Coupling Cap Switch - Pre Charging?

[exclamationmark]

I've been having a little fun making a 'programmable' gain stage with about a dozen relays, and want to have the option of changing coupling caps 'on the fly' so to speak. I've done a little transient response work on SPICE (to make sure my relays weren't going to weld themselves together), and among other things, found that depending on the capacitor size and grid leak resistor, it takes about 3/4 of a second for 0.022uf coupling cap to completely charge and not upset the bias of the preceding stage (ie, momentary blocking distortion).

Could I implement a sort of pre-charging circuit, where the capacitor to be switched would be at plate potential, but grounded to a largish (1M) resistor, then disconnected from the resistor and placed in parallel with the existing coupling capacitor when the time comes? Is the blocking distortion seen on SPICE even a problem in the real world? Normally switches would be panel mounted and would probably go unnoticed but this application is really for two separate foot switchable channels.

[Chuck H]

I might try using series caps. The sum of the series caps could equal your desired uf's and the caps would stay charged. The relays would just close the circuit over one or more caps to change the value of the string. there would still be a charge lag when adding a cap (or caps) to the string, but not when subtracting. If you use large value bridge resistors it should be silent and almost transpearent even though there will still be a lag when adding caps.

Another option would be to have several coupling caps, each followed by a high value load resistor and a high value series resistor to a common point. The relays would choose which high value series resistor to bridge and therefor "select" that cap. This also has flaws. But I don't think there's a flawless way to do this in a practical way.

[Steve Conner]

Yes, this is always done whenever a switch switches a capacitor in an audio circuit. I've seen anything from 1M to 10M used. Broadly speaking, the resistor goes across the switch contacts, so the DC voltage across them is bled down to zero before the switch is closed. And you live with any slight change in the frequency response that it causes.

[exclamationmark]

I'm thinking something like the schematic attached. When all the switched are closed, they will fully charge and not do anything tone-wise to the circuit. When the switch/switches are flipped the 10M resistor should be in parallel with the 1M grid leak resistor (leading to a value slightly lower than 1M - will not be noticeable), in addition to the added capacitance.

Edit: Slight error in schematic... Each switchable cap needs its own resistor.

[Chuck H]

Not quite. Like this:

[Chuck H]

Or this: (not even sure the resistors are needed in this one... Steve?)

[Steve Conner]

All of these should work. The last one "might" not pop without the resistors, but it certainly won't pop with them.

One subtlety in these kinds of circuits is that the grid of a tube isn't at ground potential. The tube has a few nanoamps of grid current that generates a voltage across the grid leak resistor, about 10-30mV or so. The same current that makes your guitar's volume pot scratchy in circuits that lack a first stage DC blocking cap.

I've had that voltage generate small pops in channel switching circuits, puzzling me for a long time.

[Chuck H]

All of these should work.

In emarks diagram it seems like the three caps designated for selection are in parallel. Making for only one additional selection no matter which relay is tripped.

[Steve Conner]

Oops... Of course I meant all of them except the first one.

[Chuck H]

Here's a more complicated circuit that would eliminate the charge up time (no pop and no charge up time). It has the disadvantage of causing small phase shifts above the knee requencies of the caps. That shouldn't matter unless it's used within a feedback loop.

[Enzo]

Nah, post #5 was better. This latest one assumes the output goes to something at ground. Post #5 works regardless of what voltage the output sits at. The whole point of the equalizing resistors is to match the two ends of the cap. WHether they are charged with respect to ground is irrelevant, unless ground is what one end is switching to.

[exclamationmark]

That last circuit chuck posted is the one I meant when I said "each cap needs it's own resistor" in my previous post. I'm a bit confused by the 'output at ground' thing though... Don't I want the grid of the tube it's feeding to be at ground potential (assuming it's cathode biased) at DC?

[Chuck H]

Yes, and yes. Enzo is right that the post #10 circuit assumes a 0V reference at the grid and that it would be less suitable than post #5 if there were no 0V reference. But considering that you DO have a 0V grid reference for the following grid AND you want the caps charged, post #10 should do that.

Note that I mentioned phase shifts. If you plan to use this circuit parallel to another "channel" (like "jumper-ing" channels on the same amp) there could be some phase cancellation. If you plan to use the circuit within a feedback loop it could result in instability. Other than these considerations it should work for what you want. And as a nod to Enzo, it is a specific circuit for your particular application and not necessarily the best way to switch caps for all applications.

[Enzo]

You want the caps to be at WHATEVER voltage is on that grid. Due to grid leak, often times the grid sits at a small negative voltage, not zero, with respect to ground. Wanting the grid to be zero is no guarantee it will be. Your grid return is 1 meg now, the larger that value, the more grid leak voltage will occur. Some amplifier stages even rely on it, it is called then grid-leak bias. The cathode will be directly grounded, and the grid will have like 4.7meg or 10meg resistor to ground. In this case we are discussing, the grid is not connected to zero volts, it has a 1 meg resistor down to zero volts.

With a 1 meg resistor to ground, one tenth of one microamp grid leak current would put one tenth of a volt on that grid, and that is enough to pop. Or at least tick. Maybe even tock.

Putting all those 10meg to ground resistors on those caps won;t force the grid anywhere.

[exclamationmark]

I actually have a FET muting circuit, so a little snap crackle or pop doesn't really bother me, it was just really the charging problem that eluded me. I also understand what enzo is trying to say, now that I think about it.