# (Seemingly) simple op-amp question

Discussion in 'Electronic Basics' started by Jon, Jan 24, 2004.

1. ### JonGuest

The circuit shown here:

http://www.benchmarkmedia.com/pdf/panpot.pdf

says that the total gain to one of the outputs will be .707 times the
normal gain when the pot is in the extreme position. Is this correct?

According to my calculations, the 4.87k, 10k, and 2k make a voltage
divider (because the input impedance of the inverting amplifier is
2k), which is (|| = in parallel with):

(10k || 2k)
------------------- = .255
(10k || 2k) + 4.87k

for the center position, it is:

(5k || 2k)
------------------- = .226
(5k || 2k) + 4.87k

since .255 * .707 = .180, the gain ratio of .707 (-3 dB) seems
incorrect.

the gain of the inverting amplifier is 11.8k/2k = 5.9, which,
multiplied by the voltage divider is 1.5 and 1.339, for the two
configurations. right?

If you ignore the input impedance of the amplifier, the values are
then:

10k
----------- = .672
10k + 4.87k

and

5k
---------- = .506
5k + 4.87k

and the total gains are 3 and 4, which are a ratio of .75 apart, which
is much closer. but does this mean that they forgot to include the
input impedance, or that i am in the wrong by including it?

2. ### Costas VlachosGuest

Right. Your calculations above are correct. The DC gains when the pot is at
the extremes and in the centre position are 1.5043 and 1.3382, respectively.
I don't know why the PDF says 1 and 0.707... Unless there's something else
we're overlooking...

cheers,
Costas
_________________________________________________
Costas Vlachos Email:

3. ### Tim DicusGuest

Hi Costas,

There is one minor thing you overlooked. The circuit is operating at 346KHz. I think it is the AC gain they are looking for. You
must compensate for the capacitive impedance of the 39pf cap (about 12K at 346KHz). So you are looking at a parallel reactance of
about 7800 ohms with about a 45 degree phase shift on the feedback circuit.

So it looks to me without doing exact calculations, that the gain will be somewhere between .707 and 1 with the frequency at 346KHz.

I am unable to do the exact numbers now because I must depart soon for "real work". Hopefully someone will finish this in the next
several hours.

Hope that helps and see ya later !^),

Tim

4. ### John PopelishGuest

Something wrong, here. In one extreme position, the output goes to
zero. In the other the output must be greater than when the pot is
centered. This cannot be a fractional multiple of the centered
output.
Remember that the input of an inverting opamp is a virtual ground, so
you can figure the input signal as a current entering that ground.
The output is then, that current passing through the feedback
resistor. If you figure an input of 1 volt, then the gain is the
drop across the feedback resistor.

So, for the pot at 10k, and a 1 volt input, the current arriving at
the - input is:

Solving for the voltage across the 2K resistor:

1 volt * (2k || 10k) / (4.87k + (2k || 10k) = .255 volt

So the current to the inverting node is .25 volt / 2k = .1275 ma.

This causes a drop in the feedback resistor of:

..127 ma * 11.8K = 1.504 volts

For a gain of 1.504

Repeating for the pot at center:

Solving for the voltage across the 2K resistor:

1 volt * (2k || 5k) / (4.87k + (2k || 5k) = .227 volt

So the current to the inverting node is .227 volt / 2k = .1134 ma

This causes a drop in the feedback resistor of:

..1134 ma * 11.8K = 1.338 volts

For a gain of 1.338

So the gain ratio from full to center is 1.504/1.338=1.124

If they got the ratio upside down it would be 1.338/1/504=.89

I think they screwed up.

5. ### Costas VlachosGuest

346KHz. I think it is the AC gain they are looking for. You
346KHz). So you are looking at a parallel reactance of
somewhere between .707 and 1 with the frequency at 346KHz.
"real work". Hopefully someone will finish this in the next

I see your point, but I don't think this is it... The circuit is meant for
audio signals and the filter cut-off is way too high to affect the output.
The pF caps are only there to limit the gain at very high frequencies and
are not meant to affect the actual audio signals going through.

Probably a mistake on their part. Replacing the 11.8k resistors with 7.87k
ones gives the more sensible gains of 1.00 and 0.89.

cheers,
Costas

6. ### Tim DicusGuest

Hi Costas,

You are correct. If it is an audio amp, the values were not correct. The 7800 ohm resistor should produce about the correct gain for
audio range. I see now the frequency designation in the upper right corner is Fc (cutoff), not Fo (center). My bad. Been working
with PLLs to long I guess. I see what I want to see.

Tim

7. ### JonGuest

ok good. i thought it was a mistake, but the .707 (= 3dB) difference
was what i was trying to get in another circuit, and it was very
confusing.

thanks everyone.

8. ### BalajiGuest

Hey, these guys have screwed it up completely. There is simply nothing
that they have written, do I find conforming to their circuit design
there.
I look at the Thevenin equivalent from terminal 2 leftwards. Assume
extreme position first:

Rth = 2K + (4.87K||10K) ~ 5.275K
Vth = Vin * 10 / (10 + 4.87) ~ 0.6724 Vin
Gain of inverter = 11.8K/Rth ~ 2.2369
Vth * Gain of inverter = Output ~ 1.5043 Vin

The other side is zero.

Again for middle position, we have

Rth = 2K + (4.87K||5K) ~ 4.467K
Vth = Vin * 5 / (5 + 4.87) = 0.5065
Gain of inverter = 11.8K/Rth ~ 2.64155
Vth * Gain of inverter = Output ~ 1.33872

If you see, none of these numbers is even remotely related to 0.707.
There is surely something wrong. I have a feeling this was first
designed just like that by a beginner and was not even tested before
publishing it online.

And about that fc = 346KHz, you get that by taking (1/(2pi * 11.8K *
39p)). Very funny to note that someone tried to use this configuration
for having some cutoff frequency! God knows where we get any 3 dB down
signals. The bandwidth of the inverter is limited to 346KHz, and the
input is in audio range!! Makes no real sense to me.  