Power Dissipation of An Amplifier Driving Fully Inductive Load

bkyavas

Joined Jun 5, 2012
17
Hi,

I have a question about calculating the power dissipation of an amplifier for heatsink selection. I have an amplifier (Say it is a power opamp or an audio amplifier), is driving a fully (assumed) inductive load. It is a current feedback circuit (see the figure) because of driving the inductor. The Input and output of the amplifier is a sine wave at single frequency (In KHz) without any offset.

R3 is current feedback resistor (The loading effect of R3 is neglected)
R1,R2 is voltage feedback divider

Let;
Vs is supply voltage (not shown in the figure, it is symmetrical as +Vs/-Vs)
Im Output current (amplitude) delivered from amplifier, i.e. (I(t)=Im.Sin(wt))
Vm Output voltage (amplitude) at the output of amplifier, i.e. (V(t)=Vm.Sin(wt+Phi)), here Phi is about 90 degree because of the inductive load.

What I want to calculate is; the power dissipated by the U1 amplifier. Please note that the amplifier output is in quadrature phase with the output current and the amplifier output is a typical A/B class push-pull transistors. What is the simplest way to calculate the dissipation without the time-domain complexity, including rail voltage offsets of sine over the push-pull output transistors etc.

#12

Joined Nov 30, 2010
18,223

bkyavas

Joined Jun 5, 2012
17
Thank you for your reply, dear #12, the amplifier can be an audio amplifier but the frequencies are not in audio range, typically about 50 KHz to 80 KHz. It is a VLF, RF transmitter. Anyway, I have already found out an audio amplifer to support this frequency range, thanks to the current feedback.

I will visit the suggested site you stated.

crutschow

Joined Mar 14, 2008
28,533
If no power is being dissipated in the inductor than all the power is dissipated in the amp and sense resistor, R3.
This is equal to the supply voltage times the average current from the power supplies (equal to the rectified average inductor current).
To get the amp power you just subtract the I²R power of R3 from the total where I is the RMS (not average) current through the inductor.

Last edited:

MrAl

Joined Jun 17, 2014
8,609
Hi,

I have a question about calculating the power dissipation of an amplifier for heatsink selection. I have an amplifier (Say it is a power opamp or an audio amplifier), is driving a fully (assumed) inductive load. It is a current feedback circuit (see the figure) because of driving the inductor. The Input and output of the amplifier is a sine wave at single frequency (In KHz) without any offset.

R3 is current feedback resistor (The loading effect of R3 is neglected)
R1,R2 is voltage feedback divider

Let;
Vs is supply voltage (not shown in the figure, it is symmetrical as +Vs/-Vs)
Im Output current (amplitude) delivered from amplifier, i.e. (I(t)=Im.Sin(wt))
Vm Output voltage (amplitude) at the output of amplifier, i.e. (V(t)=Vm.Sin(wt+Phi)), here Phi is about 90 degree because of the inductive load.

What I want to calculate is; the power dissipated by the U1 amplifier. Please note that the amplifier output is in quadrature phase with the output current and the amplifier output is a typical A/B class push-pull transistors.

View attachment 135771

What is the simplest way to calculate the dissipation without the time-domain complexity, including rail voltage offsets of sine over the push-pull output transistors etc.
Hello there,

The simplest way to do it is to use a time domain solution because this is not the usual linear response that we normally see when dealing with audio and such. It's a 1-sin(wt) type problem not just sin(wt).

The power is the integral of the response of V*I within the amp output stage, and we can look at the derivation if you like but the result is:
P=(2*Ip*Vs)/pi-(2*Ip*Vp)/pi+(Ip*Vp)/2

where:
P is the power dissipation in the output stage,
Vs is the source power supply DC voltage,
Vp is the peak of the 'audio' wave output voltage assumed to be not clipped,
Ip is the peak of the 'audio' wave output current,
pi is the constant 3.14159...

The reason why this is not the usual linear case is because of the 1-sinwt part which means the voltage drop in the amplifier gets less and less as the output volatge gets higher and higher, which means it's not a pure sine wave response anymore but rather a DC voltage minus a sine wave which is where the 1-sinwt part comes from.
We can also look at it in terms of harmonics, but that seems more difficult than just a regular time domain approach.

Last edited:
• bkyavas

bkyavas

Joined Jun 5, 2012
17
Thank you for detailed answer MrAl, that was the solution that I will do if no other easier idea found (that never means it is a bad solution, but somehow complex because of the inductive load and push-pull stage together). I was just looking for a practical idea that will show the problem is not so comlex at all I agree the solution will be; integrating the current flowing through and voltage dropout on the amplifier, but the quadrature phase current makes it even complex. I am thinking about how the push-pull transistors carry the current when the voltage is on the other quadrant.

Normally, in case of square wave is driven by half-bridge or full bridge MOSFETS to an inductive load; the power flows back to the supply through the internal structural diodes of MOSFET's, to a capacitor etc. will temporarily store the energy.

In the graphics (Vo assumed 10V magnitude, Vs assumed 15V) it can be said that the Integration of the red (dropout voltage) and green (load current) product will give the result, but I am confused that which driving transistor (push or pull) voltage drop-out will be sourcing the current. It seems possible to solve this (assuming ON transistor is in phase with the current). I have to admit that, I searched an simpler solution for this.

I feel lucky that distorsions can be neglected because the nature of the signals, input is a good LC tank generated sine wave, and output is linearly inductive without mutual effects on output inductor. The distorsion of the amplifier is good enough to let me get high SNR at monochromatic sine wave.

crutschow

Joined Mar 14, 2008
28,533
I think you will find that, after you do all your calculations, the answer will be the same as using my solution.

• DickCappels

MrAl

Joined Jun 17, 2014
8,609
Thank you for detailed answer MrAl, that was the solution that I will do if no other easier idea found (that never means it is a bad solution, but somehow complex because of the inductive load and push-pull stage together). I was just looking for a practical idea that will show the problem is not so comlex at all I agree the solution will be; integrating the current flowing through and voltage dropout on the amplifier, but the quadrature phase current makes it even complex. I am thinking about how the push-pull transistors carry the current when the voltage is on the other quadrant.

Normally, in case of square wave is driven by half-bridge or full bridge MOSFETS to an inductive load; the power flows back to the supply through the internal structural diodes of MOSFET's, to a capacitor etc. will temporarily store the energy.

In the graphics (Vo assumed 10V magnitude, Vs assumed 15V) it can be said that the Integration of the red (dropout voltage) and green (load current) product will give the result, but I am confused that which driving transistor (push or pull) voltage drop-out will be sourcing the current.

View attachment 135828

It seems possible to solve this (assuming ON transistor is in phase with the current). I have to admit that, I searched an simpler solution for this.

I feel lucky that distorsions can be neglected because the nature of the signals, input is a good LC tank generated sine wave, and output is linearly inductive without mutual effects on output inductor. The distorsion of the amplifier is good enough to let me get high SNR at monochromatic sine wave.

Thank you for detailed answer MrAl, that was the solution that I will do if no other easier idea found (that never means it is a bad solution, but somehow complex because of the inductive load and push-pull stage together). I was just looking for a practical idea that will show the problem is not so comlex at all I agree the solution will be; integrating the current flowing through and voltage dropout on the amplifier, but the quadrature phase current makes it even complex. I am thinking about how the push-pull transistors carry the current when the voltage is on the other quadrant.

Normally, in case of square wave is driven by half-bridge or full bridge MOSFETS to an inductive load; the power flows back to the supply through the internal structural diodes of MOSFET's, to a capacitor etc. will temporarily store the energy.

In the graphics (Vo assumed 10V magnitude, Vs assumed 15V) it can be said that the Integration of the red (dropout voltage) and green (load current) product will give the result, but I am confused that which driving transistor (push or pull) voltage drop-out will be sourcing the current.

View attachment 135828

It seems possible to solve this (assuming ON transistor is in phase with the current). I have to admit that, I searched an simpler solution for this.

I feel lucky that distorsions can be neglected because the nature of the signals, input is a good LC tank generated sine wave, and output is linearly inductive without mutual effects on output inductor. The distorsion of the amplifier is good enough to let me get high SNR at monochromatic sine wave.
Hi,

Thanks for pointing that out so clearly, i wish i could see this level of detail in all question that come to forums like this one. However, you did mean that you are using dual supplies right? That means +Vs takes the load current on positive peaks and -Vs takes it on negative peaks.
BTW we can shift the phase of the inductor current not the voltage.

It turns out that i must have confused R3 for the load because that solution is not for an inductive load it is for a resistive load.

The solution for an inductive load which is purely inductive (or with very very small sense resistor) is:
P=2*Ip*Vs/pi-Ip*Vp/pi

where:
P is the total average power in the amp output stage,
Ip is the peak current in the indcutor,
Vs is the supply DC voltage assuming a dual supply plus and minus Vs volts,
pi is the constant 3.14159...

This can be broken down into:
P=Ia*(2*Vs-Vp)

and for a single supply:
P=Ia*(Vs-Vp)

Notice that there is a difference always involved because the amp voltage drop is always the difference between the supply voltage and the output voltage.

The reason this comes out simpler is because the inductor current is 90 degrees out of phase with the voltage and that simplifies the trig functions so the time domain function for the instantaneous power comes out simpler and thus the integration is simpler also.

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MrAl

Joined Jun 17, 2014
8,609
I think you will find that, after you do all your calculations, the answer will be the same as using my solution.
Hi,

You said:
If no power is being dissipated in the inductor than all the power is dissipated in the amp and sense resistor, R3.
This is equal to the supply voltage times the average current from the power supplies (equal to the rectified average inductor current).
So you are saying that without the sense resistor we have;
P=Ia*Vs

where Vs is the power supply voltage and Ia is the inductor average current.

However, does that really make sense? Consider the case where the supply voltage is 10v and the peak voltage of the output is 9v. At that point we only have 1v across the amp output stage. When the voltage falls, the voltage across the amp will rise, but it wont be a sine wave it will be a very non sinusoidal with very sharp points at the top.

Still disagree? Do a simulation and check it out. Keep in mind though that the voltage across the amp output stage is not Vout it is Vs-Vout where Vs is the source voltage.

Of course we consider a zero centered output wave also so we either have dual supplies or we have a capacitive output coupling in order to center the wave about zero.

bkyavas

Joined Jun 5, 2012
17
Dear MrAl, thank you very much for your response, It seem s very simple, I have to understand and prove the solution In the selected frequency range XL is about 9-11 Ohms where sensing resistor R3 is 1 Ohm. L is operating 30uH @50KHz. I have intentionally neglected the R3 here.

bkyavas

Joined Jun 5, 2012
17
Dear MrAl, There should be something wrong with the formula "P=2*Ip*Vs/pi-Ip*Vp/pi" in another representation it is "P=Ip.(2Vs-Vp)/pi", (Here I assume; Vp is the peak of the sinusoidal output voltage.) because;

In case of power supply full rail (2Vs) equals to Vp (assume it is an ideal full rail output swing amplifier) the average power becomes zero with the formula. Normally the induced power cannot be recovered and stored into the amplifier, the amplifier will still consume an average power while driving an inductive load.

Am I missing something?

crutschow

Joined Mar 14, 2008
28,533
When the voltage falls, the voltage across the amp will rise, but it wont be a sine wave it will be a very non sinusoidal with very sharp points at the top.

Still disagree? Do a simulation and check it out. Keep in mind though that the voltage across the amp output stage is not Vout it is Vs-Vout where Vs is the source voltage.
If the current through the inductor is a sinewave, then the voltage from Vs to the output is also a sinewave.
What makes you think otherwise?
It's only non-sinusoidal if the op amp has insufficient bandwidth or slew-rate.

Edit: But whether the op amp output is sinusoidal or not, you can arrive at the result by simple conservation of energy. The power into the amp is its average supply current times the supply voltage.
If the output load dissipates no energy, than all that power must be dissipated in the amp.

So here's the LTspice simulation for you, of a similar circuit:
The average (rectified) inductor current is 6.37mA for an average power loss in the op amp of 10V * 6.67mA = 63.7mW, from my supposition.

The total op amp power (blue trace) is 132.8mW.

The quiescent current of the op amp is 3.35mA for an operating dissipation of 20V * 3.35mA = 67mW

Subtracting that from the total op amp power gives 132.8mW - 67mW = 65.8mW for the power caused by the output load.

This is close to the 63.7mW for the calculated power dissipation caused by the inductor current, as expected.

It would seem you are the one that needs to do some simulations, not me.  Last edited:

MrAl

Joined Jun 17, 2014
8,609
Dear MrAl, There should be something wrong with the formula "P=2*Ip*Vs/pi-Ip*Vp/pi" in another representation it is "P=Ip.(2Vs-Vp)/pi", (Here I assume; Vp is the peak of the sinusoidal output voltage.) because;

In case of power supply full rail (2Vs) equals to Vp (assume it is an ideal full rail output swing amplifier) the average power becomes zero with the formula. Normally the induced power cannot be recovered and stored into the amplifier, the amplifier will still consume an average power while driving an inductive load.

Am I missing something?
Hi,

If Vs is 10v then Vp can only be less than 10v. Vp can be say +9v down to -9v.
Thus at the positive peak, Vs=10 and Vp=9.
A the negative peak, Vs=-10 and Vp=-9.
Thus, 2*Vs=20, and 20-9=11. 11*2=22, and 22/pi=7.00 approximately.

Does that help?

MrAl

Joined Jun 17, 2014
8,609
If the current through the inductor is a sinewave, then the voltage from Vs to the output is also a sinewave.
What makes you think otherwise?
It's only non-sinusoidal if the op amp has insufficient bandwidth or slew-rate.
Hi,

Let's stop right here for a minute.

The voltage drop across the op amp output stage is the source Vs minus the output Vout.
If Vs is a DC voltage Vdc, and Vout is a sine, then Vdc-Vout is a constant minus a sine wave.
With a peak of Vp and DC voltage of Vdc, we have then across the output stage:
Vdiff=Vdc-Vp*sin(ang)

Q1: Do you agree with this ?
Q2; Do you acknowledge that this is for positive excursions of the output sine?

Next, when we have a negative excursion, the difference voltage is:
Vdiff=Vp*sin(ang)-Vdc

because now Vdc is more negative than Vout so in effect we have:
Vdiff=Vp*sin(ang)+Vdc

when we make Vdc positive and for angles (ang) pi to 2*pi.

So for angles 0 to pi we have:
Vdiff=Vdc-Vp*sin(ang)

and for angles pi to 2*pi we have:
Vdiff=Vp*sin(ang)-Vdc

Q3: agree or not with that:

because Vdc is negative for angles pi to 2*pi, or we can change that to:
Vdiff=Vp*sin(ang)+Vdc

if we keep Vdc positive. However, remembering the power supply is plus and minus 10 volts for example.
Q4: agree or not with that?

What i am showing here is that the difference across the output is not a sine.

Q5: With a voltage supply of 10v and peak output sine of 1v we have differences:
000 degrees: Vdiff=10-0=10v
090 degrees: Vdiff=10-1=9v
180 degrees: Vdiff=10=0=10v

and even without looking at the negative part of the cycle we can see that this is not a sine wave, although it may be a sine with a huge offset DC which MUST account for part of the dissipation.
Looking at the negative half, it would go 10v, 9v, and 10v again, which taken together with the positive half clearly is not a sine wave but some sort of inverted full wave rectified sine with an offset DC.

You have to realize that this has to be simulated very carefully.

• bkyavas

crutschow

Joined Mar 14, 2008
28,533
and even without looking at the negative part of the cycle we can see that this is not a sine wave
Look at the yellow waveform in my simulation.
It's the voltage difference between Vout and the V+ DC.
It's clearly a sinewave.
If you add or subtract a sinewave from a DC value, the result is still a sinewave.
I don't see how it can be otherwise. MrAl

Joined Jun 17, 2014
8,609
Look at the yellow waveform in my simulation.
It's the voltage difference between Vout and the V+ DC.
It's clearly a sinewave.
If you add or subtract a sinewave from a DC value, the result is still a sinewave.
I don't see how it can be otherwise. Hi,

Well by the time we are finished talking here you should be able to see otherwise Sorry to say i think that is why you see a problem with my analysis, because it looks like a sinewave graphically but mathematically it is not JUST a sinewave. It is actually a sinewave with a big DC offset, and you can see that right in the plot itself where the offset looks like it might be around -10 volts DC. That means that it cant be used directly in a calculation unless the DC offset is also incorporated somehow.

To illustrate, say we have a true sinewave of peak plus and minus 1 volts. We calculate the power in a resistor of say 1 ohm or something. We could end up with a result of say P1 watts. Now lets add 100 volts offset to that same sinewave. Will we get the same power dissipation of P1 watts? No way, we'll get something like 10000 watts or so So the power dissipation in the output stage follows this same idea. The DC offset counts and it could count a lot.

Also, note that the DC minus a sine is not even a sine if we dont have the voltage/current supply to support that.
10-10*sin(t) is OK as long as sin(t) is positive, but when sin(t) goes negative we could get 20 volts. That's only possible with a ground referenced load with a single supply and no output coupling cap which we cant allow in an AC circuit unless we allow a huge DC offset current in the load. This is a side point though which can be ignored i guess.

The main point is that 1-sin(t) is not a sine wave in AC circuit theory. The Fourier component analysis would show a single component which is the fundamental yes, but also an a0 component which is the DC offset. The wave sin(t) on the other hand would just show the fundamental with no a0 component so no DC offset.
The power calculations will be different for the two types of waves.
Can we even use the RMS values? We could explore that but probably not.

Here's some calculations of the output with various phase angles and the voltage differences. Note the output is always positive never going through zero.
This is with a plus and minus 10v supply and sine output with Vp=9 volts, so it goes plus and minus 9v max.

Attachments

Last edited:
• bkyavas

crutschow

Joined Mar 14, 2008
28,533
Well, I can see this is going nowhere fast. You seem to love going off on tangents to the original discussion, which are just spurious noise.
First you say the voltage between the output and the supply is not a sinewave and now you agree that it's a sinewave with offset.
That was the only point I was making.
I know that a sinewave with offset doesn't generate the same power as a sinewave without offset but that's a red herring.
What does that have to do with the original question?

You haven't shown anywhere that my simple calculation does not determine the dissipation of the amp, which is what the TS wanted.
You can go through all the arcane equations you like to calculate that value, but it won't give a different answer. • bkyavas

bkyavas

Joined Jun 5, 2012
17
If the current through the inductor is a sinewave, then the voltage from Vs to the output is also a sinewave.
Yes, but the voltage at the output transistors is not a conventional sine wave; it is a sine with DC offset, or a part of a sine wave (other part switched to the complemantary transistor that polarisation reverses on that duty).

Edit: But whether the op amp output is sinusoidal or not, you can arrive at the result by simple conservation of energy. The power into the amp is its average supply current times the supply voltage.
If the output load dissipates no energy, than all that power must be dissipated in the amp.
Yes, This is valid only for the load conditions, not the driving circuit, with an exception that the driver output is able to transfer the energy to the supply rails, It may be valid for an H-Bridge MOSFET structure because of the internal reverse diodes would do that. But for a BJT, it is not possible because it cannot transfer any current in reverse direction back to the supply rail, it can only consume that power at its output transistors.

- If the current through the inductor is a sinewave
- the voltage from Vs rails to the output is also a sinewave
- there is no structure to conduct a reverse current back to the power rails
- there is no internal power storage component within the driver

the instant power dissipated by the driver will always be positive.

My first attempts was by using a simulator to solve this, I was not sure how to consider the measurements at that.

I will sum up my message according the latest messages, I reply back with the time-domain happenings on a graph To be continued..

MrAl

Joined Jun 17, 2014
8,609
Well, I can see this is going nowhere fast. You seem to love going off on tangents to the original discussion, which are just spurious noise.
First you say the voltage between the output and the supply is not a sinewave and now you agree that it's a sinewave with offset.
That was the only point I was making.
I know that a sinewave with offset doesn't generate the same power as a sinewave without offset but that's a red herring.
What does that have to do with the original question?

You haven't shown anywhere that my simple calculation does not determine the dissipation of the amp, which is what the TS wanted.
You can go through all the arcane equations you like to calculate that value, but it won't give a different answer. Hi,

When you say sinewave in this context it is not correct, sorry. That's really what this was all about.

You're going off on a tangent by starting that i was going off on a tangent because i was calculating the power loss which is what this thread is about.

"First you say the voltage between the output and the supply is not a sinewave and now you agree that it's a sinewave with offset."
I am not agreeing with you. I agree only to 'sinewave with offset" where you said 'sinewave'. That's a different horse or else they would be exactly the same. By stating sinewave, you imply that there is an AC analysis when there isnt. By stating sinewave with offset we know there is no direct AC analysis.
I think that's the only problem really.
Also, we have the power supply ripple to consider which we did not address yet.

I'll try to do a simulation either later today or tomorrow as per your suggestion. I'll compare results.

Last edited:
• bkyavas