Parallel resistance calculation of this node

[jsteiger]

Hey all,

I have a newbie "first thread" math question involving parallel resistance calculations.

I am trying to understand the way my input modules in my desk are designed, and the heaviest potential load the opamps could be driving at different nodes.

Just looking to make sure I've figured this correctly.
Here is a quick rough sketch of part of the input channel circuit.

With Pre Fade selected and pot at full 5K value, this is what I came up with:
1 Echo's engaged, node=48.7K
2 Echo's engaged, node=29.2K
3 Echo's engaged, node=22.7K
All 4 Echo's engaged, node=19.5K

So, how did I do?

Much thanks in advance, Jeff

 

[pstamler]

Which node are you talking about?

If the echo sends are all connected then the four 39k resistors going to a (presumed low-impedance) buss are a total of 9750 ohms. That's in parallel with the echo-send pot, which is 5k when it's all the way up, for a total of 3305 ohms. (That's what the "post-fader echo" terminal will see if the switch is connected that way.) If the switch is moved to connect the 4.7k resistor to the top of the echo send pot, then that node (where the buss resistors join the pot and the 4.7k resistor) has a resistance of 3305 || 4700 = 1941 ohms to ground (assuming a low-Z source feeding the 4.7k resistor). If the pot is at less than maximum, then the node where the 39k resistors converge on the pot's slider has progressively lower Z to ground, converging on zero.

On the other hand, if you're wanting to know the input impedance (left side of the 4700 ohm resistor), that's 3305 + 4700 = 8005 ohms. That's the lowest resistance the input will see -- all echo sends active, echo send pot at max level. With the pot at minimum the input impedance will then be 4700 + 5000 = 9700 ohms.

Peace,
Paul

 

[jsteiger]

Thanks for the informative post Paul. It took me a few reads for it to make sense.

[quote author="pstamler"]... if you're wanting to know the input impedance (left side of the 4700 ohm resistor), that's 3305 + 4700 = 8005 ohms. That's the lowest resistance the input will see -- all echo sends active, echo send pot at max level. With the pot at minimum the input impedance will then be 4700 + 5000 = 9700 ohms.

Peace,
Paul[/quote]

Yes, this is what I was looking for. The heaviest load the opamp would be driving. Opamp being "left side of the 4.7K resistor" like you say.

I plan on replacing the 4 Echo on/off switches with pots and just wanted to make sure I understood, which I didn't! I was adding the 5K pot value in series with the parallel of all (4) 39K buss resistors, not parallel.

So, with only 1 echo engaged, pot a maximum (5K), from left side of 4.7K resistor would be 4700 + (5000 || 39000) = 9132 ohms. Correct?

Thanks for the patients.
Cheers, Jeff

 

[PRR]

> math question involving parallel resistance calculations.

Screw math. Brain-strain.

Back when Paul was a boy, there were Analog Computers to solve things.

> The heaviest load the opamp would be driving.

Then tack together a 4K7, a pot, and four 39K resistors. If it is active mixing, the 39K go to common (actually a few ohms, but this seems negligible against 39K). Use a Voltage/Current Ratio measurement system (an ohm meter) to find the sum equivalent resistance. If in doubt whether one or four 39K is worst-case, try each way. If in doubt what effect the pot has, turn it end to end.

It's good to know how to compute. May be quicker. Isn't "more accurate" (no accuracy is needed to figure opamp loads or mix impedances). It is also good to know how to mock-up, and to be unafraid to do so. Paper-figuring is prone to mistakes. Especially being unclear how things are parallel or series, or which case is worst-case. A mock-up either confirms your penciling or forces you to face your mistakes before you get nose-deep in a project.

> The heaviest load the opamp would be driving.

If this is the whole plan: the opamp can't see less than that 4K7 no matter what crap hangs on the far side. And since "any" audio-worthy opamp can drive 2K, we "know" it will be fine.

A point which has not been raised. On your Analog Computer, put a 9V battery where the opamp would go. Put pot full-up. Have one 39K load, and measure the DC voltage at its hot end. Then "throw switched 2 3 4" (connect three more 39K) and watch that voltage. It will drop 10%-20% (too lazy to calculate when you can Analog Compute it). That's a 1dB-2dB drop when you suddenly throw three switches mid-take. It is just-acceptable for many audio purposes.

Since the 2K opamp is underloaded now, you might consider losing the 4K7. Now with pot full up, throwing 1 to 10 of those 39K in or out makes no difference; even with pot at half one to three 39K make negligible difference. You got twice the level without the 4K7, so to stay the same you might raise the 39K to 75K... now any sane number of 75K switched on or off the pot wiper make negligible difference.

All this can be modeled on an Analog Computer consisting of a handful of resistors/pots, a 9V battery and a DMM.