power rating calculations (math inside)

[Sansteeth]

Hello there,
I needed to replace the presence knob on a Laney AOR50 and the only 25K pot I had lying around was a 16mm Alpha pot rated at 200V / .125W, so I was trying to calculate how much action that pot was seeing in the negative feedback loop of the amp, but I got stuck in my calculation and somehow I can't figure it out (when you think you know how to use Ohm's Law but you don't really)...
If you folks would be so kind to bear with me and tell me what I am doing wrong that would be much appreciated!

First off, I'm calculating the voltage coming out of this 50-watter from the 16 ohm tap at full volume:

V = √(50x16) = 28.28VAC

Which should give me a current of:

I=V/R = 28.28/16 = 1.77A (extra question (it's all about the journey): is that the current on the OPT secondary or at the speaker terminals? or both? Also, since 16 ohm is an impedance, what does 1.77A really mean since it's frequency-dependent?)

All of this should go though a classic 100k/4.7k potential divider in the NFB loop before it hits the presence pot:

Vout=R2/(R1+R2)xVin = 4.7k/(100k+4.7k)x28.28 = 1.27VAC

Now I want to know how much current will get to the presence pot. Can I use Ohm's Law on the previous resistor which happens to be the series resistor on the potential divider, what about the shunt resistor?
(these question comes mostly from my not understanding why use a potential divider which uses two resistors to drop the voltage, when a single resistor does just that)
Because if the voltage drop across the series resistor is 28.28 - 1.27 = 27.01 then the current flowing through it should be:

I=V/R = 27.01/100k = .27mA
then the power hitting the pot afterwards should be:
P=IxV = .00027x1.27 = .00034W but that does not look right.

The problem I'm having beyond that is how to measure how much voltage and current flows through a pot or a resistor when they are going to ground, since not all the signal is going to ground? (in the case of the NFB loop, some of it go back to the PI) 

I'm sorry it's a big chunk and it's not the most exciting, but I feel I need to be able to have a good idea of what is going on in every stage of an amp and except for the DMM, there's no way around cold hard math.

Would love to hear where I'm wrong!

Schematic can be found here: https://elektrotanya.com/laney_aor50h_sch.pdf/download.html

[dunner84]

1.77A looks about right. The output transformer took a high voltage, and low current on the primary side, and gave you a low voltage with more current to drive the speaker on the secondary side. The 1.77A is at the speaker terminals, and OT secondary.

Yes, the voltage is frequency dependent, and you can determine it using your existing math, and a good signal generator. Start with a low frequency 20Hz or so, and measure the voltage at the terminals, and work your way up to 20kHz...

The low power calculation at the presence pot looks about right, and a 1/4W pot should be fine. **I did not check your numbers myself though.

Simple ohms law works for AC when you are using the RMS voltage, and you know the impedance (not the same as resistance!)

[sluckey]

All you gotta do is just measure the voltage across the pot. Then put that number and the resistance of the pot into the power formula,  P = E²/R

[PRR]

> drop the voltage, when a single resistor does just that

No it doesn't! If it looks like it does, there is another resistor you are not seeing.

1.7A more or less is the current in the speaker you assumed. If it is some other impedance it will be some other current. And why are you side-tracking?? You wanted to know about a pot. 

On thumbs: 30V across 30k is 1mA and 30 milliWatts (0.030 W). Altho if you play MAX square-waves it will be 0.060W. Your 0.125W pot is adequate but not not grossly oversize.

Until you turn-down. 25K set to 2.5k will be 0.3 to 0.6 watts, smoke.

[PRR]

EDIT==  only part of the 28V appears on the 22k pot. Small part, less as you turn down.

Push-ups for the mind:

https://physics.info/circuits-r/practice.shtml

[2deaf]

This turns out to be a far more complicated problem than at first glance. 

First of all, the NFB is taken from the 4R tap, not the 16R tap.  The max rms voltage will be 14.14V, but the square wave voltage will be 20V.  We are interested in the worst case, so we will use 20V for the NFB voltage.

The voltage divider is 100K to the parallel combination of 4.7K and the presence circuit.  The impedance of the presence circuit changes with frequency and it's not immediately obvious which frequency will result in the most wattage across a section of the pot track.  It's also not immediately obvious what impedance will result in the most wattage across a section of the track when the presence pot is rotated.  In order to continue, I'm going to choose 4KHz with the presence pot set at one quarter (5.5K).

The series combination of 5.5K and 100nF at 4KHz has an impedance of 5.514K.  The parallel combination with the 4.7K resistor yields 2.537K, so the voltage divider is 100K to 2.537K.  20V at the secondary will be .495V across the presence circuit.  We're looking for the worst case and we're not sure what portion of the voltage drop is going to be across the 5.5K resistor, so we will say all of it.  If we stopped here, there would be .000044W across that 5.5K section of the pot.

The NFB current is not the only current going through the tail of the LTP.  How much current is the LTP drawing when there is a 20V square wave at the OT secondary and how much does this affect the current through the 100K NFB resistor?  Geez, I don't know.  Let's just double the current in the above paragraph and see what happens.  The resulting .99V across the presence circuit makes it .00018W across the 5.5K section of the presence pot.

[PRR]

> the voltage divider is 100K to 2.537K.

You may be right. And clearly the 22k can not have large bass across it due to the 100n cap. And even screaming solos do not have maximum power at many kHz.

But K.I.S.S. As you say, 20V distortion out, 100k:4.7k is already >20:1, so there is less than 1V here. When the 22k is dialed way down the limit is 20V/100k or 0.2mA.

1V in 22k is 0.045mW. 0.2mA times 22k would be 0.000,88W, but also 4.4V, and we can't get 4.4V due to 4.7k. So the pot power is minimal.

The longtail passes 2mA of DC current, which could be some power, but the 4.7k shunt and the 100nF series makes power trivial. There is some AC power in that longtail but much less than DC power so we can ignore it.

[PRR]

The combinations of resistance and capacitance do make a tangled frequency plot, but nothing to smoke pot.

While the "0.1" curve power is carried by only 1/10th of the pot element, it is 39 microWatts. So the whole pot must be rated 390uW, 0.4mW, 0.000,4 Watts. AND only for full power over 1kHz. You'll burn your ears out before the pot.

[Sansteeth]

Hey everyone,
thanks a lot for trying to help me out!

> drop the voltage, when a single resistor does just that

No it doesn't! If it looks like it does, there is another resistor you are not seeing.

Okay, so I really need to understand this, a single resistor does drops voltage (otherwise no Ohm's Law), but so does a potential divider. So what am I missing? 

By the way, this video on equivalent resistance is exactly the kind of stuff I need at this point PRR, thanks a lot for the pointer!

The voltage divider is 100K to the parallel combination of 4.7K and the presence circuit.  The impedance of the presence circuit changes with frequency and it's not immediately obvious which frequency will result in the most wattage across a section of the pot track.  It's also not immediately obvious what impedance will result in the most wattage across a section of the track when the presence pot is rotated.  In order to continue, I'm going to choose 4KHz with the presence pot set at one quarter (5.5K).

Yeah I didn't think of looking all the way to the pot and the cap in series to measure impedance, your post together with the video PRR posted makes a lot of sense.
I will need some more time on that for me to digest it.

Again, thanks a ton, all of this is really helpful!

[2deaf]

The longtail passes 2mA of DC current, which could be some power, but the 4.7k shunt and the 100nF series makes power trivial. There is some AC power in that longtail but much less than DC power so we can ignore it.

I think it was reasonably clear that I was talking about AC since DC won't pass through the presence circuit.

The ratio of AC current to DC current doesn't support the conclusion that AC power can be ignored.  We are dealing with fractions of milliamps through the pot resistance with the NFB current.  The LTP is running at full bore with a "constant current" resistor of only 10K.  I have the pot resistance set at 5.5K so that the 4.7K resistor isn't taking the lion's share of the current.  Under these conditions, I think that there is a very real possibility that the LTP AC current could be of significance. 

[2deaf]

The combinations of resistance and capacitance do make a tangled frequency plot, but nothing to smoke pot.

While the "0.1" curve power is carried by only 1/10th of the pot element, it is 39 microWatts. So the whole pot must be rated 390uW, 0.4mW, 0.000,4 Watts. AND only for full power over 1kHz. You'll burn your ears out before the pot.

0.25 curve power will be higher.

[PRR]

...I think it was reasonably clear that I was talking about .....

I was not trying to start an argument. I am in back-pain and not making complete sense or able to type much.

[2deaf]

I was not trying to start an argument. I am in back-pain and not making complete sense or able to type much.

I feel for you, man.  I've been in back pain since I was 16 years old.  I've never made complete sense the whole time.