I have been testing the transient response of line regulator simply with a MOS and the resistor, this method, however, cannot generate arbitrary current and the current is not accurate. To improve the test, I turn to Linear Tech's AN104 for help and the circuit ploted in Fig.1 is the solution. The question is that roles of these resistors especially those at the inverting and non-inverting node. It is mentioned in the note that "the network's resistors are arranged for minimum load current of 10mA", but how to arrange these resistors to generate current in other values?

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1 Answer 1


A1 controls the current in Q1; it controls its output so that the voltage at its -ve input equals the voltage at its positive input. For a 1V at +in, Q1 current will be 10A, so the transfer ratio is 10A per volt.

However, this transfer ratio changes to 1A per volt at the BNC terminal marked "Waveform Input" due to the voltage divider formed by the 100 ohm & 909 ohm resistors. So 1V at BNC input causes 1A at Q1.

A2 controls the DC bias current in Q1, simply by providing a bias voltage to A1 +in when the waveform input at BNC is 0V. The potentiometer allows you to adjust this dc bias from 0 to 1A.

If you want to change the range of DC bias current, or the transfer ratio of BNC input to output current, then:
Firstly, mark up the schematic so each component has an identifier, eg: R1, R2, etc.
Secondly, specify exactly what you want to change, eg: change DC bias current to be 100mA when pot is set to minimum.
I will then be able to assist further.

  • \$\begingroup\$ Thanks for your answer. I have marked each resistors. From my understanding: R6, R8 and R1, R2 form the voltage divider of each opamp, R3 is sufficiently large to eliminate the its shunt effect to R2. R5 for impedance matching in the output of waveform generator. Then what is the purpose of R4, R7, R11 and R12 all in 10K Ohm, which differ in the tolerance requirement. Another confusion is the purpose of A2, it seems that A1 can convert the voltage change in in1+ to current on R10 without A2 and R9. I plan to change the load current from 100uA to hundreds of mA (200mA probably). \$\endgroup\$
    – cc Lau
    Commented Jun 30 at 17:14
  • 1
    \$\begingroup\$ @ccLau Responding to: "I plan to change the load current from 100uA to hundreds of mA (200mA probably)". This circuit already does that. The bias can be adjusted from 0 to 1A, and the Waveform Input gives 1A per volt. Both of these parameters seem to suit your requirements, so why do you want to change this circuit? Or do you want to understand how it works? \$\endgroup\$ Commented Jun 30 at 19:30
  • \$\begingroup\$ Yes, I want to understand its principle and the confusion I have in the last comment. \$\endgroup\$
    – cc Lau
    Commented Jul 1 at 2:26

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