Summing amplifier setup

This circuit diagram was taken from one of the TI application notes for testing PSRR of LDO.
My question is about configuration of summing op-amp:

  • What's the need of resistor in between the terminals ?

  • What's the purpose of ferrite bead at the output of the amplifier ?

  • Why 100 uF at the input of inverting terminal?
    My assumption: Is it to make sure we are feeding only AC signal and no DC offset through this terminal ?

  • What's the advantage of using current-feedback summing amplifier as opposed to conventional voltage feedback amplifier for testing PSRR.


THS3120 is a current feedback amplifier (and not voltage feedback like most opamps), so the external impedance at input terminals should be low. that's why there's a 100ohm resistor at the input terminal of the device. At the output, the ferrite bead is used to handle and isolate the possible capacitive load (the input capacitance of DUT might be high). And yes the 100uF cap in front of "RF out" is used to AC-coupling and blocking any DC current which could make error for the amplifier.

More explanation: Think about the purpose of THS3120 block. It is been designed to provide a DC voltage superimposed by an AC signal at its output. Also, take into consideration that the THS device is a current-feedback amp, so it is sensitive to the difference between the "current" flowing in non-inverting and inverting terminals, furthermore, the input impedance at noninverting terminal is high but at inverting terminal is low. Now let's analyze the DC and AC situation separately. In DC analysis the input resistor of 1K is removed (because of 100uF cap). If we solve KCL and KVL equations, we approach to the result that Vo = VDC (approximately). And in AC analyzing the input DC is removed, so we reach to Vo = -1 * RFsignal (approximately). So the overall output would be: Vo = Vdc - RFsignal. So, if we remove the 100ohm resistor the output result will be something very different. The 100ohm res in fact reduces the current flowing in input terminal of THS considerably.

And about your last question: in theory we can replace this current-feedback with a voltage-feedback opamp (and also removing that 100ohm res). BUT, I suppose, they designed based on this chip because of its high-output-current capability which is much higher than conventional and popular low-cost voltage-feedback opamps.

  • \$\begingroup\$ As you mentioned input impedance of the current feedback amplifier is already very low by design. Then what's the use of the 100 ohm resistor ? \$\endgroup\$ Jun 12 '20 at 8:28
  • \$\begingroup\$ I've edited my answer to provide more explanation. Good luck \$\endgroup\$ Jun 12 '20 at 13:37

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