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Jim
 
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Default CCS for Mullard phase splitter

Hello,

Can someone explain to me how a solid-state constant current source would be
added to a Mullard-style LTP phase splitter? My amp is similar to an Eico HF89,
which is a Mullard 520 knockoff that uses a 6SN7GTB for the phase splitter (a
schematic can be seen on Ned's site at http://www.triodeel.com/eicohf87.gif).

The amp is fixed-biased, so it already has a -50V source that I could utilize
for the CCS. If the 18K cathode resistor is replaced with a CCS, I assume that
the plate resistors would need to be balanced (currently, they are 27K and 33K).
Should the grid of the 'lower' triode be grounded directly, instead of through a
..25 cap?

Thanks for the advice,
Jim

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Greg Pierce
 
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On Wed, 22 Oct 2003 12:43:15 -0700, the highly esteemed Jim enlightened us
with these pearls of wisdom:

Hello,

Can someone explain to me how a solid-state constant current source would
be added to a Mullard-style LTP phase splitter? My amp is similar to an
Eico HF89, which is a Mullard 520 knockoff that uses a 6SN7GTB for the
phase splitter (a schematic can be seen on Ned's site at
http://www.triodeel.com/eicohf87.gif).

The amp is fixed-biased, so it already has a -50V source that I could
utilize for the CCS. If the 18K cathode resistor is replaced with a CCS, I
assume that the plate resistors would need to be balanced (currently, they
are 27K and 33K). Should the grid of the 'lower' triode be grounded
directly, instead of through a .25 cap?


You don't need the -50V supply. Just change both plate resistors to 33K
and replace the 18k resistor with the CCS. I have posted a couple suitable
CCSs on alt.binaries.schematics.electronic. I prefer the MOSFET version,
but either will work. You can use any FET or bipolar as long as it has
suitable ratings, and regardless the source/emitter resistor will probably
need to be tweaked to compensate for individual device variations, even
if you use the transistors I specified (these circuits work according to
simulations) - you may want to use an 83 ohm resistor in series with a
100 ohm potentiometer instead of the fixed resistor to allow easy
adjustment. Also, the values in the resistive divider in the mosfet
versions gate circuit (NOT including the 50 ohm stopper) can be increased
by an order of 10 (i.e. use a 12k and a 470k instead of a 1.2k and a 47k)
to reduce the wattages of the larger resistor (needs to be at least 1 watt,
but preferably 2 watt for the 47k+1.2k scheme, whereas the 470k+12K
divider would only need 1/4 watt rating). A 1uf+ film cap from the divider
center to ground is a good idea, especially if the 200V supply isn't very
well regulated. I should have included it in the schematic. Incidentally,
I prefer to use the 47k+1.2K combination since it keeps the overall
impedance lower. You don't want the voltage on the gate of that MOSFET
(or the base of the bipolar) moving around. While in theory the CMRR of
the diff-pair will prevent current variations from causing much of a
problem, the reality is that any two discrete devices (tube or SS) never
perfectly match, which hurts the CMRR of the pair.

Both sources will drift with temperature variations. The coefficent of the
bipolar CCS is positive (current decreases with increasing temp), while
the mosfet will usually be negative. In either case, the amount of
variation is relatively small, being approx. 4-6uA per degree C. Thus, the
change from room temp (25C or so) to 100C (far too hot to touch - it
shouldnt ever get that hot in real life) would be a half-milliamp or less.

Hope that helps...

--
Greg

--The software said it requires Win2000 or better, so I installed Linux.

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Tom Schlangen
 
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Hi Greg,

tnx2u for your interesting explanations and the schematics
on ABSE, too.

So far, with "real" cathode resistors in such LTPs, I used to use the rule
of thump that the R value of the cathode resistor should be dimensioned so
that the voltage drop across it should equal ~1/4 to ~1/3 of B+ for good
results, resulting in a R (impedance) value high enough to act
nearly like a CSS, giving good balance for both outputs.

This R value - to be more precise the voltage drop across it - has to be
considered when drawing loadlines and deciding on operation points, because
the dropped voltage must be subtracted from the "available" voltage when
looking up operation points in the tube data nomograms.

So, I am wondering how to compute the voltage drop across such a CSS, to
further on compute the voltages available across the load resistor and the
tube rp.

Additionally, you suggested a fixed R + trimmer combo for the
emiiter/source resitors in circuits you posted on ABSE, to get some range
for exactly adjusting the bias due to data variations of the actual SS
parts used. Besides this, how to dimension the circuits you posted for
other currents as in the actual circuit, the original poster asked for?
Without having looked up the circuit he talked about, I suppose because of
mentioning the "Mullard 5-20" (or something like that) a 12AX7 was used,
which has pretty low current (probably well below 1mA per section) at its
usual operation point(s). When using, say, a 6SN7 or 6CG7 in such a LTP
instead, the needed current will be in the 5-10mA range instead. How to
adapt the CSS circuits you posted to such current demands?

Tnx2u!

Tom

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Greg Pierce
 
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On Thu, 23 Oct 2003 20:19:12 +0000, the highly esteemed Tom Schlangen
enlightened us with these pearls of wisdom:

Hi Greg,

tnx2u for your interesting explanations and the schematics on ABSE, too.

So far, with "real" cathode resistors in such LTPs, I used to use the rule
of thump that the R value of the cathode resistor should be dimensioned so
that the voltage drop across it should equal ~1/4 to ~1/3 of B+ for good
results, resulting in a R (impedance) value high enough to act nearly like
a CSS, giving good balance for both outputs.

This R value - to be more precise the voltage drop across it - has to be
considered when drawing loadlines and deciding on operation points,
because the dropped voltage must be subtracted from the "available"
voltage when looking up operation points in the tube data nomograms.

So, I am wondering how to compute the voltage drop across such a CSS, to
further on compute the voltages available across the load resistor and the
tube rp.


Well, once you have drawn the loadlines you should know what the quiescent
current per tube is. Since you have two tubes in a differential pair (LTP,
if you prefer), you need to multiply that by 2. Then you have the amount
of current that you need from the CCS.

Additionally, you suggested a fixed R + trimmer combo for the
emiiter/source resitors in circuits you posted on ABSE, to get some
range for exactly adjusting the bias due to data variations of the
actual SS parts used. Besides this, how to dimension the circuits you
posted for other currents as in the actual circuit, the original poster
asked for? Without having looked up the circuit he talked about, I
suppose because of mentioning the "Mullard 5-20" (or something like
that) a 12AX7 was used, which has pretty low current (probably well
below 1mA per section) at its usual operation point(s). When using, say,
a 6SN7 or 6CG7 in such a LTP instead, the needed current will be in the
5-10mA range instead. How to adapt the CSS circuits you posted to such
current demands?


The CCS is biased using the source/emitter resistor. You could omit that
resistor and bias it with a potentiometer in the gate circuit, but you
would end up with a very temperature-unstable CCS (unless you use
a tube, but then it would be more age-sensitive). So, you fix the gate/
base voltage and use the source/emitter resistor to set the current. While
it is theoretically easy to figure out the value of the resistor
mathematically, it assumes that the device parameters aren't variable. In
reality, it is easier to figure it out empirically.

The resistor values I used (when used with a 200V gate/base bias supply)
will give approx. 7mA of current, according to simulation with models
supplied by IR and ON Semi for the MOSFET and bipolar respectively. If
you used a 1k pot in series with a 100 ohm resistor (to set a minimum
resistance), you would probably have an adkustable range from 1mA to
15mA or more.

Note that by increasing the voltage on the gate/base (within reason - be
mindful of the voltage limit on the MOSFET) will require the use of a
larger value source/emitter resistor. This is desirable since it minimizes
variation due to device parameters and temperature. In the MOSFET
circuit shown I set the gate voltage at about 5V above ground. I have
posted an improved version in ABSE. This one sets the gate at 100V
above ground, thus requiring a 14K source resistor (1 watt). Not only
is current variation with temperature much less, but device parameters
also have much less influence. In this circuit, different samples of the
IRF820 will have little effect on the current, despite normal device-to
device variations in gm. This makes things much simpler, since it
pretty much eliminates any need of source resistor adjustment.
You will get the current very close by figuring about 4V gate-source.
Since the gate voltage is 100, the source will be at 96V Using the
good ole ohms law E/I=R to get your source resistor value is easy.
Say you need 7mA (the target for the schematic I posted), then you
would divide 96V by 7ma = 96/.007=13714, which is very close to the
14K I specified (14K will actually give about 6.8mA). Piece of cake.
BTW, If your wondering where I got the 4V figure, the answer is
experience and simulation verification. If you use a "logic level"
MOSFET, you want to reduce it to 2V. In any case, if you assume
5V, then the calculation will give you (100-5)/.007=95/.007=13571.
As you can see, the error is fairly small.

Now, the gate voltage for the above scenario is determined by how
much voltage needs to be dropped accross the CCS. In the case of
the original posters situation, the cathodes of the tubes in the diff oair
sit at 125V, so I chose 100V. Obviously you cannot run a gate voltage
higher than 125 - the CCS will simply revert to a resistor. As the source
will sit at close to the gate voltage, this means that 80% of the drop is
across the source resistor, and 20% across the MOSFET, which I
consider a good tradeoff between stability and allowances for variations
in other parts of the circuit. For example, the gate voltage (and hence
the cathode voltage) of the diff pairs tubes is set by the plate voltage
of the preceding stage which it is directly coupled to. If you put a tube
in the preceding stage which idles at a bit higher current than average,
then its plate voltage may be, say, 115 instead of 125. If vou set the
gate voltage of the CCS MOSFET to 115V, you are very close to no
longer having a CCS ( I say close because the source of the CCS will
be about 111V, and the cathodes of the diff pair tubes will be a few volts
above the grid, so you will still have 6-8V across the MOSFET). Good
engineering prudence suggests that you have adequate margins in
your circuit to accomodate such variations. Hence the 100V on the
gate.

One other thing. You will notice that I added a 10V zener diode from the
gate to source. Its purpose is to keep the MOSFETs gate from being
destroyed if you apply the 200V to the bias divider before you apply
plate voltage to the diff pair. Otherwise, without the diode, such a
scenario would put 100V between the gate and source, which is normally
limited to 20V MAX (preferably no more than 10V). In other words, it is a
disaster prevention diode :-)

--
Greg

--The software said it requires Win2000 or better, so I installed Linux.

  #5   Report Post  
Tom Schlangen
 
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Dear Greg,

tnx2u alot for these interesting and educative explanations (hitting
printer button). I will try the third schematic you posted on ABSE
somewhen during the next weeks.

Thank you again!

Tom

--
Bias them up!


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Jim
 
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Thanks Greg and Tom,

Both of your posts were really helpful to me.

Jim

  #7   Report Post  
Patrick Turner
 
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Jim wrote:

Hello,

Can someone explain to me how a solid-state constant current source would be
added to a Mullard-style LTP phase splitter? My amp is similar to an Eico HF89,
which is a Mullard 520 knockoff that uses a 6SN7GTB for the phase splitter (a
schematic can be seen on Ned's site at http://www.triodeel.com/eicohf87.gif).

The amp is fixed-biased, so it already has a -50V source that I could utilize
for the CCS. If the 18K cathode resistor is replaced with a CCS, I assume that
the plate resistors would need to be balanced (currently, they are 27K and 33K).
Should the grid of the 'lower' triode be grounded directly, instead of through a
.25 cap?

Thanks for the advice,
Jim


Here is a schematic for a whole integrated amp,
which has a CCS for the long tail pair using a transistor, which is ss,
and what you asked for.
http://www.turneraudio.com.au/htmlwe...0ulabinteg.htm

The -ve bias supply may not be the same as I have, but
if you had a bias of say -50v, allowing for sag from the rectifier
and RC filter, then juggle the values of the divider which sets the base voltage,
and then adjust the emitter resistor to set the transistor constant current.
A transistor like the MJE340 will display a dynamic input impedance to the
collector of over 20 megohms if the emitter resistor is above 1k.
This is quite a bit higher than what you may achieve with a pentode, with an
unbypassed
cathode resistor, used as a CCS.
There is no need to use a tube for CCS.
Some folks use j-fets, but they are fragile,
and I prefer the far more rugged MJE340, which
gives low noise, and this very high Rc.
The balance of the LTP output voltages remains near perfect,
and solely dependant on the anode loads of the LTP being
equal, even if different type triodes are used for either half of the LTP.
Perfectly matched triodes results in the best cancellation of 2H,
so that a well set up LTP makes extremely low thd.

Patrick Turner.




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