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Components and Circuits For discussions about component types, alternatives and availability, circuit configurations and modifications etc. Discussions here should be of a general nature and not about specific sets. |
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9th Jan 2019, 7:33 pm | #61 |
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Re: Puzzling audio circuitry
Only to the extent that I was trying to point out that, despite the lower device having no voltage gain (as measured across its input and output terminals) its transconductance is responsible for the conversion of signal voltage input into anode signal current (and the current of the series circuit that comprises cascode) and therefore the signal current in the load, regardless of the presence of the upper device (for low frequencies at least). But I agree with all your points too.
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9th Jan 2019, 9:19 pm | #62 | |
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Re: Puzzling audio circuitry
Quote:
In Cascode of course, the drive is primarily a current source (or sink) from the lower valve (or transistor) and the dynamic current signal (derived from the lower device's transconductance) is in no way "amplified" by the upper device. The dynamic current signal is converted to a dynamic voltage in the load resistance in the usual way. Thinking about it this way, it is hard to attribute any "gain" to the upper device (in the application) even though measuring across its 3 terminals (as RW pointed out) it is behaving in the usual way as a transconductance amplifier, but its current is being controlled by the lower device. |
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9th Jan 2019, 9:31 pm | #63 |
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Re: Puzzling audio circuitry
As I keep saying, the lower device has voltage gain if you use a big enough anode resistor. Just to pick some numbers at random, assume gm of 5mA/V and mu of 20. An anode resistor of 20k means a voltage gain of around 100. 20 of that comes from the upper valve, and 5 of that comes from the lower valve, approximately. 10mV in would become 50mV at the 'join' and then 1V at the output. What would you call an amplifying stage with 10mV in and 50mV out (or 50mV in and 1V out)? I would call it a voltage amplifier.
I have never said that the current is amplified by the upper device. I, and others, have said repeatedly that the voltage is amplified by the upper device. You seem fixed on the idea that a triode is a transconductance device. It is also a voltage amplifier. You need to use the most appropriate model, not just stick to just one all the time. Your simplified model of the cascode will not enable you to derive the full equations you quoted; to do that you need to take account of everything which is going on. Last edited by G8HQP Dave; 9th Jan 2019 at 9:32 pm. Reason: typo |
9th Jan 2019, 9:49 pm | #64 |
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Re: Puzzling audio circuitry
Yes I agree and the valves are not ideal transconductance amplifiers either and that is indicated by the presence of the plate resistances in the equations (and of the valves) which alter the the voltages seen on the actual anodes and affect the gain calculations accordingly.
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9th Jan 2019, 11:04 pm | #65 | |
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Re: Puzzling audio circuitry
Quote:
Also I have been pretty well stuck on the idea of a transistor as a current amplifier all my life too, though I have a lot less difficulty thinking of them as transconductance devices by programming their base-emitter voltage than I do thinking of a valve as a voltage amplifier. So maybe I'm 3/4 of the way there! |
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10th Jan 2019, 11:21 am | #66 | |||
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Re: Puzzling audio circuitry
Quote:
The lower device does not have zero voltage gain, it has unity gain, assuming the gm's of the two devices are equal. Quote:
Quote:
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10th Jan 2019, 12:07 pm | #67 |
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Re: Puzzling audio circuitry
Sorry, too late to edit - my figure of merit is rubbish, I posted without thinking it through! good transconductance device with voltage in, current out, has a high resistance at at the input, and output current hardly varies with voltage at output node, so output resistance is high too. And it's helpful to factor in the transfer parameter too. So figure of merit is Rin x Rout x gm. (The triode beats the bipolar junction transistor).
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10th Jan 2019, 1:12 pm | #68 | |
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Re: Puzzling audio circuitry
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To be able to calculate the voltage gain for cascode it is necessary to know what the impedance is looking into the cathode of the upper device (once this is known its dead easy). The best way to do it is to use the principle that the impedance between two nodes in a circuit equals the open circuit voltage divided by the short circuit current. Doing that, the upper device has an impedance of (rp+R)/(u+1) looking into the cathode. Which also shows how u effectively lowers the input impedance of the upper device. Once this is known it is all downhill easy, because in the cascode circuit you can then replace the upper device by an impedance of (rp +R)/(u+1) and replace the lower device by a voltage generator u(Vin) in series with an impedance rp. Hence the generator is u(Vin) and the total load is rp + (rp + R)/(u+1). Then to calculate the output current it is uVin/(rp + (rp+R)/(u+1) ). Then, multiplying this current by the load resistance R to get Vout, yields their exact equation for Vout/Vin. |
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10th Jan 2019, 3:47 pm | #69 |
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Re: Puzzling audio circuitry
Hugo,
That's brilliant. Deriving the formula in the book yourself from first principles gives an insight, and also verifies what is written! I have seen the occasional statement in a book which is wrong. Not often, but it happens. And then, the reader is misled - especially if he just takes it as gospel. |
10th Jan 2019, 4:08 pm | #70 | |||
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Re: Puzzling audio circuitry
Quote:
Quote:
Quote:
So to derive the equations you quoted you assume that the valves are voltage amplifiers with finite impedances. |
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10th Jan 2019, 9:59 pm | #71 |
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Re: Puzzling audio circuitry
We are in agreement on this. The thing though that perhaps I was a bit obsessed with, shown in the final step that is required to complete the task of the equation solution for the overall gain, it is required that the signal current in the series circuit of cascode arrangement, is multiplied by the external load resistance to calculate Vout. I was more concerned about the way the signal current came about by the lower device (focusing on transconductance) and the role of the lower device generating it, vs the upper device. But now I know better, thanks !
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11th Jan 2019, 1:27 pm | #72 |
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Re: Puzzling audio circuitry
Yes, you can analyse it either way - voltage amp or transconductance. You can even use a mixture of the two. For some reason it took me a long long time to realise that
mu = Ra x gm applies just as much to a circuit as it does to a device. |
11th Jan 2019, 6:21 pm | #73 | |
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Re: Puzzling audio circuitry
Quote:
The gm of the upper device does have a major impact on how 'still' the join is. And the signal at the join makes a big difference to the Miller fed-back capacitance. |
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14th Jan 2019, 12:53 pm | #74 |
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Re: Puzzling audio circuitry
The lower triode cathode node also has complexity from the feedback voltage and current. It may be remiss to just assess that as a voltage or current feedback node.
Morgan Jones (3rd Ed) had a go at assessing that node in the similar, but probably simpler Williamson input stage - its not a simple set of calculations. |
15th Feb 2019, 6:29 pm | #75 |
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Re: Puzzling audio circuitry
I came across the circuit looking through an old Radio Constructor magazine! See attached.
There's no other explanation other that what is shown in the photo, but hopefully all clear. Interesting to see is used in a power output stage! |
18th Feb 2019, 2:40 am | #76 |
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Re: Puzzling audio circuitry
It looks like a somewhat more complicated version of the familiar Philips totem pole output stage.
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18th Feb 2019, 3:46 pm | #77 |
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Re: Puzzling audio circuitry
It is complicated, but that is largely because it uses pentodes with the complication of a screen-grid supply, suitable decoupled, AC isolated, etc.
If you replace the pentodes with triodes, it simplifies. Then consider removing the upper triode, and have just the lower one. For ease, think of the load as connected to HT+. Then load current flows through the cathode resistor of the (removed) upper valve, creating a voltage drop across it as in the sketch: |
18th Feb 2019, 4:17 pm | #78 |
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Re: Puzzling audio circuitry
And then, plug in the upper valve, noting that it gets exactly the drive it needs to deliver an equal current into the load.
It looks as though the lower valve operates into a load RL+Rk and the upper valve RL, but this is not actually the case. |