I ran a sim for a BTL configured op-amp. I expect that the instantaneous peak current from the supply will equal the instantaneous peak current through the bridged load.

Ignoring quiescent current and other second order non-linearities if I have X amount of current in the load that same current must be supplied by the bench supply.

So my sim pretty much shows this to save my sanity. It is a little off but the concept is there.

The problem? I do not see this in real life. On my bench the peak current from the supply is half the peak of what is measured through the load.

What gives?

It is a LeCroy Waverunner and I have two current probes. One is clamped around the cable connecting the load and the other probe is clamped around the cable from the DC supply.

It has to be an oscope setting problem or a setup issue on the bench.

Thoughts?

 

The peak currents are approximately the same. Monitor the current through V2 to see the current for the other peak.
Your amplifier has a class AB output stage.

 

I know the sim is right, but this is not what is showing up on the bench in real life.
With a real part, the supply current is approximately half of the current peak through the load and this does not make sense.

 

My bad. For some reason, I thought the waveforms were from your scope. It was late.

 

Here are actual scope captures from my bench. Again, I have a current probe on VCC and a current probe across the speaker wire to the load. In this case the load is purely resistive.


The Class AB BTL current waveforms make perfect sense and what I would expect. The load current (ideally) would be equal to the current from the supply.

The Class D BTL current waveform is counter intuitive. Given the exact same setup, the load current is almost twice as high as the supply current!

Now, unless I am breaking the laws of physics, this cannot be. I am having trouble working it out in my head, but it must be the LC filter on the output of the Class D amplifier that is providing the additional current once the inrush current charges the inductor on power up.

 

And here is the LC filter for the Class D amp.

 

Do you have a schematic for the class D amp?

Generally with switchers you have a power-in = power-out situation, so if the output voltage is less than the input voltage the input current is lower than the output current. It's evidence of the great efficiency you get with a switcher compared to a linear circuit.

 

Do you have a schematic for the class D amp?

Generally with switchers you have a power-in = power-out situation, so if the output voltage is less than the input voltage the input current is lower than the output current. It's evidence of the great efficiency you get with a switcher compared to a linear circuit.

I do not follow.

I understand that the efficiency of a Class D is supposed to be, theoretically, 100%, but there are conduction losses and switching losses that reduce this efficiency.

Apparently the input current will be less than the output current (I have proof!), but I am having trouble understanding why that is.

In comparison to the class AB H-bridge even if you drive the appropriate pMOS and nMOS pair into saturation the load is connected between VCC and GND so the currents from the supply through the load must be equal. This is proven in my screen shot.

 

Yes the efficiency is less than 100%.
Say it's 80% - if your output power is 10W your input power is 12.5W.

If the output voltage is half the input voltage, say 5V from 10V, your output current is 10W/5V = 2A vs an input current of 12.5W/10V = 1.25A.
You have this situation for average power (instantaneous will vary):
Efficiency*Vin*Iin = Vout*Iout

The reason is that the load is not connected from Vcc to Gnd... you were right, it is the LC elements that mediate the exchange. They are energy storage devices which allow you to decouple the input and output currents.

If you had a more complete schematic we could walk through it.

 

Yes the efficiency is less than 100%.
Say it's 80% - if your output power is 10W your input power is 12.5W.

If the output voltage is half the input voltage, say 5V from 10V, your output current is 10W/5V = 2A vs an input current of 12.5W/10V = 1.25A.
You have this situation for average power (instantaneous will vary):
Efficiency*Vin*Iin = Vout*Iout

There is a bad assumption being made here. The output current will always be equal to the input current if quiescent current of the device is ignored.
Your math is technically correct, but it has no physical meaning, at least for a linear amplifier. If there is 2A through the load that same current must be coming from the supply. 2A*10V is 20W of input power. This also means that efficiency is far lower than 80% which also makes sense for a linear amplifier. Theoretical is 73% I believe?

Now depending on where one is operating, in a linear amplifier, the greatest power dissipation internally is when the outputs are at Vdd/2.
Even here what current is coming in will equal the current going out again ignoring quiescent current.
There is no way, for a linear amplifier, that the current from the supply will be lower than the current through the load.

If you had a more complete schematic we could walk through it.

Don't have a schematic handy of my board, but let us take a look at the TPA3100 EVM (attached). Figure 3, pg. 5. The only difference is the values for the L's and C's.

Again I am using 680nF and 22uH.

 

There is a bad assumption being made here. The output current will always be equal to the input current if quiescent current of the device is ignored.
Your math is technically correct, but it has no physical meaning, at least for a linear amplifier. If there is 2A through the load that same current must be coming from the supply. 2A*10V is 20W of input power. This also means that efficiency is far lower than 80% which also makes sense for a linear amplifier. Theoretical is 73% I believe?

Now depending on where one is operating, in a linear amplifier, the greatest power dissipation internally is when the outputs are at Vdd/2.
Even here what current is coming in will equal the current going out again ignoring quiescent current.
There is no way, for a linear amplifier, that the current from the supply will be lower than the current through the load.

I was describing the Class D switcher amp... 80% was a made up number.

Like I said, in a switcher the load is connected to Vcc and Gnd through energy storage devices which changes everything.
I'm not sure where you figured I was talking about a linear amp...