What is the AC Voltage Across a POT?

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Discussion Overview

The discussion revolves around the AC voltage across a potentiometer (POT) connected to a 26VAC signal through a diode, resulting in a half-wave rectified output. Participants explore the relationship between the measured voltage at the wiper and the theoretical calculations, particularly focusing on the average output voltage derived from the rectified signal.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions the interpretation of the output voltage being calculated as 26√2/π, suggesting it appears to be the average output of an RMS signal rather than a maximum output.
  • Another participant clarifies that the average value of a half-wave rectified signal is indeed (26√2)/π, while the RMS value is (26√2)/2.
  • Concerns are raised about the impact of a -13VAC signal, which is 180 degrees out of phase, on the calculations, suggesting that previous calculations did not account for this phase shift.
  • A participant proposes that a capacitor may not be necessary for obtaining a DC value, as it only improves ripple, indicating a potential misunderstanding of circuit behavior.
  • One participant describes the circuit as a summing circuit where the output voltage from the POT is nulled by the 13VAC signal, presenting a set of equations used for calculating the summing resistors.
  • Another participant agrees with the described operation of the circuit but notes that the feedback path of the op-amp is not shown, leaving some uncertainty about its processing of the signals.
  • A later reply challenges the correctness of the voltage calculations across the resistors, indicating discrepancies in expected versus calculated values, yet acknowledges that the circuit functions as intended despite these concerns.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of the output voltage and the calculations involved. There is no consensus on the correctness of the voltage calculations or the impact of the -13VAC signal, indicating ongoing debate and uncertainty.

Contextual Notes

Some assumptions about the circuit configuration and the role of the components, such as the diode and the phase relationship of the signals, remain unresolved. The discussion also highlights potential limitations in the calculations due to the lack of a complete schematic.

hubbat
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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 [itex]26\frac{\sqrt{2}}{\pi}[/itex].
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!
 
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[itex]26\frac{\sqrt{2}}{\pi}[/itex] , just fixing your latex so I know what the equation was without thinking too hard
 
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).


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

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

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


Vlad
 
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
[itex]25.6\frac{25.6}{\pi}*\frac{180}{1001.5} = V1[/itex]
[itex]13 = V2[/itex]
[itex]\frac{V1}{V2} = \frac{R1}{R2}[/itex]
where R1 and R2 are the two summing resistors.

Does this make sense to anyone?
diagram.jpg
 
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
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.
 

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