Grid current

[loudthud]

The photo below is a Voltage (Horizontal) vs Current (Vertical) trace of the input grid of a 12AX7 with a bypassed 1.5K cathode resistor. The center of the screen is zero Volts, zero current. You can see that no current flows until the Voltage reaches about + 3V. The current reaches a peak of just over 4mA. Once current starts to flow, the grid looks like about 800 Ohms. Because significant grid current is flowing, the operating (bias) point shifts somewhat. The second photo is the same setup but the grid current is limited and current starts to flow at just over +2V.

[lowell]

So the first photo has no stopper and the second does?

[Malcolm Irving]

A common-cathode triode stage amplifies whatever voltage appears on the grid (relative to the cathode). That amplification is quite linear even when Vgk is positive. But when grid current starts to flow (roughly speaking when Vgk is positive) it becomes harder to make the voltage on the grid go more positive. The source signal voltage might go more positive, but there is greater voltage drop across the source impedance and across the grid stopper (if there is one) due to the grid current, so the voltage on the grid cannot follow the signal voltage. This is grid-limited clipping.
With a very low source impedance and no grid stopper, the grid voltage can track the signal voltage, but a high source impedance and/or grid stopper causes the grid voltage itself to be clipped (and this is then amplified to give a clipped output from the stage).
Bigger grid stoppers do reduce the amount of grid current that can flow, but also increase the voltage drop between the source signal and the grid voltage. Bigger grid stoppers therefore cause ‘harder’ clipping due to grid-limiting. Smaller grid stopper (and low source impedance) give softer grid-limited clipping.

[lowell]

That all makes perfect sense to me. Thanks. Well, the one thing that doesnt make sense is that the common suggestion of increasing grid stopper reduces blocking distortion. Which maybe now I get because it reduces blocking, but not grid current flow. Correct? Blocking is a symptom of grid current.

[Malcolm Irving]

While grid current is flowing, the decoupling cap charges up to a higher voltage (across the cap). Once grid current stops flowing, the discharge of the cap is much slower (via other paths than the grid - e.g. via the grid leak resistor) and the higher voltage across the cap persists for some time. This extra voltage across the cap corresponds to a more negative bias on the tube, which may cause the tube to cut-off (hence the term 'blocked').

A high value grid stopper reduces the grid current and increases the time it takes for the cap to charge up. The bias therefore shifts less over a given time interval and there will be less blocking (within that time interval). However, if overdriving the input to the stage lasts for long enough, blocking will still occur.

The grid current also charges up the cathode bypass cap, if there is one. This also makes the bias more negative and contributes to blocking.

[lowell]

Coupling charges NEGATIVELY on the grid side though is my understanding. Doesn't make sense if it charges up positive. Just to be clear.

[Enzo]

The grid side of the cap is negative with respect to its other end at a plate. That doesn't mean the grid end cannot be charged up to a more positive potential thaqn it is at rest.

[Malcolm Irving]

Let’s say the stage we are considering (call it stage 2) is being driven by a similar previous stage (stage 1). Under normal (non-overdrive) conditions there is a quiescent plate voltage of (say) 150V on stage 1 and 0V quiescent on the grid of stage 2. The AC signal is superimposed on the average voltage of 150V across the decoupling cap.

Now overdriving stage 2, grid current into stage 2 charges the cap up to (say) 160V. When the cap is charging, the grid of stage 2 stays near 0V (it is clamped by the grid current).

While that extra voltage across the cap persists, when the stage 1 plate voltage is near its quiescent value of 150V, the grid of stage 2 is already at -10V and stage 2 is effectively cut off for small signals.

[g1]

This is where the terminology gets ugly again with regard to coupling, decoupling, bypass, and filter caps.
I think I know what you mean by calling them decoupling, but I can see where someone could find it all very confusing.

[Malcolm Irving]

I'm talking about a cap that connects the signal from one stage to the next. It 'couples' the AC signal between the stages and 'decouples' the DC between the stages. Both terms are widely used. I'm not sure which terminology is more standard.

[g1]

Yeah, I know what you mean, there was another thread recently that got into the same issue. In my experience, decoupling usually refers to power supply type caps, coupling for interstage caps. But as you say, both terms are technically correct as they do one thing for AC, another for DC.
Around here, farmers are currently running their "combines". What they do is separate the wheat from the chaff.
But I guess they do a combination of jobs so I imagine that's where the term comes from.

[Enzo]

You may be describing the physics, but in use a coupling cap couples the stages for signal transfer. decoupling caps are placed on the power supply to prevent unwanted coupling through the power connections. We would say it blocks DC, rather than decouples it.

[Malcolm Irving]

Hi Enzo,

OK. Fair enough. I'll try to remember to use the term 'coupling cap' for connecting the signal between stages. But the power supply caps are usually called 'filter' caps, are they not? Their primary purpose is to filter out ripple in the DC. They do also 'decouple' in the sense of preventing motor-boating, but I don't think they are often referred to as decoupling caps.

The motor-boating issue arises because the power supply is usually a series chain. If the supplies to various stages (or pairs of stages) are parallel, which is used to some extent in some amps, then the filter caps are still filtering but would not be so obviously 'decoupling'.

[lowell]

While that extra voltage across the cap persists, when the stage 1 plate voltage is near its quiescent value of 150V, the grid of stage 2 is already at -10V and stage 2 is effectively cut off for small signals.

This doesn't make sense to me... especially right after pointing out that the Grid is "clamped" at 0v. I must be looking at this the wrong way. Here's what's confusing. The grid typically sits at 0v with no signal. Yes it's ~1.5v more negative than the grid (-1.5vgk), but 0v nonetheless. When the upswinging AC signal hits ~+1.5v the grid conducts and clamps it there even if the AC signal is larger than 1.5v peak. So we could say the grid is now clamped at 1.5v or 0vgk. From here, how does it become more negative and cause "cut-off", AND pull the coupling cap negative? Technically it's at +1.5v.... right?

[Malcolm Irving]

Hi Lowell,

Sorry, if I’m being confusing. It’s always difficult to describe circuit behaviour in words.

The clamping of the grid to near 0V is only while grid current is flowing (when Vgk is slightly positive roughly speaking). This occurs while the input wave to stage 2 is positive (above its quiescent voltage). While grid current is flowing, the time constant for charging up the coupling cap is faster (because the internal resistance in the tube, from grid to cathode, is only about 1k during grid current). So, while grid current is flowing, the stage 2 side of the coupling cap stays near 0V (clamped) while the stage 1 side of the coupling cap charges up to 160V in my example. (In reality it might take a few cycles of the input wave to charge up this far.)

Now, when the input wave to stage 2 is going negative (or rather still positive but below its quiescent value, at the stage 1 plate), the extra voltage across the coupling cap (extra 10V in my example) stays there, because the time constant for it to decay is much longer (the discharge must go via the 1M grid-leak resistor - the grid is no longer conducting). This extra 10V is putting a ‘bias’ of -10V onto the signal.

A large overdriving signal is needed to ‘set up’ the block. While the large overdriving signal continues it still ‘gets through’ although the duty cycle of the wave is greatly altered. It is only once the large overdriving signal finishes and is followed by a small signal that we see the full effect of the blocking, i.e. the small signal can’t get through stage 2, until that extra 10V has discharged.