What's the load?

[Gaz]

I feel like I should know the answer to this. Is the load on the following gain stage 500K or 1M? My confusion stem from the bypassing of the top leg of the divider. Thanks!

[tubeswell]

Assuming your voltage divider is coupled to a signal source from the preceding stage, and the only other load on the preceding stage's plate is the plate load resistor, then the AC load on the preceding stage is Ra||Rg

Assuming a fully bypassed inverting triode gain stage, the gain is therefore:

[u(Ra||Rg)]/[(Ra||Rg)+ra]

where:

u = the amplification factor of the triode
Ra = the plate resistor
ra = the (internal) plate resistance of the triode
Rg = the grid load for the following stage (in this case your voltage divider)

Say we has a 12AX7 with an assumed plate resistance of 62k5 and a plate load resistor of 100k and a fully bypassed Rk and your 1M voltage divider for Rg, then the gain for that stage would be:

[100x(100k||1M)]/[(100k||1M)+62.5k]

[100x(90.9k)]/[(90.9k)+62.5k]

[9090909]/[153409] = 59.26

[Gaz]

Thanks Tubeswell, I appreciate the detailed answer, which makes my naive question look a lot more post-worthy. My question was really just asking if bypassing the top leg of the voltage divider changes the load at all. From you answer, I don't think that it does.

[tubeswell]

Yes, without the bypass, the voltage divider functions as normal and Rg is calculated across the 1M (for the purpose of calculating the AC load on the previous stage).

Depending on the bypass capacitance, the top leg of the voltage divider is progressively selected out of the voltage division (with smaller capacitances acting as a high-pass filter), and the Rg component of the AC load for the bypassed frequencies is calculated based upon the lower leg of the divider. Because the unbypassed frequencies are reduced by the ratio of the voltage divider output divided the the voltage divider input, then those (lower) frequencies are attenuated by a factor greater than the bypassed frequencies, giving the impression of a high frequency boost.

[martin manning]

My question was really just asking if bypassing the top leg of the voltage divider changes the load at all. From you answer, I don't think that it does.

It does, but it is a function of frequency. For this circuit fragment you can just find the reactance of the cap and parallel it with the 500k resistor, then add that to the 500k pot element:

at 10Hz it's 970k
at 100Hz it's 807k
at 1kHz it's 569k
at 10kHz it's 508k

[Gaz]

Thanks for the responses. My main reason for asking is because I have a two channel design where both channels are fed from a single coupling cap, and switched by a relay so that they are not both connected at the same time. For the clean channel There is just a 1M pot, whereas the dirty channel has the bypassed 500K, then the 500K pot in the schematic I posted.

My thinking was they would both have a total of 1M load, so that the bass response of the RC filter formed by the coupling cap the pot/divider and the would be the same for both channels. Does that make sense?

[martin manning]

So you have a coupling cap from the plate of the output stage of each channel, one of which is connected to the top of a 1M pot, and the other is connected to the circuit you showed above? From a load on the output stages that should be fine, and no one said the loads or the coupling caps have to be the same value. You want to have a different tone from each, right? So then you are using a relay to feed the output from the wiper of one or the other pot to the next stage or PI? I'd set up the relay to ground the output of the channel not in use and send the other one through.

[Gaz]

So you have a coupling cap from the plate of the output stage of each channel, one of which is connected to the top of a 1M pot, and the other is connected to the circuit you showed above? From a load on the output stages that should be fine, and no one said the loads or the coupling caps have to be the same value. You want to have a different tone from each, right? So then you are using a relay to feed the output from the wiper of one or the other pot to the next stage or PI? I'd set up the relay to ground the output of the channel not in use and send the other one through.

Hey Martin, here's a moer detailed schematic of what I'm doing. You're absolutely right about the channels being different. I just would like to keep the bass response the same for both channels (as a starting point, at least), and am also interested from purely technical point of view because I seem to have a very hazy idea of what's actually happening.[/img]

[tubeswell]

Looking at that schematic, the AC load is the sum of the 2M2 in parallel with the 220k and whichever of the '1M' is switched into the circuit.

So the total AC load on the driving stage is ~167k on the 'ordinary' channel side, and ~166k on the 'high-boost' side (discounting the effect of the bypass cap). If you take account of the frequencies bypassed in the .002uF bypass cap (when switched to that side), the AC load on the driving stage is ~ 143k.

[Gaz]

Thanks, Tubeswell! I always get more knowledge than I bargain for with these simple questions.

[martin manning]

I just would like to keep the bass response the same for both channels (as a starting point, at least), and am also interested from purely technical point of view because I seem to have a very hazy idea of what's actually happening.[/img]

Ah, so its the frequency response you are interested in rather than the load on the input stage, and the input stage is common to both channels (not what I had assumed above).

The Bass response will be different since you have a high-pass shelving filter on the upper one. A quick estimate puts the low frequency corner of that shelf at 170Hz (calculated as 1/(2piRC) with R=470 and C=0.002uF), so on the upper path you will have a 6dB lower response (from the 2:1 voltage divider) below the corner.