Vague voltage question, need an EE/techie's help

[cd]

Help me settle an argument: I say current capability of an opamp has very very little relevance in a typical guitar effect.  Friend says you always need tons of current to drive the input of tube amps hard.  My reasoning is:

- typical guitar effect uses a 9V supply w/4.5V bias, theoretically there's 9V peak/peak for the signal, but due to the internal working of the opamp (say in a non-inverting configuration) it's really around 7.5V peak/peak
- if we take this maximum 7.5V peak/peak signal, and we want to drive a tube amp with an input impedance of 1M, that means going by Ohm's law:

I=V/R, I=7.5V/1Mohms, I = .0075mA.  Power = I*V = .0075*7.5 = .05625mW, which is hardly anything

Friend says current should not be calculated with R = 1Mohms, but the R of the cable (!) driving the load (which makes no sense to me) so according to him:

I=V/R, I=7.5V/1ohm (or less), I = 7.5A.  Power = I*V = 56A!

Obviously I tried to point out the flaw in his reasoning, comparing speakers to amp inputs and whatnot, but he's stubborn.  Plus I'm a non-technie type of person so I don't know if I got the math right.  So who's right (or since I'm obviously right, how is the 1ohm reasoning wrong?)

[R.G.]

You're right.

The resistance of the cable (1 ohm or less) appears in series with the 1M ohm input of the amp, so the opamp has to drive 1M plus 1 ohm - which is trivial as you say.

Tell your semi-techie friend the following, by way of one-upmanship:

The ugly dark side is that cable has capacitance. The hard thing to drive for the opamp is that capacitance. The cable will have capacitance of 10 to 30pF per foot. In a 10 foot cord, that's up to 300pF (or more for bad cables) for the opamp to drive in parallel with the 1M + 1 ohm resistance.

That means that below a frequency of F = 1/2*pi*R*C = 530Hz, the opamp is driving the capacitance instead of the resistance - which it can do OK, it's just why cables cause treble loss. The capacitive load does not get down to as low as 10K until a frequency of 53000Hz, well over audio.

[Transmogrifox]

You're right.  The cable's resistance is in SERIES with the input, so it's now actually 1,000,001 ohms  :lol: .

If you even put 10 A through a guitar cable, you would more than burn it up, let alone 56 A!  You're stressing it at 1A (a reason guitar cables should not be substituted for speaker cable).

EDIT: Looks like RG and I were answering at the same time

[cd]

Thanks for the help/clarification guys.  I knew there was something about capacitance but I couldn't remember how it affected the signal.  What's the formula for figuring out how to drive a capacitive load?  Same as F = 1/2*pi*R*C?  Correct me if I'm wrong here:

If I have a really, really bad cable, which has a capacitance of .1uf, and I want it to drive 40Hz with no loss, the equivalent resistance would be R=1/6.28*40*.1u, or approx 39k ohms... but the 39k is in series with the 1M ohm input impedance of the amp (which is a trivial increase) so the opamp can drive it no problem: driving 7.5V into 39k directly, that would be .19mA or 1.44mW.

Wow, you CAN be an EE just by knowing Ohm's law!

[Transmogrifox]

{EDIT}
ummmmm...that formula you were using and what you were calculating provided the correct results, but it was interpreted incorrectly.  What you actually calculated was the required resistance seen by a capacitor to create a pole at 40 Hz, or in the case of a low-pass filter, what resistance would be needed to put a 3-dB cut-off at 40 Hz.

What you need to know is the resistance seen by the line capacitance, which is the op amp's output impedance (for all practical purposes).  Then you calculate what the magnitude of the impedance (of the capacitor) is at 40 Hz, which is an impedance that will appear in parallel with the input impedance of the amp.  This can be calculated as:

Zc = 1/jwC

|Zc| = 1/(2*pi*f*c)

You notice this is the same as what you used above, the problem is that what you used above came out of an application of the capacitive impedance with a resistive impedance circuit, but mathematically what you solved for coincidentally was the same thing in this case--it's just that I don't think you knew what you were actually solving for.

so using your example:
1) line is C = .1 uF

2) then assume output Z of op amp driven circuit is 470 ohms since .1uF cap load would make some op amps go unstable (just put in series with output, see explanation below to see why).

3) You want to know the equivalent magnitude of impedance at 40 Hz:

|Z| = 1/2*pi*.1E-6
= 39.8 kOhms...  

so that part you got right.  The issue is that 39k is has little effect compared to 470 ohms output impedance...but what if we are using a 100k pot on the output of our circuit?  Then 50k and 100k look quite substantial compared to 39 k--you're losing over half your signal level at that frequency, magnitude decreasing at 20 dB per decade.  So at 5 kHz, the capacitive impedance is now:

|Z|= 318 ohms

this would cause significant high-frequency signal loss on the 470 ohm output amp, let alone the 100k pot output!  Even worse, with a pot, this roll-off is changing with the pot volume level.
{/EDIT}

That said, here's the real issue:

usually, driving capacitive loads for op amps is a stability issue as opposed to a signal loss problem.  Like previously stated, the cable has a resistance of about 1 ohm so the capacitance has to be real big to make a signal loss problem on its own.

The "R" in the equation, therefore, is the output impedance of the pedal (driving circuit) plus the line impedance (1 ohm) all in parallel with the amp's input impedance.  

Since the amp's input Z is usually several hundred k-ohms, and usually much larger than one ohm, the R in the equation reduces practically to the op-amp's output Z.

This is generally on the order of 25 ohms, but significantly decreases with feedback, therefore a high gain op amp circuit will have more difficulty maintaining bandwidth on a long cable line than a unity-gain op amp.

With the 25 ohm output impedance range, losing bandwidth due to cable capacitance is not a problem.  Op amps make good line drivers.

What you lose with large load capacitance is op amp performance, and ultimately stablility when it gets too large.  Datasheets usually specify an output capacitive load handling capacity for maintaining specified performance levels in the op amp.

You see some circuits add a series resistance on the output--this is part of the reason.  The only problem is that now the added resistance reduces bandwidth, however, it's not significant when considering 30 pF with 470 ohms.  It can become a consideration when you have a long cable and a 100k pot on the output.

Is this enough techie-geek talk to make your friend spin?

[niftydog]

just ask him to show you an op amp sourcing 56A and you've won the arguement!