# Valve (tube) amplifier schematic help

Discussion in 'General Electronics Discussion' started by Solidus, Aug 22, 2012.

1. ### KrisBlueNZSadly passed away in 2015

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Yes, conductance is the reciprocal of resistance. The formula for conductance (G) is G = I / V (which is the reciprocal of V / I, the formula for resistance).

TRANSconductance is also calculated as I / V but the current and voltage used in the calculation are not in the same circuit. In the case of a valve (or FET), the current is the anode (or drain) current, and the voltage is the grid (or gate) voltage.

Valves and FETs are voltage-driven, that is the grid/gate responds to voltage, but the output is a current that flows into the anode/plate. (This current is converted back into a voltage by the anode/drain resistor.)

The transconductance figure therefore tells you the relationship between the input voltage and the output current, expressed as output_current / input_voltage.

When we're interested in signal amplification, we use lower-case parameter names, i.e. 'g' instead of 'G', to indicate that these figures are not absolute numbers; they relate to small CHANGES, which represent the signal being amplified, and which appear ON TOP OF the DC bias conditions that put the device into the part of its characteristic where it will amplify a signal.

Let's take a typical small-signal voltage amplifier using a triode valve. It has an anode load resistor of 100k and a 200VDC supply. Generally you want the anode to sit at about half the supply voltage, which is 100V, so there will be 100V across the load resistor. That means the quiescent (no-signal) anode current will be 1 mA. These figures aren't directly relevant to transconductance; I'm just trying to set the scene

A valve is biased into its linear region by applying a slightly negative voltage to the grid (at least, that's true of the valves that I've dealt with). In this circuit, a typical triode might want its grid to be biased at -2V (relative to the cathode) to give an anode current of 1 mA.

Now if you apply a small AC voltage to the grid (in addition to the static bias voltage that sets the quiescent DC operating conditions), the anode current will vary in a corresponding way. This will cause the voltage across the load resistor to vary, and therefore the anode voltage will vary. This is amplification.

Say that a gate signal of v = 0.1V RMS causes the anode current to vary by i = 0.03 mA RMS. With a 100k anode load resistor, 0.03 mA RMS corresponds to 3V RMS. So the circuit has a voltage gain of 30 (vout = 3V RMS output for vin = 0.1V RMS input).

The valve's transconductance in this circuit is i / v which is 0.03 mA / 0.1V which is 0.0003 Siemens (also sometimes called 'mho' - 'ohm' written backwards). That is 300 µS (microsiemens). (Different from 300 µs, which is microseconds.)

Well, you asked for it ;-)

Last edited: Aug 31, 2012
2. ### CDRIVEHauling 10' pipe on a Trek Shift3

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May 8, 2012
I can't speek for Solidus but I liked it!

Chris

3. ### Solidus

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Jun 19, 2011
That was actually not as bad as I thought.

Of course you explained it very well - if you can lay down the basics for someone like me in one shot, you've done very well!

So let me apply what I've gathered to this circuit and you tell me if I'm on the right track.

The input is chained to the grid. So if I play a guitar into the amp input, the grid bias will change proportionally according to the signal. It will modulate the anode current accordingly, and so you are receiving at the anode a larger signal which is an amplified "carbon-copy" of the input grid bias.

If I read and understand correctly, what you're saying is you're not making the input signal any larger - but you're using it as a guide and model to shape the signal coming out of the anode.

Do I have any hope with this stuff?

Last edited: Aug 31, 2012
4. ### CDRIVEHauling 10' pipe on a Trek Shift3

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May 8, 2012
That's a good enough understanding as any.

Chris

5. ### KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
Thanks And thanks Chris.

I think you've got it pretty well. I have a few corrections to your terminology.
Yes, "coupled" to the grid, normally through a coupling capacitor, which couples the AC signal while leaving the DC bias unaffected.
"Bias" refers to the static conditions (voltages) that put the valve into its linear (amplifying) region by setting up the quiescent, DC conditions. When the signal is applied "on top of" that bias, the bias (the average DC voltage) itself doesn't change, but the voltages vary slightly; these variations are the signal itself, superimposed onto the DC bias.
"... amplified "carbon-copy" of the grid SIGNAL". Yes, that's right.

The signal at the anode is actually upside-down compared to the grid signal - this is an "inverting" amplifier. It sounds exactly the same, but if you look at the signals with an oscilloscope, they are upside-down relative to each other. This is because when the grid voltage increases, the valve conducts more current, which pulls the anode voltage downwards; when the grid voltage decreases, the valve conducts less current, and the anode load resistor pulls the anode voltage upwards.
That's right. An amplifier doesn't CHANGE the signal at its input. It amplifies the input signal by creating a copy of that signal at the output with a higher amplitude (higher voltage, in this case).
Yeah I think you've got it.

This amplifying configuration is called "common cathode", because the cathode is common to both the input and the output signals. There are two other configurations: common grid and common anode (also called cathode follower). They use the valve in slightly different ways, and have different characteristics.

FET and transistor amplifiers also use these three topologies, with different names of course:

1. Valve: common cathode; transistor: common emitter; FET: common source. This circuit. Used for general amplification, and switching (in digital applications).
2. Valve: common grid; transistor: common base; FET: common gate. Has a low-impedance input and is used mostly in high-frequency circuits.
3. Valve: common anode aka cathode follower; transistor: common collector aka emitter follower; FET: common drain aka source follower. Has no voltage gain but has high current gain. Used for signal buffering to increase drive capability.

Configurations 1 and 3 are the most widely used. Look them up on Wikipedia or google if you're interested.

Last edited: Aug 31, 2012
6. ### CDRIVEHauling 10' pipe on a Trek Shift3

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May 8, 2012
Besides the excellent theory of operation that Kris has given you, including the fact that the plate signal is ideally an image of the grid but inverted, you inadvertently touched on what I was trying to convey earlier. What I'm saying is that the signal at the Plate derives its power from the power supply, not the input signal.

Kris, I can't type as much as you can without nodding off or getting wrist cramps!

Great job!

Chris

7. ### KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
Yes, that's a fair way of looking at it. The input signal just controls the active element; all of the energy in the output signal comes from the power supply.

A common cathode Class A amplifier can't be more than 50% efficient, but as a small-signal voltage amplifier, it isn't designed for efficiency anyway. That's why I flagged the talk of input and output power, and "efficiency", as not really relevant.
Thank you Chris I'm lucky I'm a fast typist. And I have far too much time on my hands. I REALLY need a job! LOL

8. ### Solidus

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Jun 19, 2011
So this means that the issue of relative gain is hidden in the transconductance figures?

If this is indeed how transistors (basically) function, you have no idea what ideas you have just given me for things. I struggled to understand how transistors functioned, what physical principles made them do the things they do, but it makes a lot more sense now.

It bears saying that I like the notion of valves better - the idea of electrons flying through grids in vacuums sits better with my brain than the idea of electrons traveling through differently-doped regions of a silicon wafer.

So the term "bias" refers to the initial voltaic conditions that must be set for the unit to work? (essentially overhead costs for the valve?)

I'm working on finishing the enclosure now, with any luck, should be done in a couple hours.

You guys will make an electronics maniac out of me with that lesson in about a month or so close to everything in my room will probably have a valve or transistor on it to amplify it just for kicks.

Last edited: Sep 1, 2012
9. ### Solidus

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Jun 19, 2011
Also, Kris, when you say the input is coupled to the grid with a coupling cap, is that the purpose of the 47nF cap between the input line and the grid pin #2?

10. ### KrisBlueNZSadly passed away in 2015

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Yes, the voltage gain of the stage depends on the transconductance and the anode load resistor. The transconductance determines how the anode CURRENT varies in response to the grid voltage, and the anode resistor converts that varying current into a varying VOLTAGE.

The voltage gain is equal to the transconductance multiplied by the anode resistor. In my example, transconductance was 300 uS and the load resistor was 100 kilohms. The product of 300e-6 and 100e3 is 30, which is the voltage gain of the stage.
Great! That's brilliant.
Yeah. To be honest I've never really cared how transistors work internally, and I've been designing with them professionally for 20 years!

As long as you're familiar with their behaviour in reasonable detail, I believe it's fair to consider them to be black boxes that have that specific behaviour and can be modelled using a set of simpler approximations that are relevant to how they are used in practice. I'll probably cop some flak for saying that here. You can only really understand their behaviour by understanding their internal operation, but unless you're going to be designing ICs, or microwave radio circuitry, you just need to know the behaviour in enough detail for the application you're designing.

BTW, FETs are closer to valves than transistors are. FETs and valves are both voltage-controlled, whereas transistors are current-controlled. The depletion-mode N-channel JFET is the closest semiconductor equivalent to a triode.
Yes, that's right. The DC, static, quiescent conditions.

If you look at a "mutual" characteristic graph for an active device, which graphs the voltage or current at the control terminal (grid, gate, base) (on the horizontal axis) against the current at the output terminal (anode, drain, collector), you'll see there is part of the graph where the line is (roughly) straight, and at a certain angle. This is the "linear" region - the region that can be used for amplification, and if the device is being used as an amplifier, it is biased around the middle of that region.

Outside that region, the line is horizontal, so any small signal you couple onto the control terminal will have no effect on the output terminal, so it won't amplify. Those parts of the characteristic are used when the device is used as a switch, when you only want it to be ON or OFF.
Yes. It couples the AC signal from the input onto the grid, without affecting the DC conditions on the grid.

Last edited: Sep 1, 2012
11. ### Solidus

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Jun 19, 2011
Now the 12A7 valve series are dual-triode, so does that mean the signal goes through two different stages of amplification?

12. ### KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
Yes, in that schematic that you linked to in your original post. The two triodes are shown as separate objects, though only one has the heater shown because the heater pins (4, 5, 9) supply the heaters for both halves. The two triodes are cascaded, as you can see if you follow the signal from left to right in the diagram.

13. ### CDRIVEHauling 10' pipe on a Trek Shift3

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May 8, 2012
I believe this to be a modern incarnation that began with logic blocks and the way they're drawn separately, despite the fact that they're all contained in a single chip. I also believe that the advent of Spice modeling had a hand in this. Back in the day dual triodes were never drawn like this.

Chris

14. ### Solidus

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Jun 19, 2011
Now is it possible to make this a multi-stage unit by feeding the output into another identical tube circuit?

15. ### KrisBlueNZSadly passed away in 2015

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Sure. But isn't this circuit intended to be a distortion unit? The triodes are overdriven so the anode voltage is clipped (because it can't exceed the supply rail), supposedly resulting in a "valve sound"-style distortion. If two cascaded triodes provide so much gain that the second one clips, then feeding that clipped signal into a third triode probably won't do much to the signal (apart from inverting it again).

Also, consider what happens when your guitar strings are muted. Even though there's not enough signal from the guitar to drive the amplifier into clipping, there is hum picked up by the pickups, noise due to microphonics in the cables and movement in the plugs and sockets, and noise from each triode stage as well. The more gain you put in your distortion unit, the more these signals will be amplified. You'll get to the point where the distortion unit won't ever be quiet - while you're not playing anything, you'll get a mains buzz, radio stations, hiss, and thumps and bumps!

But of course, feel free to try it out! The best way to learn is by experimentation.

16. ### CDRIVEHauling 10' pipe on a Trek Shift3

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May 8, 2012
In addition to what Kris said it won't make much sense with a 9V B+ either..

Chris

17. ### Solidus

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Jun 19, 2011
This is true. Amplification could get so intense minute vibrations caused by air currents and static would go through the channel as well.

For learning purposes, could one of you show me how this circuit would be wired with transistors instead of a valve?

Also, I don't mean to start up a firestorm that will consume the universe with this, but what is the difference between a bipolar-junction transistor and a field-effect transistor? Can they be used for the most part interchangeably?

Kris, you said you work with these things, what are the practical considerations of each?

You two know not what you have done you have not only increased my confidence with this stuff a thousand-fold, but you have given me endless curiosity about it!

18. ### CDRIVEHauling 10' pipe on a Trek Shift3

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May 8, 2012
MOSFETs are primarily used for switching because they have a very narrow linear region. JFETs are preferred for amplifier applications. Their operation closely mimics a vacuum tube and have a broad linear region.

Chris

Last edited: Sep 5, 2012
19. ### Solidus

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Jun 19, 2011
Switching - as in operating as an automatic switch?

You're going to have to explain that one to me...amplification makes perfect sense, but I can't see how it would operate like a switch.

20. ### CDRIVEHauling 10' pipe on a Trek Shift3

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May 8, 2012
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