Grid current

[Malcolm Irving]

Clarification and correction:

When I said 'clamping of the grid to near 0V', I was thinking Vgk near 0V. So if we have cathode bias of Vk = +1.5, Vg would be clamped to just above 1.5V.

So, I should have said 'the stage 2 side of the coupling cap stays near +1.5V (clamped)'.

I should also have mentioned that, for this example, I am assuming there is no grid stopper, so the coupling cap is connected directly to the grid of stage 2.
If there is a grid stopper, Vg is still clamped near +1.5V but there is a voltage drop across the grid stopper (while grid current is flowing) to consider before we get back to the voltage at the stage 2 end of the coupling cap.

[Malcolm Irving]

There's a great description and explanation of blocking and how it relates to grid current etc., including schematics and graphs, in Merlin's Preamp book:
'Designing Tube Preamps for Guitar and Bass', Second Edition, Merlin Blencowe, pages 100-104.

In his example, he also has 150V on the stage 1 plate. (It's a coincidence, honestly, I haven't read that section of the book recently!) But in his example the cathode bias is 1V instead of 1.5V.

[Chuck H]

If you drive the grid with a 50 Ohm signal generator direct with no stopper, you won't see much clipping. If you have any kind of reasonable stopper like 10K or more, the clipping will be quite sharp.

I'm responding to this as read in the saga so far, so forgive me if this has been covered below...

Thanks LT, for that excellent observation. I hope to expound on it for clarity. The reason for a lack of clipping in the above case is due to the low impedance signal source allowing the grid to conduct without clamping, yes? Too little voltage can develop on the grid to shift the bias relationship significantly if the source impedance is that close to 0V.

[Chuck H]

In my own low tech and very mechanical (so possibly flawed) understanding of the operation, when a grid reaches the cathode voltage it would need to draw current in order to track the signal operation. It can't draw this current through a typical high impedance source so it clamps. The grid at this point is trying to draw current through the coupling cap of the previous stage. This causes the voltage build up we've called grid loading, where a negative charge is developed and the bias is shifted. This is relative to the time constant of the coupling cap and the grid load resistor, which is always imperfect. In the case of a low impedance signal source the elimination of grid resistance allows for a lower time constant and reduces voltage building up on the grid. In this case adding grid resistance will increase the time constant so the grid can't unload as fast. In the case of a higher impedance signal source adding grid resistance imbalances the time constant between the grid and the coupling cap. So now the coupling cap is feeding a lower impedance circuit than the grid sees. This somewhat allows the capacitor to unload faster than the grid, which is trying to draw current, is able to load it. So in this case we're just back to plain ol grid clamping and reducing grid loading.

Corrections accepted.

[lowell]

I unfortunately lost that awesome book a couple years ago. I'm still not totally clear on this but I'm gonna reread all these responses again and see if it come together. I'm a super visual person so words don't tend to work well at first. BUT once I have a visual understanding in my (slow) brain it'll stick with me. 😆

[66 Kicks]

Thanks LT, for that excellent observation. I hope to expound on it for clarity. The reason for a lack of clipping in the above case is due to the low impedance signal source allowing the grid to conduct without clamping, yes? Too little voltage can develop on the grid to shift the bias relationship significantly if the source impedance is that close to 0V.

The 50 ohms refers to the type of coax cable used for the generator lead, not the output impedance. For the length of lead typically used, the resistance is zero for as close as we can measure it.

The reason that there is no clipping without a grid-stopper is because the function generator forces the grid to follow the signal. That generator will put out the same signal whether it sees a 1M or a 1K load.

The grid doesn't clamp all by its self, it just conducts more current as the voltage on it becomes more positive. Its what you put in front of that grid that makes it look like it is clamped, a grid-stopper for instance. With a resistor between the voltage source and the grid, the grid current will create a voltage drop across the resistor. What gives it the clamped look is the exponential rise in current with increasing Vgk and the resulting exponential increase in the voltage drop across the resistor. The grid voltage will continue to rise with increasing source voltage, but the rate of change for the grid will decrease with increasing source voltage.

[Chuck H]

The 50 ohms refers to the type of coax cable used for the generator lead, not the output impedance. For the length of lead typically used, the resistance is zero for as close as we can measure it.

Ok. The old 50 ohm coax signal generator lead thing. Got it

The reason that there is no clipping without a grid-stopper is because the function generator forces the grid to follow the signal. That generator will put out the same signal whether it sees a 1M or a 1K load.

Right! Ok. So the impedance has nothing to do with it. It's just that the generator signal is so forceful. Got it

The grid doesn't clamp all by its self, it just conducts more current as the voltage on it becomes more positive. Its what you put in front of that grid that makes it look like it is clamped, a grid-stopper for instance. With a resistor between the voltage source and the grid, the grid current will create a voltage drop across the resistor. What gives it the clamped look is the exponential rise in current with increasing Vgk and the resulting exponential increase in the voltage drop across the resistor. The grid voltage will continue to rise with increasing source voltage, but the rate of change for the grid will decrease with increasing source voltage.

Is this SGM.?. Someone is messing with me. Where's Alan Funt!?!

[Enzo]

Malcolm, yes you can call all of them filter caps, and everyone will understand and not argue. But in a tube amp, the ripple should be gone by the screen node, and the filtering action in the preamp nodes is more about signal than ripple.

[Malcolm Irving]

Hi Enzo,

Thanks for making that point. I do see what you mean, but I'm not totally convinced. Just as a 'back of the envelope' calculation, let's say we have an RC power supply filter stage of a 16uF shunt cap with an 8k series resistor. The impedance of the cap at 100Hz is 99.47 ohms. So the 100Hz ripple is reduced (approximately) by a factor of (8000+99.47)/99.47 which is a factor of 81. That is not very different from the gain that we would get from an amplification stage.

As we go down the filter chain (towards the preamp) the ripple is being cumulatively reduced, but the cumulative gain of the amplifier stages is increasing, so that a very small ripple at preamp stage one is greatly amplified by the time it gets to the speaker output.

I need to do some bench measurements of ripple before I am convinced either way.

[Enzo]

The gain of the amp may increase through the stages, but that gain is not placed on the B+. The noise, be it hum, signal,or whatever, on the plate of a tube might leak into the B+ at the top of the plate resistor, so the big cap there prevents that. remember, the tiny ripple left at the last node is not applied to the grid, it is applied to the B+ for the stage, so AFTER the amplification of that stage.


I have never bothered to math out the B+ string, but I believe your number. But that means ripple there is 1/81 of whatever ripple was left from the previous nodes. I know from experience I will see a few volts at the reservoir cap, then the screen cap will have little or none, but what if it were a half a volt at the screens? Two more B+ nodes downstream, and we have 6mv ripple (a 1/81 reduction) , if I have two such reductions I get something like 0.07mv. I call that none. Even if only one stage, that 6mv isn't shabby. but what is it up against?

If I had a 100k plate load resistor and your 8k B+ series resistor, and did not have the filter/decoupling cap, I figure at the cap node we'd have 8k/108k of the signal riding there. If we have oh say 30v of signal, that means about 2.2v of signal trying to escape to other stages on that node. That is a hell of a lot more than a couple millivolts of ripple. More signal than that makes it even worse. Adding the decoupling cap stops it cold.

[Malcolm Irving]

I agree with what you are saying there. I'm certainly not suggesting that we don't need the filter caps. I'm just discussing whether or not they perform a useful ripple filtering function all the way back through the preamp. When we are concerned about buzz from an amp we are really thinking about the silent bits in the guitar playing, so a few millivolts of ripple near the front of the preamp can be important. The ripple voltage modulates the plate current of V1 and from there affects the voltage on the grid of V2.

[Enzo]

Sorry I was unclear, of course we need the cap. I re,moved it from the circuit only to calculate what it had to do, what it was up against. With it gone we could determine the several volts of signal crosstalk we needed to decouple, compared to the much smaller ripple there. The cap when added kills the crosstalk as well as any residual ripple.

The buzz/hum you hear when muting your guitar strings but the pickup volume up is not usually from ripple. Much more of that comes from radiated fields picked up by the guitar, from grounding issues and ground currents in the amp, as well as heater noise. In fact when I hear ripple in the first stages (120Hz instead of 60HZ, or 100/50 across the pond), it is way more likely to be from improper ground paths than the ripple itself. Ripple currents ground return instead of voltage if you will.

The important consideration is the ripple as compared to the noise floor.

[Malcolm Irving]

Agreed. But if we (hypothetically) remove the 16uF cap, why leave the 8k resistor there? Leaving the 8k there (without the cap) is like changing the plate load from 100k to 108k, but being supplied from a filter cap further up the chain?

I think we are actually in good agreement and we are starting to split hairs!

As far as buzz is concerned, I'm thinking about a guitar amp in a studio where the sound engineer puts a mike up against the speaker and monitors it through headphones (before the guitarist even plugs in) and says 'I can hear a buzz from that amp. Get me another one!'

[Chuck H]

Maybe pedantic at this point. Take it for what it's worth as you like.

There are a few guitar amp designs that mimic Malcolm's notion by using the last node in the chain to supply more than two series stages and sometimes several parallel stages as well. Some of them are fairly high gain amps too and don't have any reputation for ripple hum. They do, however have a reputation for developing oscillations due to amplifier crosstalk on the last power supply node when the filter degrades. That's the reason we commonly call the first node in the B+ chain "reservoir cap" and we often call the following stages "decoupling caps". They are often all called "filter caps". I think because of this they are sometimes all considered to be doing the same function. Which isn't necessarily the case.

[Malcolm Irving]

OK guys - I'm going to concede on this one! I think we all understand what the caps are doing and its been an interesting discussion.