Opamp Question: Supply current vs Frequency!

I am using a high frequency power operational amplifier (PA107DP) in the inverting configuration, without an external load as shown here:



According to my measurements, when I increase the frequency, the supply current (drawn from high voltage supply) increases too, but it has a pick near 5 MHZ (for all input amplitudes) and then, it decreases.

Can anyone explain to me why we see this extremum point? In other words, why the supply current increases before an specific frequency and then decreases after that?
The figure below, shows the supply current vs. frequency for different input amplitudes (Peak-Peak):



You may see the PA107DP datasheet here: (**broken link removed**)

Any help would be appreciated...

The amp has frequency/gain product of 180 MHZ, so with a gain of 30 (as above), gives its Frequency of 6MHZ (-3dB). The reason that the current falls of at higher frequencies is that the output voltage falls. Its all because of the output stage requiring more drive as the frequency increases because of the FETs big capacitive input impedance.
FWIW In the 1960s, germanium audio class B power amplifiers amplifiers use .5A at 10W @1KHZ and 1.5A @10KHZ!
Frank

chuckey said:

The amp has frequency/gain product of 180 MHZ, so with a gain of 30 (as above), gives its Frequency of 6MHZ (-3dB). The reason that the current falls of at higher frequencies is that the output voltage falls. Its all because of the output stage requiring more drive as the frequency increases because of the FETs big capacitive input impedance.
FWIW In the 1960s, germanium audio class B power amplifiers amplifiers use .5A at 10W @1KHZ and 1.5A @10KHZ!
Frank

Click to expand...

Thank you Frank for your quick response!

Still, I have some questions. Could you please explain more about the need for "more drive" at output stage? or do you have any useful online text regarding this issue? (I couldn't follow exactly this point)
Another question is that why does the supply current (or power dissipation) increase when the frequency increases before getting to 5 MHz? Is it related to the stray capacitors inside the Opamp which become smaller at higher frequencies?

And finally, is there any formula to explain the power dissipation of the Opamp based on the output voltage and frequency, especially when there is no load at the output? Do you know any text/paper explaining more about this?

Thanks,
Vahid

The input impedance of the gate of the output FETs, is in the order of 500 PF, which at 5 MHZ represents an impedance of 60 ohms, so a few volts of drive into this impedance represents a current of 2/60 = 30 mA.
More drive?, as the output voltage is controlled by the negative feedback as the gain decreases, the internal voltages increases. Vout = G X Vdiff. where Vdiff is Vin-Vnfb, so if G falls, for constant Vout, Vdiff increases, as Vin is the same, the Vdiff must be higher, i.e. the actual voltage being amplified inside the amp.
I would suppose the current consumption, has three main components, 1, the steady DC consumption of the DC bias of the early stages, then the DC losses of the output stage driving a load (capacitive output currents must be considered). I think this is the maximum output current limit. Then the AC losses, every stage will run out of gain at some frequency, but I think its the single time constant of the output stage that puts in more losses (as above) and effectively determines the amplifiers frequency limit.
I have never seen any data on this, I suspect that it is accidently referred to in IC data as " able to deliver 1 V into 50 Ohms +500 PF at 10 MHZ" . From what you have found, this sort of spec is as likely to be determined by the IC dissipatation as much as its slew rate.
Frank