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Old 7th Jan 2019, 2:18 pm   #41
kalee20
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Default Re: Puzzling audio circuitry

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If one regards the upper valve as a voltage follower, its cathode attempting to follow the grid voltage and the grid is at signal ground, if it were perfect in that application the plate voltage of the lower valve would not move at all. But since it has a source resistance, looking into the cathode of the upper device (valve), of around a few hundred Ohms typical, I agree it will move. In transistor cascode though, the collector voltage of the lower device is pretty well rock solid voltage wise, so I may have been thinking more of that than the valve case.
Err... no, sorry.

Consider two decent NPN transistors (hfe = 100) in a cascode configuration, with the base of the upper one fixed, and 1mA collector current flowing through the lower one. All are at room temperature.

Then the lower one will have a gm at room temperature of 40mA/V; the internal emitter resistance will be 25Ω, and the input impedance will be 2.5kΩ.

A 1mV signal will give rise to a collector current signal of 40μA.

The upper transistor will be fed with 1mA quiescent emitter current, so its emitter resistance will be about 25Ω. The 40μA signal current will therefore give rise to an emitter voltage variation of 1mV - the same as the input voltage to the stage!

This will always be the case, whatever the current through the pair. It's hardly 'pretty rock solid' if it's the same magnitude as the input! In fact it's comparable with the valve version. Miller effect in the transistor will kick in, feeding back effectively 2 x Ccb.

Whether valve or transistor, if you have two similar devices in cascode, they'll have the same gm. So the voltage on the anode/collector of the lower device will be approximately equal to the input voltage. Why? Because the lower device will give a current variation equal to Vin x gm, and the upper device will have a 'looking-in' input impedance approximately 1/gm. So the voltage here is Vin x gm x 1/gm = Vin.

Last edited by kalee20; 7th Jan 2019 at 2:40 pm. Reason: Added last sentence
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Old 7th Jan 2019, 11:56 pm   #42
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The 40μA signal current will therefore give rise to an emitter voltage variation of 1mV - the same as the input voltage to the stage!

This is correct, but by omission it makes the very statement that I was trying to make.

Firstly though about the "rock solid" remark, hyperbole I agree, but the fact that the collector voltage of the lower device only moves about equal to the small signal input voltage is pretty "rock solid" compared to what it would be if there was the usual impedance in the collector (R or L etc) which allows its voltage to move 10 to 100 times more, so by comparison to that, I can say it is sort rock solid, but not zero though.

Secondly, in your analysis, you will note that your calculations did not have to refer to the gm of the upper transistor, just saying it was equal to the lower one.

You only needed its input impedance (25 Ohms) derived from its collector current, which ignoring base current, is its emitter current and depends on the lower device's collector current.

And then you calculated that the 40uA of signal current (generated by the lower transistor's gm) creates the emitter voltage variation in the upper transistor.

But this completely neglects the fact that nearly all (99%) at least, of the upper transistor's signal collector current (where the output signal of cascode is derived from) is from the lower transistor's signal collector current and perhaps only 1/100 being the upper transistor's base current, contributing to the lower transistor's collector current.

This means you could double the upper transistor's gm (or hfe) and the gain of the cascode circuit remains the same (all that this does is lower the upper transistors base current).

But the gain of the cascode circuit is directly linearly proportional to the gm of the lower device, which is the point I was making about where the gain in the cascode circuit originates, rather than saying the lower device has a "fixed gain of 1" which it only appears to have across the input and output terminals of the lower device.
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Old 8th Jan 2019, 12:26 am   #43
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Default Re: Puzzling audio circuitry

We seem to have wandered a long way from the original post having flipped from a circuit which is not the traditional cascode (though at first glance it can look like it) into the trad cascode, and then having switched from valves to transistors.

I wonder if this ought to be in a different thread?

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Old 8th Jan 2019, 7:41 am   #44
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Default Re: Puzzling audio circuitry

Getting back to the original shunt-regulated circuit, my understanding is that this was originally developed as an economical way to drive TV transmitter video modulator stages. These presented a very difficult load, generally of quite low and varying (frequency-dependent) impedance. When the load conditions were such that the drive output voltage might be pulled down, the shunt valve provided the additional current required to maintain the voltage to the modulator.

That being the case, why would it be used in a simple audio circuit? A single-ended 6L6, presumably always operating in class A, would not seem to be a particularly difficult proposition for the driver stage.


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Old 8th Jan 2019, 9:56 am   #45
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... in your analysis, you will note that your calculations did not have to refer to the gm of the upper transistor, just saying it was equal to the lower one.

This means you could double the upper transistor's gm (or hfe) and I the gain of the cascode circuit remains the same (all that this does is lower the upper transistors base current).
That's true, but based on the assumptions I stated, the gm's will be the same (to within 1% as the emitter currents are equal to within 1%, the hfe being 100 and both devices at room temperature).

The only way to double the gm of a transistor, that I know of, is to double the operating current (impossible in a series-connected cascode), or run the transistor at double the absolute temperature. You can't choose another type with a different gm (unlike valves). A cascode comprised of a BC107 and a 2N3055 will still display equal gm's and have the same signal voltage on the collector of the lower device as on the input. It's not a massive signal, but it's not rock-steady either!

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Old 8th Jan 2019, 10:11 am   #46
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The only way to double the gm of a transistor, that I know of, is to double the operating current (impossible in a series-connected cascode), or run the transistor at double the absolute temperature. You can't choose another type with a different gm (unlike valves). A cascode comprised of a BC107 and a 2N3055 will still display equal gm's and have the same signal voltage on the collector of the lower device as on the input. It's not a massive signal, but it's not rock-steady either!
Quite true, but it seems counter-intuitive to people not familiar with the Ebers and Moll equation and onwards. You could also change the Gm by having one transistor in a different material with a different band-gap. Can't think why I'd want to do this though.

If flat-out gain isn't required, emitter degeneration will artificially lower the Gm of the lower device only, and the swing at its collector will decrease (assuming Iq isn't changed) And the Miller pole moves up correspondingly.

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Old 8th Jan 2019, 10:29 am   #47
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We seem to have wandered a long way from the original post having flipped from a circuit which is not the traditional cascode (though at first glance it can look like it) into the trad cascode, and then having switched from valves to transistors.
We have somewhat, but it's been fascinating!

We explored whether the original circuit is cascode or not; explored whether there's any signal voltage at all on the lower-device anode or collector, with 'ideal' devices, and found that there IS a small signal voltage, which would be easy to 'scope, although as Hugo says it is not massive, and moreover it is (maybe surprisingly) sensibly independent of device type or even whether valves or transistors are used!

So the 'puzzling audio circuit' is fundamentally different, having much higher signal voltage at the lower device anode.

The original circuit, in its application by the OP, does not appear to have any technical advantages over a conventional RC coupling, but it could be claimed by the designer to be a novel use of the SRPP and have audiophool appeal. Certainly the performance will be no worse! It's interesting that as soon as a load is put on the stage, some current feed-forward action takes place to maintain the level of output voltage.
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Old 8th Jan 2019, 6:16 pm   #48
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The only way to double the gm of a transistor, that I know of, is to double the operating current (impossible in a series-connected cascode), or run the transistor at double the absolute temperature. You can't choose another type with a different gm (unlike valves). A cascode comprised of a BC107 and a 2N3055 will still display equal gm's and have the same signal voltage on the collector of the lower device as on the input. It's not a massive signal, but it's not rock-steady either!
Quite true, but it seems counter-intuitive to people not familiar with the Ebers and Moll equation and onwards. You could also change the Gm by having one transistor in a different material with a different band-gap. Can't think why I'd want to do this though.
I don't think this will matter! If you have two transistors in series, they'll have the same current passing through them (to within a couple of percent, if their hfe's are high). You could have one silicon, one germanium. The gm's will still match at the same current and temperature, so (gm lower) x (re upper) will still equal 1. And that means that (signal voltage on collector of lower) = (input voltage on base of lower).

Adding emitter degeneration by an added resistor will certainly change things, and the Miller pole will move up in frequency, as you say. That's moving the goalposts, of course!

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Secondly, in your analysis, you will note that your calculations did not have to refer to the gm of the upper transistor, just saying it was equal to the lower one.

You only needed its input impedance (25 Ohms) derived from its collector current, which ignoring base current, is its emitter current and depends on the lower device's collector current.
Yup. But input impedance (25Ω) into the emitter of the upper transistor is 1/gm of that device. By knowing one, you know the other.

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If one regards the upper valve as a voltage follower, its cathode attempting to follow the grid voltage and the grid is at signal ground, if it were perfect in that application the plate voltage of the lower valve would not move at all. But since it has a source resistance, looking into the cathode of the upper device (valve), of around a few hundred Ohms typical, I agree it will move. In transistor cascode though, the collector voltage of the lower device is pretty well rock solid voltage wise, so I may have been thinking more of that than the valve case.
That's the bit I'm querying! In a valve cascode (with identical valves), the anode voltage of the lower valve will move with the same amplitude as the input. It has to, as long as (gm lower) x (1/gm upper) = 1. And with transistors, the same applies (the transistors don't even have to be identical as long as you don't pair a bipolar with a JFET). There ain't no difference!

If Argus's concept of the upper valve acting as a perfect voltage follower were true, then with the grid at signal ground the cathode would indeed also be at signal ground and so would the anode voltage of the lower valve. But to get a perfect follower, the gm of the upper valve must be infinite (Rout = 1/gm). And as the valves are matched, the gm of the lower valve would also be infinite, so the minutest input signal would give infinite current into the upper cathode. And we know what happens with infinity times zero, we get anything we like - which in this case is 1.

Last edited by kalee20; 8th Jan 2019 at 6:33 pm. Reason: Clarity
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Old 8th Jan 2019, 9:57 pm   #49
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If Argus's concept of the upper valve acting as a perfect voltage follower were true, then with the grid at signal ground the cathode would indeed also be at signal ground and so would the anode voltage of the lower valve. But to get a perfect follower, the gm of the upper valve must be infinite (Rout = 1/gm). And as the valves are matched, the gm of the lower valve would also be infinite, so the minutest input signal would give infinite current into the upper cathode. And we know what happens with infinity times zero, we get anything we like - which in this case is 1.
But it is just a conceptual thing, there is no point in heading to infinity and nothing is perfect. For the transistor case, Horowitz and Hill (pg 103) describe the upper transistor (they label as Q2) being interposed in the collector circuit of the lower transistor (Q1) to prevent Q1's collector from swinging (thereby eliminating the Miller effect) and in their words (which I believe): "passing the collector current through to the load resistor unchanged". So I remarked "pretty well rock solid"

They (H&H) didn't say how well the upper transistor prevented swinging, but clearly the voltage swing will be a fraction of what it will be across the load resistor.

You have pointed out, that this swing will be about equal to the signal input voltage. Also, that it appears then, the voltage gain of the lower transistor is -1.

However, my main point was that if the signal current in the load is generated by the lower transistor, and passes unchanged through to the load, then overall, the voltage gain of the cascode circuit, is attributable to the lower transistor, not the upper one, so I simply disagreed that the lower transistor contributed no voltage gain.

Also, this means if you wanted in a cascode transistor circuit, you could replace the upper transistor with a super beta transistor with an hfe of 1000 and the gain of the circuit would stay the same. So the gain (overall) is determined by the lower device.

Last edited by Argus25; 8th Jan 2019 at 10:25 pm. Reason: add remark
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Old 8th Jan 2019, 10:11 pm   #50
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No, the transconductance gain of the circuit is attributable to the lower transistor/valve/FET. The voltage gain comes from both, except perhaps for a BJT cascode where all the voltage gain comes from the upper device.

Think about a valve cascode. Doubling the anode load resistor will roughly double the overall voltage gain. In some circumstances (e.g. fairly high anode load resistor) doubling the resistor will nearly double the signal voltage at the lower anode/upper cathode, while the voltage gain of the upper valve remains relatively unchanged - it cannot exceed mu. So most of the doubling of gain is contributed by the lower valve.
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Old 8th Jan 2019, 11:39 pm   #51
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No, the transconductance gain of the circuit is attributable to the lower transistor/valve/FET. The voltage gain comes from both, except perhaps for a BJT cascode where all the voltage gain comes from the upper device.
If the transconductance gain is due to the lower device (which is what I have been saying too) The only thing that converts transconductance gain to a voltage gain is the load resistor, not the upper device, the voltage gain is -gm.R where the gm is the lower device's gm and R the load resistance.

So can you explain then, how the upper device can increase the voltage gain beyond that provided by the lower device's gm and the load resistance product, that is the part I'm having trouble understanding. I accept that measured across the lower device's terminals the gain looks like -1, but it is still the lower device's gm that does the amplification, its current merely passes through to the load (as its a series circuit), making it appear as though the large signal variations there are due to the upper device.

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Old 9th Jan 2019, 12:05 am   #52
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There are more than two nodes moving in voltage, therefore there are more than one transfer functions or gains at play

The transfer function of the bottom device base voltage to the upper device collector in a cascode circuit is the overall gain of the amplifier. This is of interest because it is what the circuit is required to achieve.

The transfer function of the bottom device base to the bottom device collector (== top devive emitter) is also of interest because this is the particular gain which sets the amount of Miller effect and thus dominantly controls the upper limit of the frequency response of the whole amplifier. It may not seem like an interesting node because we are not doing anything with it, but it is certainly doing things with parameters affecting the nodes we are using.

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Old 9th Jan 2019, 3:55 am   #53
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There are more than two nodes moving in voltage, therefore there are more than one transfer functions or gains at play

The transfer function of the bottom device base voltage to the upper device collector in a cascode circuit is the overall gain of the amplifier. This is of interest because it is what the circuit is required to achieve.

David
I can see your point about the nodes.

Also I can see, that in Cascode, that one could regard the upper device as a grounded grid or grounded base amplifier, with the signal drive to it being equal to the input signal to the lower device (but inverted) applied to the emitter or cathode; then the circuit voltage gain with that reasoning, could be attributed to the upper device.

But it is interesting that the upper device could be completely removed, allowing the plate voltage of the lower device to move around, with no great overall change in the voltage gain (at low frequencies where Miller effect wasn't dominant). So I find it difficult, viewing the circuit as a working whole, to ascribe its voltage gain to the upper device, even though that is a reasonable way of looking at it too.
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Old 9th Jan 2019, 5:37 am   #54
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The lower device looks at the voltage swings on its grid, cathode and anode. It hasn't got a clue what goes on at nodes it's not connected to. Grid goes up 1 volt, cathode stays still, anode to cathode current goes up Gm mA, anode falls 1 volt. Within that triode's own frame of reference, it sees itself as having a gain of -1. These are the voltages across Cga. Cga sees twice the input voltage across it so we could replace Cga with a capacitance twice as big from grid to ground. Miller factor =2.

The upper triode looks at the voltage swings on its grid, cathode and anode. It too hasn't got a clue what goes on at nodes it's not connected to. It sees its cathode current go up by Gm mA, and its grid stay still, so it has to let its Vc fall by 1v. so that it can carry that current. It pulls Gm mA out of whatever impedance the circuitry is connected to, Ranodeload. So the anode voltage falls by Gm milliamps times Ranodeload. The top triode, as far as it can see is a grounded grid amplifier giving unity current gain, for it cannot do otherwise and it sees signal voltages on its cathode and anode of 1v and Gm*Ranodeload volts. Within the only frame of reference available to it, it thinks it has a personal gain of Gm*Ranodeload. Its Cga acts to load the anode to ground, and Cgc acts to load the cathode to ground. At high frequencies these rob some off the cathode current and some of the current into the load. We have three poles in the response.

If we dump the top triode and connect the anode of the bottom triode to where the anode of the top triode went, quite right, the overall circuit gain at low frequencies is the same as it was with the cascode. And we now have only a single pole, but that pole is at a much lower frequency because it now sees the input signal voltage across it times (1+overall gain) Lorts of Miller factor.

So the cascode is a deal with the devil. It costs you an extra valve structure AND needs cathode-heater isolation AND either low cathode-heater capacitance (or chokes) and it gives you more poles in the response. All you get for these costs is that each of those poles is usefully higher in frequency than the single pole of the plain single triode amplifier would have been.

If you're after high frequency amplification, it's not a bad deal.

If you're not, you don't have to spend the money. AND you can get more anode swing for the same HT.

The cascode really is a 2-stage amplifier. Each stage has clear demarcation of its input and output. We can calculate all the signal voltages and currents. Two divisions and we can calculate the voltage gain in each stage. These are real gains, measurable with an AC voltmeter or scope. They look like ducks, they quack like ducks, and by golly they are ducks.

Yes the bottom stage alone, with the top one removed would give the same LF voltage gain. But in the cascode circuit, the bottom valve is NOT alone. It has its friend above it. The anode voltage swing of the bottom triode is modified by the presence of the top triode. In the cascode circuit, what gain the bottom valve would have had in a different circuit doesn't happen. The bottom valve is in the cascode, not in that other circuit.

************************************************** *********************

Now, cascodes really come into their own with bipolar transistors. Cbc is the C of a reverse biased diode junction. It is a varactor diode! Ugh! The damned thing is non-linear. So the Miller modified pole isn't just low in frequency, it moves with the signal voltage. Each signal voltage component phase and amplitude modulates each other. Ooooh, Narsty!

So cascoding to shove the pole up out of the way reaps extra benefits.

A few decades ago I designed an audio amp for ultra-low distortion even at full welly, and across the full audio band - both at the same time. Cascoding played a big part in pulling off this stunt. It wasn't needed because of particularly high frequencies being handled, it was needed for its benefits in intermod and harmonic distortion at pedestrian frequencies. This amp was ridiculous... a truly stupid amount of overkill and almost completely pointless, but doing it was great fun and it's sat in my lounge for 38 years. Not quite pointless because I learned a bit doing it and that knowledge got re-used all over the place.

A hobby is also a place where you can play with things which could not be financially justified, or any other sort of justified, at work.

David
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Old 9th Jan 2019, 10:43 am   #55
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However, my main point was that if the signal current in the load is generated by the lower transistor, and passes unchanged through to the load, then overall, the voltage gain of the cascode circuit, is attributable to the lower transistor, not the upper one, so I simply disagreed that the lower transistor contributed no voltage gain.
Agree! Overall voltage gain of the cascode is gm (lower) x load resistor (upper). The gm of the upper device does not feature in the calculation.

More accurately, the above formula should be multiplied by:
(hfe (upper) -1)/ (hfe (upper)) to account for base current of upper device. But as that's so nearly 1, we can forget that - Argus25 you made that point.

Where gm of the upper device DOES come into play, as we have seen, is determining the magnitude of the (small) voltage swing at the midpoint. And this small voltage is what we have been debating. Its significance is that it has quite a big impact on Miller fed-back capacitance. I've just been playing with paper and pencil the last couple of days, but David RW has obviously 'been there, done that.'

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For the transistor case, Horowitz and Hill (pg 103) describe the upper transistor (they label as Q2) being interposed in the collector circuit of the lower transistor (Q1) to prevent Q1's collector from swinging (thereby eliminating the Miller effect) and in their words (which I believe): "passing the collector current through to the load resistor unchanged". So I remarked "pretty well rock solid"
H & H is a super book, full of great engineering approximations and tips which work well in practice. But one does do well to challenge things, because that leads to greater insight. Miller effect in a cascode is vastly reduced, but not entirely eliminated. Where I see 'pretty well rock solid' my thought then is, 'how "pretty well" exactly?' and a bit of analysis has shown that it's 'pretty well compared to output, but not compared to input', being equal - and equal whether valves OR transistors.

If the middle node could be TOTALLY rock solid, then Cin = Cgk + Cga (capacitances of lower device) but with the middle node moving as we have found, Cin = Cgk + 2Cga. If Cga = Cgk = 1.5pF (typical values for triodes), then you get Cin = 3pF with the invalid assumption that the middle node is rock solid, and 4.5pF if you do it properly. That's a 50% increase!! And the same with transistors.
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Old 9th Jan 2019, 11:48 am   #56
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Here's the Cascode folks with a practical demo:

https://www.youtube.com/watch?v=Op_I3Ke7px0

Lawrence.
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Old 9th Jan 2019, 12:10 pm   #57
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Agree! Overall voltage gain of the cascode is gm (lower) x load resistor (upper). The gm of the upper device does not feature in the calculation.
The best analysis of valve Cascode I have seen was done by Millman & Halkias where they point out that the voltage gain of the Cascode circuit is:

Gain = -(u(u+1)R)/(R + (u+2)rp).

rp is the plate resistance and R the load resistance.


And as pointed out in a post above, if (u+2)rp >>R and if u>>1 then the gain is simply:

Gain = A1 = -uR/rp = -gmR

Also, they outline what happens when you apply a variable voltage to the grid of the upper valve and then the lower valve acts as an impedance of magnitude rp in the cathode of the upper valve and the voltage gain, A2 for the signal applied to the upper valve is:

A2 = -uR/(R+ (u+2)rp)

So then if you apply two sinusoidal signals V1 and V2 to the lower and upper grids respectively, the output voltage, Vo is:

Vo = A1V1 + A2V2.
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Old 9th Jan 2019, 12:31 pm   #58
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If the transconductance gain is due to the lower device (which is what I have been saying too) The only thing that converts transconductance gain to a voltage gain is the load resistor, not the upper device, the voltage gain is -gm.R where the gm is the lower device's gm and R the load resistance.
You are confusing two things:
1. a simple model to calculate approximately what the circuit does
2. an explanation of how the circuit works
The simple model says that the upper device merely passes on the current from the lower device, so overall voltage gain is -gm x Rload.
The explanation of how the circuit works means looking at the voltage gain of a grounded cathode stage with a low impedance load, followed by a grounded grid stage with an input impedance modified by its anode load and an output impedance modified by its source impedance.

Most of the voltage gain in a cascode comes from the upper device. Almost all of the transconductance gain comes from the lower device. These two statements are not contradictory, because they are saying different things.

Quote:
I accept that measured across the lower device's terminals the gain looks like -1,
No. It doesn't "look like -1"; it actually is around -1, or somewhat more if a high value anode load is used.

You have actually been arguing for two contradictory positions:
1. the lower device provides all or almost all the gain (it has the transconductance, the upper stage merely passes on the current)
2. the lower device provides almost no gain (it has negligible output voltage)

The truth is that the lower device provides most of the output current and the upper device provides most of the output voltage. As I said earlier, if you boost the gain by having a really big anode resistor then some of this gain has to come from the lower device because the upper device cannot have a voltage gain exceeding mu.
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Old 9th Jan 2019, 3:28 pm   #59
John10b
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Default Re: Puzzling audio circuitry

Lawrence that you for the video link, I thought it was an excellent demonstration.
Cheers
John
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Old 9th Jan 2019, 3:38 pm   #60
ms660
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Default Re: Puzzling audio circuitry

No problem, that bloke does some excellent videos.

Lawrence.
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