# AC Voltage Across a POT

1. May 25, 2012

### hubbat

Hi,
I have a circuit with a pot connected to an 26VAC signal through a diode and to ground... i.e. there is a half wave rectified signal across the pot. I'm then measuring the voltage off the wiper.
I'm looking through some peoples work, and they write the max output voltage is $26\frac{\sqrt{2}}{\pi}$.
Can anybody explain this? To me, this looks like the average output of an RMS signal, but the calculations they make are assuming it's the max output.
Thanks!

Last edited by a moderator: May 30, 2012
2. May 25, 2012

### DragonPetter

$26\frac{\sqrt{2}}{\pi}$ , just fixing your latex so I know what the equation was without thinking too hard

3. May 25, 2012

### blighme

Hello

You might want to take a look at wikipedia's Rectifier (sorry, it won't let me insert links unless I have 10 posts).

4. May 25, 2012

### hubbat

Thanks. I've seen that equation, but from what I can tell, that is really just the average DC voltage.
$\int_0^{\pi}sin(t)*\sqrt{2}=\frac{\sqrt(2)}{\pi}VAC$

The problem with this is the signal is being subtracted from a -13VAC (180 deg phase shift) signal. The calculations other people used did not do the same conversion for this signal, so that would mean they are subtracting the AC signal from the average signal.

5. May 25, 2012

### truesearch

Assuming 26VAC means 26Vrms then the peak voltage is 26√2 volts.
A single diode produces 1/2 wave rectification, as you say.
The average value of 1/2 wave rectification = (26√2)/∏
The rms value of a 1/2 wave rectification is (26√2)/2
hope this helps.
I am not certain what you mean regarding the -13V and the 180 phase shift!!

6. May 25, 2012

### Staff: Mentor

Indeed, that may well be true—is there is a capacitor somewhere in the circuit to do the averaging?

7. May 27, 2012

### blighme

hubbat,

truesearch put it plainly to you, if Wikipedia couldn't. If not, can you tell us what your reading values are, RMS and DC? Also, what did you expect to see at the output and what, in fact, you get. Even better, can you provide a simple schematic? They say a picture is worth a thousand words...

NascentOxygen, there's no need for a capacitor to have DC value, that only improves the ripple.

8. May 30, 2012

### hubbat

Ok. There should be an attachment included now. Basically, I'm trying to work backwards from some other peoples work. This is a summing circuit. V0 goes to an op-amp, but for this case the output voltage is zero.
In these guys work I'm using, they convert the ac voltage off of a pot at a known position to a DC voltage using the equations discussed above. The 25.6VAC signal is coming from one power supply.
Another 13 Volt power supply is used to create a signal which is 180 degrees out of phase with the 25.6V signal.

At the summing junction, they are saying the DC voltage output from the pot is nulled by the 13VAC signal. Their calculations for the two summing resistors are
$25.6\frac{25.6}{\pi}*\frac{180}{1001.5} = V1$
$13 = V2$
$\frac{V1}{V2} = \frac{R1}{R2}$
where R1 and R2 are the two summing resistors.

Does this make sense to anyone?

9. May 31, 2012

### Staff: Mentor

It looks legit. The waveform fed to the op-amp summing junction via the "upper" resistor is the negative half-cycle of 26VAC, but reduced in amplitude by the pot. Co-incident with this half-cycle, and fed via the "lower" resistor, is a positive half-cycle of 13VAC. These will cancel when the pot setting is 50% (assuming the two unlabelled resistors are of equal value); for other settings the waveforms will sum to give a half-cycle sinusoid of predictable amplitude. We can't say how the op-amp processes this, as its feedback path is not shown.

BTW, this appears to be related to homework, so should have been posted in the homework sub-forum.

10. May 31, 2012

### hubbat

Unfortunately, I don't think this is correct...
The voltage across the top resistor is -9.6VDC...
The voltage across the bottom resistor I would think is 5.8VDC, however, the original circuit was calculated using 13VAC across the bottom resistors...

I would say this calculation was completely wrong, however the circuit seems to work as designed and has been used for a number of years.