Trim pot: Magic Smoke has escaped! ...But why?

[monomatic]

I was diddling around on my breadboard with a bazz fuss trying to learn a little something about biasing and decided to get rid of the 10k resistor and throw a little 100k trim pot from V+ to collector of the MPSA13.
Leg 1 and 2 were jumper'd together and routed to V+. Leg 3 was connected to the Collector. Trimmer was set at 50%.

Everything sounded Ok until I started to turn the trim up - then the Magic Smoke revealed its self.

So I figured, Id give it another go, but this time jump the wiper to leg 3, and NOT use a metal screw driver to turn the pot. (incase it touched the visible metal down inside the slot or something?)
This time with a nylon probe (spudger) I turned the trimmer counter-clockwise, and when I reached a point near the end of the turn...With a brief flash of light, the Magic Smoke was expelled from the trimmer once again.

Is there some fundamental concept that I should be aware of that may be causing this phenomenon? I'd like to think of this as a lesson learned, but Im not sure I understand what the lesson is.
I mean, unless the lesson is that Magic Smoke easily escapes 100k 6mm trim pots. And I don't like that lesson at all.

Here is a link to the trim pot I was using, in case that is of any use: http://www.taydaelectronics.com/100k-ohm-trimpot-variable-resistor-6mm.html
And I was starting with Chromesphere's lovable little 7min Fuzz rendition of the bazz fuss: http://www.diyguitarpedals.com.au/shop/boms/7_Min_Fuzz_Schematic.pdf

[armdnrdy]

The trimmer is only rated for 1/10th watt.

[Transmogrifox]

For this circuit 1/10W is plenty.  The thing to consider is at what point in the turn does this thing exceed 1/10th watt?

The transistor as it's biased through the diode will take as much current as you give it -- has all to do with that resistor value.

Power = V^2/R

In this case, V is about 8.3V assuming 9V supply.

Where is Power >= 0.1W?

R = V^2/Power

R=8.3^2/0.1 = 689 ohms

In other words, when the turn of the pot got down below 689 ohms it was in danger of smoking.

Put a 500 ohm resistor in series with the pot and try again.  That should keep it pretty much safe from smoking.

[electrosonic]

I don't think that's quite right (and I hesitate to disagree with you given that I have learned a lot from your posts)

My understanding is that a 0.1 watt pot can dissipate its full rating over the entire length of its resistance. If it is wired as a rheostat then the unused portion is not contributing to the total dissipation so the max power must be derated.

I would calculate the maximum current the pot can dissipate. P=VI= I^2 * R.   P=0.1 W, R- 100k, therefore Imax=1mA. At most 1mA can travel along any section of the pot.

If the pot is wired as a rheostat, once the value from wiper to the end gets below 8.3V/1mA = 8.3k then the pot is being used outside its rating.

Andrew

[monomatic]

Ok, this makes sense to me now.
I mean, the math part is intimidating, but the general concept seems to be:
Turn the trim too far down and exceed either what the trimmer is rated to handle, or the amount of resistance is too low for smoke captivity.
Either way, I see that trimmers don't work exactly as I theoriezed (based on little science, mind you) at their extremes.

Thanks so much for your answers!

Also, bias is determined by the diode across the collector and base? So I wasn't really even adjusting bias then, I guess. Damn. Well, ya can't win 'em all. Back to the breadboard!

[antonis]

I mean, the math part is intimidating, but the general concept seems to be:

Power is the product of Current & Resistace BUT they aren't straight proportional because Current is to the power of 2 (quadratic) so increasing one of them and deceasing the other one by the same amount doesn't sustain their product..

Also, bias is determined by the diode across the collector and base? So I wasn't really even adjusting bias then, I guess.

You guessed right.. You were adjusting Gain ( Rc/re, where re = 26mV/Ie )
(actually, you could adjust bias only in case of the resistance was TOO HIGH to "restrict" the appropriate current amount for diode forward biasing and resulted to a BJT "OFF" state..)

If the diode was replaced by a resistor in this way of bias (collector feedback) you would affect bias by trimpot variation...

[Transmogrifox]

My understanding is that a 0.1 watt pot can dissipate its full rating over the entire length of its resistance. If it is wired as a rheostat then the unused portion is not contributing to the total dissipation so the max power must be derated.

That is a good point I failed to consider, and you are probably correct.  If that is the case then the danger zone is happening somewhere between 6k and 9k.

That would sort of limit your ability to experiment much below the 10k value.

Also, bias is determined by the diode across the collector and base? So I wasn't really even adjusting bias then, I guess.

Depends what is meant by bias.  If you mean the voltage at the collector, then yes, it's pretty well clamped at 1.4 ish volts.

This resistor will change the bias CURRENT, and that will make the following changes:
high current -> lower input impedance but higher gain
low current (large resistor) -> lower gain but higher input impedance.

There is still some value in tweaking this with a pot, only, get a pot with a higher power rating to test settings below 10k (and use a 10k pot)

With your 100k pot, per electrosonic's correction, probably put a 10k in series to limit power dissipation in the pot.

[armdnrdy]

So would a 1/2W trimmer work for the OP in this situation?

[Transmogrifox]

So would a 1/2W trimmer work for the OP in this situation?

Using 100k trimmer, using power at full resistance:
P=I^2*R
P=0.5
R=100k
I = sqrt(P/R) = sqrt(0.5/100k) =  2.24 mA max through the pot

Now to find the resistance you need to add in series to keep a 1/2 watt pot safe:
V = I*R
Where,
V= 8.3V
I= 2.24 mA (from above)
R = unknown min resistance

R=V/I = 8.3/2.24mA = 3.7k

So the OP would want a 3.7k resistor in series with the pot to prevent exceeding 1/2 watt rating.

Now let me condense all of that down into one single neat expression:

Code Select Expand

Vmax == Max voltage you will apply across the pot
P == Power rating of the pot
Rp == Nominal Resistance value of the pot (100k pot, then Rp = 100k)
Rs == Series resistance you want to add to the pot to keep it safe

Rs = Vmax/sqrt(P/Rp)

For example, if you are using a 10k pot rated for 1/4 watt in this circuit at max voltage of 8.3V,
Rs = 8.3V/sqrt(0.25/10k) = 1.66k

You would want to put a 1.66k resistor in series to prevent the pot's max current from flowing and causing destruction