, and thd under 0.2% at 37 watts, and at 2 watts, expect about 0.02% thd, and even slightly lower Rout. Operation with triodes is possible with the same load values as I like to use with UL. As for the PT rating, follow the ideas in the last detailed post I sent; I hate having to repeat myself. Always use a far bigger VA rating than the actual VA likely to be drawn. Bean counters pair everything down to save a penny here and there and everyfukkinwhere, but such folks could never find employment at this establishment. The last thing the diyer/hobbiest person needs is to allow penny pinching ideologies to stuff the righteousness of a design. So don't think like the guys in the office at Quad, Leak, Radford, Dynaco et all; Think BETTER. Spread the wings of your mind and fly!. Patrick Turner. Thanks, Iain Anyone can end up with something by simply throwing money at it, perhaps even quality. A real designer knows how to do that while keeping cost under control. Working inside a cost target is something achieved only by professionals. JLS |
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"John Stewart" Anyone can end up with something by simply throwing money at it, perhaps even quality. ** That would be a fair definition of what "boutique" audio businesses are all about. A real designer knows how to do that while keeping cost under control. Working inside a cost target is something achieved only by professionals. ** Didn't President Herbert Hoover once say ..... " an engineer can do for 50 cents what any damn fool can do for a dollar " . ................ Phil |
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Snip what I said Iain said except :- The last thing the diyer/hobbiest person needs is to allow penny pinching ideologies to stuff the righteousness of a design. So don't think like the guys in the office at Quad, Leak, Radford, Dynaco et all; Think BETTER. Spread the wings of your mind and fly!. Patrick Turner. Thanks, Iain Anyone can end up with something by simply throwing money at it, perhaps even quality. A real designer knows how to do that while keeping cost under control. Working inside a cost target is something achieved only by professionals. JLS The hobbyist places no monetary value on his creative time, which might be otherwise wasted by watching tele, going to the pub. So should he resolve to never come down the lowest common denominator like much of the "entry fi " junk of the past, then he sure can fly very well, and cheaply. He would only have the parts costs to consider. He should not go overboard with overly expensive parts, lest he risk a possible divorce, should he never take his wife and childeren for a holiday and instead choose silver wiring in the OPTs, which may not make much difference to the sound. A sensible man with a sensibly constructed life but one where he isn't a high wage earner and who has spare time can cobble together a superb system which functions identically in comparison to whatever is cionsidered the best from CJ, ARC, Ongaku, etc. Balance, and the middle path in life has its plusses, and dosn't imply compromise on audio parts has to be made, but may mean patience and slow progress, until funds are saved. Some how the theme in society's consciousness is to make everyone feel guilty for having spare time, because the wowsers would say hobbies and time off work is sloth, and as evil as bonking the bosses daughter when you are married, which is lust, or doing all manner of other things that could be considered to be vice-ridden activities. We are made to feel fearful about the future, threatened by terrorists, and insecure, and the only cure for our background emotional disquiet and anxiety is to pay up like a good little consumer and never say boo to authorities and big business. We are expected to believe the relentless advertising, but I won't. I have never been a slave to societal conventions, believing in give and take, and society will take more than it gives, should you let it, and I don't, and If I wanted to spend 6 years cycling, so be it, rather than building someone else's house, and if I chose to give up the challenge of this for what I consider a lowly paid career in analog electronics, so be it. In many other societies, I wouldn't have such freedoms. Should my righhts be infringed, I will be there mixing it with the partisans. So I have always had the right to snub mainstream products and culture. I never go to Mc Donalds for food, and I have never had a need or desire to buy a Hammond tranny, but for those who wish to, then good luck. I cannot remember when I have eaten a sausage, for they are full of **** as AJ pointed out. Meat is USD $6 per Kg, (2.2lb) where I am in the supermarkets, largely fat free, and $12 worth of *real* meat is enough per week for me, along with a lot of other fresh and non processed food stuffs. Sausages are fake nuitrition. It might be cheaper to live some other way, but I don't care to. So, the message is, self reliance, low debts, insistence on quality, don't waste time on tele, ( I am amoung 1% of Oz ppl who does not watch TV ) don't drink more than a glass of wine a day, never smoke, and never waste anything. And don't marry a woman who has the opposite ideas about the basics. I tried several times, but never minded when they left. Self reliance needs stickatedness. Its extremely unromantic, by today's absurd standards. Between 23 and 26, I saved $11,000 from my pay as an employee, which is about USD $125,000 in today's money. When I bought a house I bought a small one which I then needed a small loan to pay off a small balance, because I established high equity at once. Since I was a builder then, it made sense that I construct additional rooms to give me the room more suitable to enjoy myself in. So unlike nearly everyone else, I finished up with a very comfy workshop and house which I had fully paid for by age 37. I worked my butt off between 18 and 38, to make damn sure it all happened. And I constructed the additions to my house so I could rent out 1/2 the area to give me a little extra $$ to fund time to dabble in hobby persuits, without always worrying about finding enough income. I also built all my own furniture, and saved another 10 grand, today's value. I have lived very cheaply, but very well, with few worries, and continue to do so. All the rest of those with less self reliance have paid a lot more for their cost of living, because every time you pay another to do something for you, you pay his profits, his taxes, and oncosts, and the interest on the loans needed to pay him, stamp duties, etc, etc, etc. The attitude transferred to the electronics for my music, and a hobby grew into a business, when i realised that at 50 I was not competitive with younger builders, and that I was sick and tired of building after 32 years. Ladders and barrow fulls of concrete began to cause episodes of knee pain and illness, so I quit building. I avoided paying twice for my life by avoiding an expensive divorce. Ppl around me could allways have the freedoms they craved, except the freedom to **** up my life. While others may discuss cars and other OT matters, I'd prefer to drift to life philosophy, because life choices will get you further than any motor vehicle, imho. Audio gives me an easier life than building, yet is far less well paid. But I meet more people, and there is a personal reward. One needs personal satisfaction from keeping ppl happy, not just oneself. Work makes the soul thrive..... If I'd had the expectation that I may have been valued by the offspring I never had then I may have been seriously dissapointed, judging by the dysfunction in family lives I see all around me, so I feel no guilt about my choices, philosophy, or lifestyle. I knew an old guy who built a CD player from scratch, with a gold plated double sided board. That's dedication to self reliance. I don't have the time for everything I want to do, but lazing around isn't on the list. Patrick Turner. |
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"Ned Carlson" wrote in message news On Tue, 15 Mar 2005 10:49:41 +0200, Iain M Churches wrote: So, should I rate the transformer Generally speaking, power transformers are rated based on their temperature rise at full load over a long period of continuous use (like between 8 to 24 hours). For shorter periods or on peak demand they can normally supply somewhat more than the rated current, how much more depends on how conservatively they're rated, ie: what temperature rise is considered acceptable at rated load. 320 ma is 160 watts of plate input power, sounds like plenty of headroom for a 50 watt amp! I posted a correction to this. The total current draw by one channel B+ at full power is 240mA at 500V. Iain |
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