Power dissipation when we add a transistor or resistor ...

Thread Starter

cspwcspw

Joined Nov 8, 2016
72
I was intrigued by an analysis of power dissipation done at http://electronics.stackexchange.com/questions/75140/how-to-calculate-the-power-dissipation-in-a-transistor, An analysis at the end by MikeTeX puts it into a little theorem I'd not seen or realized before:

"Theorem: the power dissipated by the transistor is not larger than 1/4 of the power that would be dissipated by the two resistors R3 and R4 if they were directly connected." [R3 and R4 are Load and Emitter resistors in that circuit.]

I simplified this to just think about resistors in series. If one adds a new resistor in series with others, its power dissipation can never exceed a quarter of what the rest of the resistor chain dissipates without that resistor. Here is a screenshot of my LTSpice simulation:

Power_dissipation_1.png

It abuses a voltage source to create a time-varying resistance (idea pinched from http://www.instructables.com/id/LTSPICE-Voltage-Controlled-Switch-and-Resistor/) where I use a sine wave to vary the resistance from 0.01 ohm (LTSpice won't deal with 0 resistance) to 2000 ohms. In the five panes we see the voltage (but interpret it as resistance), the power dissipation of R1 (400 watts peak when R2 is zero) the voltage across R2 as its resistance varies, the current in the circuit, and the power dissipation of R2 over time.

Sure enough, the R2 power dissipation peaks at 1/4 of the power peak of R1. It twin-peaks when R2 = R1 = 100 ohms.

Is this a reliable and generalizable rule-of-thumb? If you need to add a component, its power dissipation won't exceed 1/4 of the dissipation in the loop if that component were short-circuited?

If I have confused any of this, please set me straight or give me a counter-example.

Thanks
Peter
 

Dodgydave

Joined Jun 22, 2012
9,304
Resistors in series or parallel always obey Ohms Law, where the power is its voltage squared divided by its resistance, or its current squared times resistance.

W= VxV/R or W =IxI R
 

crutschow

Joined Mar 14, 2008
25,242
Here's an LTspice simulation of your observation, and also the Maximum Power Transfer Theorem mentioned by Albert.
It shows the variation in power for R1 and R2 as R2 is varied from 0.1Ω to 200Ω.
The maximum R1 power is 1W and drops to 1/4W for both when the two resistances are equal.

upload_2016-11-22_9-10-43.png
 

MrAl

Joined Jun 17, 2014
7,747
I was intrigued by an analysis of power dissipation done at http://electronics.stackexchange.com/questions/75140/how-to-calculate-the-power-dissipation-in-a-transistor, An analysis at the end by MikeTeX puts it into a little theorem I'd not seen or realized before:

"Theorem: the power dissipated by the transistor is not larger than 1/4 of the power that would be dissipated by the two resistors R3 and R4 if they were directly connected." [R3 and R4 are Load and Emitter resistors in that circuit.]

I simplified this to just think about resistors in series. If one adds a new resistor in series with others, its power dissipation can never exceed a quarter of what the rest of the resistor chain dissipates without that resistor. Here is a screenshot of my LTSpice simulation:

View attachment 115758

It abuses a voltage source to create a time-varying resistance (idea pinched from http://www.instructables.com/id/LTSPICE-Voltage-Controlled-Switch-and-Resistor/) where I use a sine wave to vary the resistance from 0.01 ohm (LTSpice won't deal with 0 resistance) to 2000 ohms. In the five panes we see the voltage (but interpret it as resistance), the power dissipation of R1 (400 watts peak when R2 is zero) the voltage across R2 as its resistance varies, the current in the circuit, and the power dissipation of R2 over time.

Sure enough, the R2 power dissipation peaks at 1/4 of the power peak of R1. It twin-peaks when R2 = R1 = 100 ohms.

Is this a reliable and generalizable rule-of-thumb? If you need to add a component, its power dissipation won't exceed 1/4 of the dissipation in the loop if that component were short-circuited?

If I have confused any of this, please set me straight or give me a counter-example.

Thanks
Peter

Hello there,

The maximum power transfer theorem deals with power in the source and load, but i dont think that applies here because we are dealing with two DIFFERENT topologies. To graph this using one fixed resistor you'd have to use two different circuits where one has just the fixed R1 and the other has both R1 and variable R2, then graph the ratio of the two powers (in R1 and then in R2) from the TWO different circuits as R2 is varied.

The first is what power we have with one resistance, the second is what we have with two resistances in series, and we want to know the ratio of the two powers.
We can look at this simply by finding the power in R1 in a circuit with just a single resistor R1, and then finding the power in R2 in a second circuit with R1+R2 (two in series) and then find the ratio when the power is maximum or just when the ratio is maximum. Also ideally we want to let R1 vary too so we know this works for all R1 and R2 or at least all positive non zero values or R1 and R2.

So starting with the first circuit where we have just R1 and a constant voltage source we have:
i1=V/R1

and with the second circuit where we have both R1 and R2 in series we have:
i2=V/(R1+R2)


The power in R1 in the first circuit is:
P1=i1^2*R1

and the power in R2 in the second circuit is:
P2=i2^2*R2

The ratio, which we are told is 1/4, we find with:
y=P2/P1=i2^2*R2/(i1^2*R1)=(R1*R2)/(R2+R1)^2

We then take this last expression (R1*R2)/(R2+R1)^2 and take the first derivative with respect to R2:
d[(R1*R2)/(R2+R1)^2]/dR2

and that comes out to:
dy/dR2=R1/(R2+R1)^2-(2*R1*R2)/(R2+R1)^3

We then set that equal to zero:
R1/(R2+R1)^2-(2*R1*R2)/(R2+R1)^3=0

and solve for R2 and get:
R2=R1

To ensure we have a max here, we change the ratio of powers equation to:
(R1*(a*R1+R1))/(a*R1+2*R1)^2

and that is just after replacing R2 with R1+R1*a where 'a' is a small number. We then get:
ym=(a+1)/(a+2)^2

Taking the first derivative of this and setting that equal to zero, we get as solution for a:
a=0

and note this result will help us meet one of our goals of showing that the overall result will be independent of both R1 and R2.

To check this (a=0 result) we graph it and we see we get a max when a=0 and less when 'a' is either less than zero or more than zero, so we do have an absolute max when R2=R1. Also, a second derivative test shows it is concave downward.

Knowing this for sure now, we go back to:
y=(R1*R2)/(R2+R1)^2

and set R2=R1 and get:
y=(R1*R1)/(R1+R1)^2

and after simplifying we get:
y=1/4

So there we proved mathematically that the maximum power in the transistor is 1/4 of what it would be in the single resistor (R2 was the transistor resistance here and R1 is the combined value of the collector resistor and the emitter resistor).

Note this is different than the maximum power transfer theorem which results in a ratio of 1/2.


Also, the base emitter current may or may not matter in the power scheme depending on the Beta of the transistor and the mode it is being used in the circuit. For a high Beta transistor it wont matter as much, but as the Beta gets down near around 10 the base current is 1/10 of the collector current and the base emitter diode drop could be as high as 1v or slightly higher, so instead of say 10 watts we could end up with around 11 watts, or instead of 100 watts 110 watts for example. Not that much but may be worth thinking about especially if the transistor base is over driven.
 
Last edited:

WBahn

Joined Mar 31, 2012
26,029
I was intrigued by an analysis of power dissipation done at http://electronics.stackexchange.com/questions/75140/how-to-calculate-the-power-dissipation-in-a-transistor, An analysis at the end by MikeTeX puts it into a little theorem I'd not seen or realized before:

"Theorem: the power dissipated by the transistor is not larger than 1/4 of the power that would be dissipated by the two resistors R3 and R4 if they were directly connected." [R3 and R4 are Load and Emitter resistors in that circuit.]

I simplified this to just think about resistors in series. If one adds a new resistor in series with others, its power dissipation can never exceed a quarter of what the rest of the resistor chain dissipates without that resistor. Here is a screenshot of my LTSpice simulation:

View attachment 115758

It abuses a voltage source to create a time-varying resistance (idea pinched from http://www.instructables.com/id/LTSPICE-Voltage-Controlled-Switch-and-Resistor/) where I use a sine wave to vary the resistance from 0.01 ohm (LTSpice won't deal with 0 resistance) to 2000 ohms. In the five panes we see the voltage (but interpret it as resistance), the power dissipation of R1 (400 watts peak when R2 is zero) the voltage across R2 as its resistance varies, the current in the circuit, and the power dissipation of R2 over time.

Sure enough, the R2 power dissipation peaks at 1/4 of the power peak of R1. It twin-peaks when R2 = R1 = 100 ohms.

Is this a reliable and generalizable rule-of-thumb? If you need to add a component, its power dissipation won't exceed 1/4 of the dissipation in the loop if that component were short-circuited?

If I have confused any of this, please set me straight or give me a counter-example.

Thanks
Peter
As already mentioned, it can be viewed as a consequence of the Max Power Transfer Theorem. Let's focus on real-valued resistive components for the moment. With a fixed source resistance, the max power that you can get into a load will be when the load resistance equals the source resistance. A consequence of this is that the power dissipated in the load is equal to the power dissipated in the source. Also, half of the voltage appears across the source and half across the load. If you short the load, then the total resistance is cut in half so the current doubles. But, in addition, now the full voltage appears across the source resistance, so it doubles as well. Since P = V·I if you double both the voltage and the current you will increase the power by a factor of four. Hence the max power in the load is no more than 1/4 of the power in the source resistance with the load shorted.

But, no, this is NOT a general rule that applies to all components. It applies to resistors strictly and may apply in other situations, but you need to look at them case by case.

A counter example would be a source and load impedance that have reactive components. The source impedance might have a relatively large reactance that limits the current when the load is shorted resulting in low power dissipation. But the load might have a reactance of the opposite type that cancels out the source reactance allowing significantly more current to flow such that the power dissipated in the load (and the source) impedance is greater than when the load is shorted. the same that is being dumped into the source impedance. The Without the source impedance
 

MrAl

Joined Jun 17, 2014
7,747
As already mentioned, it can be viewed as a consequence of the Max Power Transfer Theorem. Let's focus on real-valued resistive components for the moment. With a fixed source resistance, the max power that you can get into a load will be when the load resistance equals the source resistance. A consequence of this is that the power dissipated in the load is equal to the power dissipated in the source. Also, half of the voltage appears across the source and half across the load. If you short the load, then the total resistance is cut in half so the current doubles. But, in addition, now the full voltage appears across the source resistance, so it doubles as well. Since P = V·I if you double both the voltage and the current you will increase the power by a factor of four. Hence the max power in the load is no more than 1/4 of the power in the source resistance with the load shorted.

But, no, this is NOT a general rule that applies to all components. It applies to resistors strictly and may apply in other situations, but you need to look at them case by case.

A counter example would be a source and load impedance that have reactive components. The source impedance might have a relatively large reactance that limits the current when the load is shorted resulting in low power dissipation. But the load might have a reactance of the opposite type that cancels out the source reactance allowing significantly more current to flow such that the power dissipated in the load (and the source) impedance is greater than when the load is shorted. the same that is being dumped into the source impedance. The Without the source impedance

Hello there,

I have a complete proof of the factor of 1/4 in post #5 above, and also stated that the maximum power transfer theorem does not apply here directly.

The MPTT shows that 1/2 the power is in the source and 1/2 in the load. That's what that is for. It doesnt care that there is 1/4 of the total power in the load, that's a coincidence. The main idea is that we can only get 1/2 of the power into the load. We might be able to use this fact to calculate other things, but a more direct proof is more enlightening i think.

Of course you will disagree, and i will be willing to listen to your side, but i think i already stated that side :)
 

dannyf

Joined Sep 13, 2015
2,197
"Theorem: the power dissipated by the transistor is not larger than 1/4 of the power that would be dissipated by the two resistors R3 and R4 if they were directly connected." [R3 and R4 are Load and Emitter resistors in that circuit.]
It is basically a power transfer question: what's the load resistance that allows the maximum transfer of energy from an less-than-ideal source to a load? it happens when the load resistance is equal to that of the source resistance.

When that happens, the power dissipated on the source resistance is the same as the power dissipated on the load.

From that you get your answers.

The original set-up of that question is unnecessarily complicated.
 

WBahn

Joined Mar 31, 2012
26,029
I have a complete proof of the factor of 1/4 in post #5 above, and also stated that the maximum power transfer theorem does not apply here directly.
And I provided a proof of the factor of 1/4 that starts from the maximum power transfer theorem. Note that stating that something doesn't apply doesn't make it so -- and, yes, stating that something does apply doesn't make it so, either; but I showed that it DID apply precisely because I was able to use it as the starting point for the proof.

The MPTT shows that 1/2 the power is in the source and 1/2 in the load. That's what that is for. It doesnt care that there is 1/4 of the total power in the load, that's a coincidence. The main idea is that we can only get 1/2 of the power into the load. We might be able to use this fact to calculate other things, but a more direct proof is more enlightening i think.
Sometimes. But always going back to basics also hides the beauty that so many concepts are interrelated and that there a many results that can be arrived at from many different directions -- and by leveraging prior results, the new results can be obtained in almost no time compared to going back and re-inventing the wheel from scratch on a problem that is only slightly different.

Of course you will disagree, and i will be willing to listen to your side, but i think i already stated that side :)
Out of disagreement often comes enlightenment for all concerned.
 

MrAl

Joined Jun 17, 2014
7,747
And I provided a proof of the factor of 1/4 that starts from the maximum power transfer theorem. Note that stating that something doesn't apply doesn't make it so -- and, yes, stating that something does apply doesn't make it so, either; but I showed that it DID apply precisely because I was able to use it as the starting point for the proof.



Sometimes. But always going back to basics also hides the beauty that so many concepts are interrelated and that there a many results that can be arrived at from many different directions -- and by leveraging prior results, the new results can be obtained in almost no time compared to going back and re-inventing the wheel from scratch on a problem that is only slightly different.



Out of disagreement often comes enlightenment for all concerned.

Hi again,

Ok so we agree, you just dont realize it yet :)

However, you had shown that the source resistance gets 4 times the power when the load is shorted, and then state that it also means that the max power in the load must be 1/4 but do not show any reason for that 1/4. A partial quote:
". But, in addition, now the full voltage appears across the source resistance, so it doubles as well. Since P = V·I if you double both the voltage and the current you will increase the power by a factor of four. Hence the max power in the load is no more than 1/4 of the power in the source resistance with the load shorted."

So we double the voltage and double the current when we short the load, and we get a power increase by a factor of four. That make sense, but i think you still have to show your reasoning for how this actually equates to "max power in the load no more than 1/4 of the power in the source resistance with the load shorted". In other words, if the load is shorted we cant have any power in the load so that statement makes no sense as of yet, but you probably mean that the load power is 1/4 of the power in the source resistance *IF* we shorted the load.
But it would be good to state the correction and also exactly the reasoning behind what shorting the load does in how it helps us find that 1/4.

So what doesnt seem to follow is how you got 1/4 in the load from 4 times power in the source. You might want to show that more explicitly.

BTW, as always, thanks for your interest in this too. Always gets more interesting to hear different views.
 
Last edited:

WBahn

Joined Mar 31, 2012
26,029
Hi again,

Ok so we agree, you just dont realize it yet :)

However, you had shown that the source resistance gets 4 times the power when the load is shorted, and then state that it also means that the max power in the load must be 1/4 but do not show any reason for that 1/4. A partial quote:
". But, in addition, now the full voltage appears across the source resistance, so it doubles as well. Since P = V·I if you double both the voltage and the current you will increase the power by a factor of four. Hence the max power in the load is no more than 1/4 of the power in the source resistance with the load shorted."

So we double the voltage and double the current when we short the load, and we get a power increase by a factor of four. That make sense, but i think you still have to show your reasoning for how this actually equates to "max power in the load no more than 1/4 of the power in the source resistance with the load shorted". In other words, if the load is shorted we cant have any power in the load so that statement makes no sense as of yet, but you probably mean that the load power is 1/4 of the power in the source resistance *IF* we shorted the load.
But it would be good to state the correction and also exactly the reasoning behind what shorting the load does in how it helps us find that 1/4.

So what doesnt seem to follow is how you got 1/4 in the load from 4 times power in the source. You might want to show that more explicitly.

BTW, as always, thanks for your interest in this too. Always gets more interesting to hear different views.
Read what was written earlier in that paragraph:

"With a fixed source resistance, the max power that you can get into a load will be when the load resistance equals the source resistance. A consequence of this is that the power dissipated in the load is equal to the power dissipated in the source."

I don't see what is so unclear about this. With a fixed source resistance, the maximum power that you can get into any load will occur when the load resistance is equal to the source resistance. Since this happens when the load resistance is equal to the source resistance, this happens when the power being dissipated in the load is equal to the power being dissipated in the source. This also means that the voltage across the source resistance is equal to half of the voltage that will appear across it if the load is shorted. Doubling the voltage across a fixed resistance increases the power dissipated in it by a factor of four. Hence the maximum power that can be dissipated in the load is one-fourth of the power that will be dissipated in the source if the load is replaced by a short (and, remember that this is only applicable for a purely resistive source and load -- whether it applies to anything else has to be considered separately).
 

MrAl

Joined Jun 17, 2014
7,747
Read what was written earlier in that paragraph:

"With a fixed source resistance, the max power that you can get into a load will be when the load resistance equals the source resistance. A consequence of this is that the power dissipated in the load is equal to the power dissipated in the source."

I don't see what is so unclear about this. With a fixed source resistance, the maximum power that you can get into any load will occur when the load resistance is equal to the source resistance. Since this happens when the load resistance is equal to the source resistance, this happens when the power being dissipated in the load is equal to the power being dissipated in the source. This also means that the voltage across the source resistance is equal to half of the voltage that will appear across it if the load is shorted. Doubling the voltage across a fixed resistance increases the power dissipated in it by a factor of four. Hence the maximum power that can be dissipated in the load is one-fourth of the power that will be dissipated in the source if the load is replaced by a short (and, remember that this is only applicable for a purely resistive source and load -- whether it applies to anything else has to be considered separately).
Hi again,

Yes, but isnt that just a repeat of the maximum power transfer theorem (MPTT) ?

That tells use that WHEN the two resistors are equal we get equal power, and then we can extract some consequences. What it does not state is what happens when the resistors are NOT equal.
This is a problem where one resistor is allowed to vary over a complete range, and all we know from your proof so far is that the max power is when the two resistors are equal, and that when one is shorted we get a different power amount which is then blindly applied to the first theorem (MPTT).

Let me try to illustrate this point and see if we can come to an agreement, which i think we can.

Lets say we have two 1 ohm resistors R1 and R2 in series and some DC voltage source 2v, and R1 is on top and R2 is on bottom. MPTT says that when they are both equal, we get max power in R2 (that's the one we are using to emulate the transistor). So we know that max power in R2 is (2/2)^2*1=1 watt.

We've used MPTT so far, that's all so far.

Next, in your proof, we short R2 and then look at the power in R1. So R2=0 and R1 still is 1 ohm.
Now with 2v power source, we of course have 2^2/1=4 watts.

Now your proof states (indirectly BTW) that 4/1 means that since the power in R2 with MPTT is 1 watt and the power in R1 is 4 watts with R2 shorted, that the ratio of these two is 1/4, which clearly it is.

So the first problem is that it does not look like you had shown why we get that 1/4, or even how.

Now you might see the other problem already though...

The second and more important problem is that in that proof we ONLY know the power for two distinct topological states given (R2=1 or R2=0), and although that's not too bad in itself, we only know the maximum power for ONE of those states, and have no proof that the other power level is appropriate for comparison. Thus what we really need next is some way to show that when we short R2 that is somehow relevant and appropriate for the coming comparison.
What if we get MORE power (more than 4 watts) for a different value of R2, such as 0.1 ohms, 0.2 ohms, 0.3 ohms, etc. ? Then we would not truly finding the maximum.

This is a subtle point i know, but in the very least we should show some proof or idea how we can know that the max power in R1 is when we short R2. You and i both know it is true, but we havent shown that yet in your proof. Not all proofs like this are so well behaved.

A more direct point though is when finding min and max values in proofs it is more customary to show a curve that is concave downward and the peak is the max. Can they get away with using MPTT in a book for this question in a course? Probably, although the better course will show the full derivation.
 
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