Output Stage Current Requirements for Electrostatic Headphone Amps

[JimL]

There seems to be some consensus on the voltage requirements for electrostatic headphone amps.  Nearly all of the commercially available designs put out between 1000 and 1600 volts peak-to-peak, a range of about 4 dB.  The legendary Stax SRM-T2 was specified to put out a bit more, close to 1800 volts peak-to-peak, which is 1 dB higher.  This would be equivalent to almost all loudspeaker amplifiers putting out between 50 and 125 watts, with the T2 being like a 160 watt amplifier.

 

However, there hasn’t been much discussion on the current demands for electrostatic headphone amps.  Output stage currents in commercial amplifiers have run between 2 mA/channel (Koss ES950) and 36 mA/channel (Blue Hawaii).

 

Back in 1978, Nelson Pass published in The Audio Amateur (issue 4, p. 12) some measurements he had done on the slew rate of music signals.  He tried out various cartridges and LP records, and using a 100 watt amplifier with a 30 volt/microsecond slew rate, reported that the highest slew rate he found with music signals was 1.5 volts/microsecond up to clipping levels.  The late Peter Baxandall also published some years ago in Wireless World that music signals required an amplifier slew rate sufficient to drive a 6 kHz sine wave to clipping with low distortion, which works out to pretty much the same thing.

 

A 100 watt amplifier has a peak-to-peak output of 80 volts.  The Blue Hawaii, to take a current state of the art amplifier, has a peak-to-peak voltage at clipping of close to 1600 volts, which is 20 times higher, so the fastest music signal would have a slew rate of 30 volts/microsecond when the Blue Hawaii is driven to clipping.

 

So how much current does an electrostatic headphone amp need to produce a slew rate of 30 volts/microsecond?  A typical electrostatic headphone approximates a load of about 100 pf - Stax specifies most of their current models between 94 and 120 pf.  The amount of current required for 30 volts/microsecond into 100 pf would be 3 mA.

 

This is the amount of current that the amplifier has to supply to the headphones alone in order to play the fastest music signals up to clipping.  Since amplifiers don’t sound their best at the very limits of their capability, for any real amplifier, there should be additional capacity in both slew rate and current over the bare minimum required. John Broskie has suggested on his TubeCAD website that for low distortion the maximum signal current demand on a tube be a fifth of the standing current.

 

This calculation also assumes that the amplifier itself does not consume any signal current.  But that is not always true.  

 

Take the Egmont, a basic, inexpensive tube electrostatic amp circuit.  It uses 66k resistor loads in its output stage.  With +/- 260 volt supplies the output stage runs at 7.9 mA current.  If we drive the headphones to 1000 volts peak-to-peak using our fastest music signal the headphone consumes 1.9 mA, but the resistor consumes 7.6 mA, using all the current the output stage is theoretically capable of supplying.  The reason that an amp with a total current of 7.9 mA can supply both 1.9 mA to the phones and 7.6 mA to the resistor loads is that the current to the headphone is approximately 90 degrees out of phase with the current to the resistors – remember the geometry of a right angled triangle? 

 

The headphone and resistor compete for the available current, and since the resistor is lower impedance than the headphone, the resistor hogs most of the current and the headphone is left with the scraps.  Furthermore, the amount of signal current soaked up by the resistor depends on the magnitude of the signal, whereas the amount of current going to the headphones depends on the speed of the signal, so the ratio of 1.9 mA to the phones and 7.6 mA to the resistor is even worse almost all of the time.

 

In fact, this is a problem for any electrostatic headphone amp that uses resistor loads in the output stage since the resistor sets both the standing voltage and the standing current.  Massively increasing the voltage and current so that no user will ever come close to reaching its limits doesn’t really solve the problem, it just pushes it farther away.  And then, a further problem is that devices and components which can withstand that amount of voltage, current and power are expensive, which rather defeats the goal of an inexpensive design.

 

Now, take my revision of the Stax SRX tube design using current loads.  The output stage runs at a higher current and voltage: 14 mA current with the power supplies run at +/- 325 volts. More importantly, the cascoded current loads on each plate measure over 160 megohms impedance, thus requiring a mere 4 microamps to drive them to clipping, so 99.9% of the total standing current is available to drive the headphones.  The maximum current required to drive the headphones at clipping is about 2.4 mA, less than a fifth of the current available. 

 

To further illustrate the value of a good current source, let’s go back to the Egmont.  With the output tubes in that design delivering the same peak signal current of 2.4 mA, it would produce about 300 volts peak-to-peak with about 2.3 mA going to drive the resistors and 0.6 mA to the headphones.   For the same signal voltage into the headphones, the Egmont output tubes have to produce 4 times as much signal current.

 

Now these are “back of the envelope” calculations.  But at least, now we have a reasonable estimate of how much signal current an electrostatic headphone needs to faithfully reproduce the fastest music signals.  And, it is clear that replacing resistor loads with current sources is a much more efficient method.

 

Finally, let me make a brief comment about a related matter.  It is sometimes said that electrostatic headphones require voltage but no power.  This is false.  It is true that electrostatic headphones resemble capacitors, and with a capacitor, the drive voltage and current are 90 degrees out of phase so that no power is consumed.  However, remember that a capacitor is a simplified model of a stat headphone.  In fact, electrostatic headphones have to consume energy, because we can hear the sound they produce!  Sound is a form of energy, and by the law of conservation of energy, one of the most fundamental laws of physics, that means the headphones have to consume energy.  

[JimL]

Another quick example.  This time, comparing the SRX revised with 6SN7GTA, +/- 325 volt PS, 14 mA/channel with current sources vs the ESX with triode strapped EL34, +/- 400 volt PS, 16 mA/channel with 50 kilohm plate resistors.  The EL34 stage dissipates 40% more power than the 6SN7 stage.  Identical circuit except for PS voltages,  output stage tubes and resistors vs current sources.  The more powerful stage should win, right?

 

Driving both amps to 800 volts peak-to-peak with the fastest music signal, producing 106dB measured into an Omega 2 (per Kevin Gilmore),

 

the ESX requires 8.1 mA signal current, > 50% of the idle current

 

the SRX revised requires 1.5 mA signal current, 11% of the idle current.

 

A current source load can allow a smaller, weaker output device to outperform a bigger, stronger output device with load resistors.

[PretentiousFood]

I'm having a bit of trouble following your calculations. Could you elaborate on how you got the current requirement from the slew rate and capacitance?

 

I've ususally seen the current requirement calculated based on the bandwidth and voltage swing. Morgan Jones has a short section on it in his chapter on preamps. For Stax, it would work out to something like this:

Zc = (jCw)^-1

Zc = (j 120*10^-12F 2*pi*20000Hz)^-1

Zc = 50K j^-1

Knowing the plate load, we could then draw a dynamic loadline. Regardless of the load, the transducers' worst-case current draw can be calculated by Ohm's law.

I = V R^-1

I = 800Vpp / 50000

I = 16mApp

This would mean that the combined signal current of the output stage to drive a 120pF load to 800V would be about 16mA, if it were to remain strictly class A. 

 

I think the slew rate would also need to be greater. 800V at 20kHz works out to about 100V/us. I might have an extra zero in there somewhere.

[JimL]

Sure.

The equation relating capacitance, charge and voltage is:

V = Q/C where V = voltage, Q = charge and C = capacitance

Taking the time differential of both sides we get:

dV/dt = (1/C)*dQ/dt where dV/dt = voltage change/time = slew rate, dQ/dt = change in charge/time = current = I

So:

dV/dt = I/C, or C*dV/dt = I

For my example, dV/dt = 30 V/microsecond, C = 100 pf, plugging everything in gives I = 3 mA for 1600 volts peak-to-peak.

I used to be a physics major in a previous life. )

The reason you got a different answer for slew rate is that you are calculating for a full-power, 20 kHz signal. However, music does not contain full-power signals at 20 kHz. As Nelson Pass and Peter Baxandall found, the music power spectrum rolls off above around 5-6 kHz at a rate of approximately 6 dB/octave. Because of this, for a 100 watt amp that can swing 80 volts peak-to-peak the fastest slew rate with a music signal is 1.5 volts/microsecond up to clipping, whereas a sine wave 20 kHz signal at clipping would slew at 5 volts/microsecond for the same amp. The fact is that music signals are not that fast compared to some test signals.

Multiply those results by 20 for a 1600 volt peak-to-peak signal at 20 kHz gets you to 100 volts/microsec, which is what you calculated, or 30 volts/microsecond for music signals, which is what I got. You'll get the same answer as me if you plug in a full-power signal at 6 kHz, which is what Baxandall said was needed to reproduce music.

Now, the calculations in the second post assumed a signal of 800 volts peak-to-peak, which is 400 volts peak, whereas your calculation was for 800 volts peak, hence your answer is twice as high as mine. My number also includes the 1.5 mA current into the capacitative load of the headphone, and because that current is 90 degrees out of phase with the current to the resistor, the overall answer 8.1 mA (rounded off) rather than 8 mA for the resistor load alone.

[PretentiousFood]

Neat! Thanks for taking the time to explain. I've been nervous about how cold most gain stages were biased for the same reasons-- they do need to drive the Miller capacitance, as you pointed out in the other thread. Looks like it's less of an issue than anticipated, especially when it's wrapped in a feedback loop.

Edited April 8, 2015 by PretentiousFood

[JimL]

Well, the Miller capacitance can definitely be an issue.  For example, with the SRX, the first stage has a very high output impedance because it's a cascode, so the impedance is nearly as high as the plate resistor.  In combination with the Miller capacitance of the output tubes, with a 6SN7 output tube the open loop response is about 3 dB down at 11 kHz and the closed loop response is about 3 dB down at 53 kHz - that's measured with the SRX Plus (that's my name for the revised SRX).  With a triode connected EL34, I measured the grid-to-plate capacitance at around 10 pf, so the open loop response is about 3 dB down at 8-9 kHz and with  the lower maximum gain of around 10.8 in the output stage, the calculated closed loop response just makes it to 20 kHz before it starts rolling off if you use a current load - that's what Dr. Gilmore also reported on measurement.

 

Also, the SRX input stage has  puny standing current of about 0.55 mA/side.  With a peak voltage of 16 volts into the output stage which should drive it into clipping, the 6SN7 has a Miller capacitance of about 84 pf, requiring 0.16 mA/side for a 20 kHz sine wave, but with music signal the max current needed drops to about 0.05 mA/side, which is more than acceptable.

 

When you analyze the SRX circuit you can see where the engineering compromises were made, and how well chosen they were to produce a result that is "just good enough" but works very well in practice.

Edited April 8, 2015 by JimL

[JimL]

Another example, let's take the ES-X but with 10M90S current sources instead of the 50 kilohm resistor load.  I measured the dynamic impedance of a single 10M90S current source at about 170 kilohms, so for 800 volts peak-to-peak as in the second post, the worst case current demand is about 2.8 mA, about 17.5% of the total idle current of 16 mA/channel.  This is much better than the >50% demand using the resistor load. 

 

But now, we can see another benefit of using current loads.  There is no reason why we need to keep the idle current at 16 mA.  We can set the idle current at any level we wish, within the limits of tube dissipation and heatsink size.  For example, if we increase the idle current to 28 mA (like the T2) or 36 mA (like the BHSE), the worst case current draw is only 10% (T2) or < 8% (BHSE) of the idle current.  Aside from increasing the current reserve, which should decrease distortion, why would we want to do this?  Well, running the EL34 at 8 mA per tube is really a pretty low current, where the tube linearity is not as good as it could be. Running the tube at a higher current puts it into more linear region, further decreasing distortion.

[congo5]

After building the SRX Plus I decided to try the CCS on an Egmont.

Copying Kevin's layout I drew up this:

 

Having two Egmont's I can A/B them to compare.

It is a huge improvement, more volume and much less distortion.

Much more of a usable amp now. Noticeably cleaner bass.

like Night and Day.

 

With 320v rails and the resistor values of the SRX CCS the tube bias is adjustable

up to about 5.55 ma, currently running them around 5ma.

 

The R& L channel offset is up to -40v so I lowered the 580v bias a bit to compensate.

Also tried 3 sets of 5965 output tubes and results are similar.

May try to raise the cathode resistance a little.

Add a hundred ohms?

 

I have an old BK scope and I will try to figure out what point the

two amps go into distortion.

 

A Big Thank You to All for posting your work ....

[kevin gilmore]

I like it.

[spritzer]

Mess around the with the cathode resistance a bit and you will null the offset.  I just picked a value that was close enough for that plate load.  100R should be plenty. 

[JimL]

Doh!   I deleted my last post because it was totally wrong.  Sorry about that.

 

So, the problem right now is that the cathode-to-plate voltage is too low, thus the output voltage is negative with respect to ground.  In order to increase the cathode-to-plate voltage we need to increase the cathode-to-grid voltage difference.  You can see this if you look at the tube curves, going to a more negative grid voltage along a constant current line increases the plate voltage.

 

In order to increase the cathode to grid voltage difference we have to increase the cathode resistor.  Since we are using current sources with a total of 10 mA into the cathode resistor, the calculation is relatively easy.  A 100 ohm change in the cathode resistor will result in a 1 volt change in the cathode-to-grid voltage.  Looking at the tube curves for the 5965, as we run along the 5 mA current line, decreasing the grid voltage by one volt will result in about 30-40 volts increase in the cathode-to-plate voltage.

 

So, for each channel, take the average offset between the + and - outputs, and substitute a cathode resistor that is more by the calculated amount.  

 

One caveat, the offset needs to be checked when the amp is warmed up with the cover on.

 

Oh, and good job, pongo5!  It's always gratifying when practice confirms theory.

Edited April 26, 2015 by JimL

[JimL]

Those who are interested in finding out more about constant current sources should read the two part articles by Walt Jung which were published in AudioXpress in 2007, with addendum in 2009.  The first and second parts are available on the web, just do a search for Walt Jung, the title is: "Sources 101: Audio Current Regulator Tests for High Performance", which shows a number of different current source designs with measurements of performance. 

 

Interestingly, while browsing through the first part, I found (Figure 3A)  a low voltage version of a design that is identical to the constant current source used in the output section of the eXstatA, except with high voltage NPN transistors instead of low voltage PNP transistors that Jung used in the article.  Jung measured its dynamic impedance as roughly equivalent to a 45 kilohm resistor.  He comments that "the performance of this circuit is easily bettered by many others, both more simple and cheaper.  For these reasons, plus the stability caveat, this circuit isn't recommended.  Except as an example to avoid, perhaps."  If the high voltage eXstatA design is no better then it is not clear the advantage that 6 parts plus heatsink has over a high power resistor.

 

By contrast, the low voltage version of the transistor plus LED current source that is commonly used in Dr. Gilmore's designs, for example, has a measured dynamic impedance of around 170-180 kilohms which is similar to the results with a single 10M90S current source.  One can do even better substituting a reference diode such as an LM336Z2.5 which bumps the dynamic impedance up to around 3 megohms, although Jung didn't address the noise performance of the substitution. 

 

Anyway, for those interested in the nitty-gritty of design, well worth reading IMHO.

[insanity]

Very interesting for me to read. Thank you!

[JoaMat]

^{I do agree with insanity. Interesting reading indeed. Thanks a lot.}

[kevin gilmore]

he also did not address the temperature issues when you use a fixed reference.

the reason to use a red led is that it exactly matches that of the transistor
so the thermals don't cause runaway conditions.

important because at 1kv, 1ma difference is a big deal.

there is a way to bump up the impedance by adding one more transistor
and another led in series.

its also no surprise that the exstata and the idiot that designed it have no clue.

for fun you can analyze the llmk1,llmk2,llmk3... given the high capacitance mosfets
that replace the transistors, has to be really bad. schematics were posted.

Edited April 29, 2015 by kevin gilmore

[JimL]

Thanks for the comment, Dr. Gilmore, that's a very good point.  One thing about the cascode CCS (constant current source) on my SRX Plus (modded SRX) is the output voltage drifts about 40-50 volts as it warms up, also the final value may vary by a few volts with different turn-on/turn-off cycles, presumably due to thermal issues.  Thermal stability is very significant for practical amplifiers.

 

In terms of the LL, my copy of the prototype schematic courtesy of Dewey, Cheatem and Howe  , uses the same topology as figure 3A discussed in my previous post, but substitutes a MOSFET for a BJT as the device that has the large voltage swings in the output CCS.  Here is what Jung has to say a few paragraphs earlier in his introductory remarks in the section discussing the one VBE current source shown in figure 3A:

 

"Of course, higher-voltage parts should be used when appropriate.  While exotic and super-high-gain parts aren't necessary for very good performance from these circuits, LOW CAPACITANCE DEVICES DEFINITELY ARE PREFERRED (<10pf), A CRITICAL POINT IF SUBSTITUTING [emphasis added]."  

 

The reason is that high capacitance devices cause the effective impedance to decrease at high frequencies, degrading the performance of the current source.  Given that MOSFETs have a relatively high shunt capacitance, this means that performance of the LL output current source is likely to be even worse than the BJT version that Jung cited as a design to avoid.  Now the "fixed" version of the LL uses a low-capacitance BJT for the MOSFET, which is better, but.... the SRM323, which shares a very similar topology, uses the LED and BJT CCS on the output, which is significantly better than the CCS of the "fixed LL."  You can't say that Dr. Gilmore and spritzer didn't warn us about this when the LL came out, although they didn't go into it in the infinite gory detail that I just did.

 

Now, one of the advantages of a cascode MOSFET CCS such as the SRX Plus uses, is that the upper device shields the lower device from voltage variations, so that the lower device, which sets the current, sees nearly constant voltage regardless of the voltage variations across the cascode pair.  This means that the effective capacitance of the cascode is very low, so its performance is preserved to high frequencies.  For example, Pimm measured a cascode DN2540 pair as having an effective capacitance of < 0.2 pf!  By comparison a single DN2540 measured about 32 pf.

 

ADDENDUM:

I got a look at the production schematic courtesy of spritzer's LL Mk I thread elsewhere on this site.  The output current source is identical to the Dewey, Cheatem and Howe version with the exception of a protection zener added to the MOSFET in the production version, so same lousy performance. 

Edited April 30, 2015 by JimL

[JimL]

Let me get back to something I discussed briefly in my first post but haven’t addressed since then, and that is that stat headphones must consume power because they produce sound.  Now, all the calculations I’ve done up to now are based on a simple model of the electrostatic headphone as a capacitor.  However, even though capacitors use require current from the amplifier to swing voltage, they don’t use up power because the voltage and current are out of phase.  And of course, we don’t listen to capacitors because they don’t make a sound (at least in theory).

 

Since headphones do make sound, that means they must use extra current from the amplifier beyond what we have previously calculated.

 

So let’s take our previous example of 800 volts peak-to-peak.  This is 400 volts peak, and since Dr. Gilmore has stated that the SR007 can consume one watt at peak output, let’s say the peak current is 2.5 mA, which works out to that one watt peak.  If we add this current demand to our previous results, we get:

 

ESX 50k load resistors:                      10.6 mA = 66% of 16 mA standing current

 

ESX 10M90S current source:               5.1 mA = 32% of 16 mA standing current

            = 18% of 28 mA standing current (T2)

            = 14% of 36 mA standing current (BHSE)

 

SRX Plus cascode current source:      2.9 mA < 14% of 14 mA standing current

 

Again, these are “back of the envelope” calculations so don’t take the absolute numbers too seriously but the relative relationships should hold good.  We can now see why pongo5's Egmont sounds much better with constant current source loads.  The Egmont with 60k load resistors is similar to the ESX with load resistors in terms of current demand, whereas the Egmont with cascode current sources is similar to the SRX Plus.  The basic circuit draws 3x as much signal current for the same signal voltage because the load resistors waste so much of it swinging the voltage in the resistor.

[dsavitsk]

Could you comment on the influence of the plate impedance on your calculations. Or specifically, when the output device itself is a constant current source. Several of the triode based amps all have a lowish impedance. But some of them use a cascode of some sort which would produce more pentode/transistor like plate curves with a very high Z. Does it matter?

Edited May 3, 2015 by dsavitsk

[Laowei]

Good reading JimL. Shouldn't the current losses of the global NFB be included somewhere?

[JimL]

dsavitsk:

So if you have a current source on a cascode or pentode stage the benefit is very high gain, but the down side is very high output impedance.  Loading it with a resistor will decrease the gain, and loading it with a tube, with its Miller capacitance, will result in rolled-off frequency response.  This is why I didn't put current sources on the input stage cascode of the SRX - with the 300 kilohm resistors in the basic circuit, the open frequency response rolled off with a -3dB at around 11 kHz, and substituting a current source would make things worse.

 

For an amp output stage you need to have a relatively low output impedance to supply signal current to the transducer, whether it be a speaker or a headphone.  Most speakers are designed to be driven by voltage sources, so most speaker amplifiers have an output impedance of 1 ohm or less, often very close to zero.  

 

Now, a triode connected EL34 such as in the ESX has a plate resistance of around 1 kilohm, and a 6SN7GTA such as in the SR Plus has a plate resistance around 7 kilohms without feedback.  Feedback does help, cutting the effective output impedance down to around 400 ohms for the EL34 in the ESX and around 1400 ohms for the SRX Plus.  That sounds fairly high, but remember that a 100 pf cap has an impedance around 80 kilohms at 20 kHz, which rises at lower frequencies.

 

Laowei:

Good point.  With the exception of a couple designs which have no neg feedback, most stat amps have feedback resistors around 200k to 250k, which with 400 volts peak amounts to another 1.6 to 2 mA required from the output devices.  So now, including the current going to the NFB loop we have:

 

ESX 50k load resistors:                12.6 mA = 79% of standing current

 

ESX 10M90S current source:        7.0 mA = 44% of 16 mA standing current

                                                                   = 25% of 28 mA standing current

                                                                   = 19% of 36 mA standing current

 

SRX Plus cascode CCS:               4.4 mA = 31% of standing current

Edited May 4, 2015 by JimL

[dsavitsk]

So if you have a current source on a cascode or pentode stage the benefit is very high gain, but the down side is very high output impedance...

 

This much I am familiar with. And in essence you are basically saying that a CCS has a high impedance that forms a voltage divider with the load with the lion's share of the current going to the lower impedance portion. Were it practical to do so and if we didn't care about efficiency, an amp with a 100KV supply could use a 10M resistor to much the same effect.

 

I guess the question is that in a common cathode triode, the lowish plate impedance drives the plate load in parallel with the load. A pentode is the opposite and has a Zout that is PS's Zout in series with the resistor load with the tube modulating the PS. So I am wondering if that resistor load on the pentode is a reasonable and somewhat unexplored way to drive these things or if your above calculation contains something to suggest otherwise.

[JimL]

The main point of my previous calculations is to show that a constant current load is a much better way to drive a stat headphone than a resistor (or a plate inductor for that matter), because using a really good current load basically removes it from the equation - all the driving triode sees is the headphone load.  Much better efficiency in terms of getting the signal current to the headphones.  So for example, the SRX Plus can do about 2/3rds the output for the same ratio of signal current to standing current as the BHSE, with less than 25% of the power in the output stage.  A good big 'un can beat a good little 'un but the little one can come surprisingly close by working smarter, not harder.

 

The problem with a pentode or cascode as an output stage is its high impedance, which means it doesn't vary its current much as the output voltage changes.  This is what we DON'T want in an output stage.  For example, if we were to combine a cascode or pentode with constant current load, the output voltage could vary all over the place (high voltage gain) but the output signal current would be very low.  Think of the reductio ad absurdum where the cascade or pentode is modeled as a constant current source, into a constant current load.  The voltage would move up and down but there would be no signal current output, and you need both signal voltage and signal current to generate power,  By the law of conservation of energy, no power, no sound output  As the old saying goes, "it couldn't drive its way out of a wet paper bag." 

 

Now, the problem of using a resistor load for the cascode/pentode as the driver is that you need to have a low impedance resistor so the current can easily go to the headphones.  So, let's pick a 5 kilohm resistor for the plate of our output stage to provide a sufficiently low impedance driver for the headphones.  If we want to swing 400 volts peak, using Ohms law, we have to modulate the current in the resistor by 80 mA, which means we need at least 80 mA standing current in the resistor, which is 32 watts of power - we need a BIG resistor, and also a BIG tube because that also has to burn up 32 watts.  Now, if we have a differential output stage that is 128 watts/channel power dissipation, not counting the heaters for the tubes.  Note this is to swing a mere 800 volts peak-to-peak, if we want to swing 1200 volts P-T-P  increase everything by 50%.  Kinda makes a BHSE with its 40 watts/channel standing power dissipation  seem like a paragon of energy efficiency. And since the resistor is so much lower in impedance than the headphones, 99.8+% of the signal current is wasted in the resistors.  So, a long-winded way of saying, not a good idea.

Edited May 4, 2015 by JimL

[Laowei]

Hey JimL, have you considered using the Mu output to power the ESL earspeakers instead of the plate? Some theoretical advantages, such as lower impedance and ability to drive capacitive loads. I have used it to good results in the past on preamp outs to drive long Goertz ribbon interconnects. But that was single ended, and I am not sure how this might be applicable for balanced outputs.

[JimL]

Gary Pimm discusses using the mu output in the section on constant current sources on his website.  Basically, this involves taking the output signal off the junction between the lower and upper device in the cascode current source.  In the cascode, the upper device takes nearly all of the voltage variation, isolating the lower device, which runs at very close to constant voltage, and hence constant current.  This means the output tube also runs at constant current, which is its lowest distortion mode, and the output signal current comes from the upper device rather than from the output tube.  Since the lower device is running at a constant voltage, like a battery, the mu voltage output closely follows the plate voltage.  The upper device is low impedance and so able to drive large capacitative loads better than the tube plate could.  

 

However, as Pimm points out, if you take the signal current from the tube plate which is loaded by a constant current source, the power supply is only responsible for providing a pure DC current to the circuit, and the signal current only circulates in the active circuit. The current sources isolate the active circuit and the signal current from the power supply.

 

If you take the signal current from the mu output, since the output tube is now running at a constant current, the signal current must come from the power supply, which means that the supply, with its non-ideal electrolytic capacitors, is now in the signal pathway.  This is a significant disadvantage. OTOH, the capacitative load of a stat headphone is quantitatively similar to that of a grid resistor and output tube, so the lower impedance and improved drive capability of the mu output does not provide a significant benefit in exchange.  

 

This can be partially alleviated by using a shunt regulator fed by a constant current source.  This would remove the passive power supply capacitors prior the regulator current source from the signal pathway, however the shunt regulator itself would still be in the signal pathway.

 

In the SRX Plus, my goal was to maintain the integrity of the original circuit while optimizing its function. The best way to do that was to  replace the output resistors in the original circuit with constant current loads.  The additional benefit was isolating the active circuit from the power supply.

[bobkatz]

On 4/7/2015 at 4:54 PM, JimL said:

The equation relating capacitance, charge and voltage is:

V = Q/C where V = voltage, Q = charge and C = capacitance

Taking the time differential of both sides we get:

dV/dt = (1/C)*dQ/dt where dV/dt = voltage change/time = slew rate, dQ/dt = change in charge/time = current = I

So:

dV/dt = I/C, or C*dV/dt = I

For my example, dV/dt = 30 V/microsecond, C = 100 pf, plugging everything in gives I = 3 mA for 1600 volts peak-to-peak.

I used to be a physics major in a previous life. )

SNIP

@kevin gilmore

@JimL

Dear JimL, please forgive me for reviving your excellent 8 year old thread! I was definitely not a physics major in a previous life, but I slowly can handle your math.

 

To review, common wisdom thinks that 100 pF is a very high impedance bridging load. But what is the theoretical load based on the current?

Can I solve for Z load based on current?

I'm trying to use your information to calculate the effective reflected impedance load on the primary (low impedance) side of the transformer in order to know what kind of low voltage amplifier I'm going to need. I'm also trying to expand this to a higher output voltage requirement of 700 volts sine wave RMS = 1980 v p-p at max. Which would be nominally 116 dB SPL peak max with a Stax.

Let's round that up to 707 volts sine wave to make the math look easier, which brings us to a round 2000 volts p-p at max with sine wave. 
My thinking is Z = V / i

Next, if i for 1600 v p-p is 3 ma., then i for 2000 v p-p would be 4.6 ma. p-p. Let's round that up to a whopping 5 ma. and see what effective Z load we get:

Z = E/i 

Z = 2000 / 0.005 = 400 k ohms!

So that's the nominal load impedance, it seems.

Reflected load, if it was nice linear algebra, would be, I suppose    Z load x the turns ratio. Turns ratio is 33:1 in my transformer.

So, I calculate  400 * 10 eex 3   *    30.3 * 10 eex -3 =  12 kohms!

Source voltage = 2000/33 = 60 volts p-p.  My push-pull opamps run on +-24 volts, so they can certainly deliver 60 v p-p into 12 k ohms.

How does my calculation look to you math experts?

 

 

 

Edited August 1, 2023 by bobkatz
clarify the equation