![\"Original](\"http:\/\/tangopardo.com.ar\/2cf7\/wp-content\/uploads\/2014\/02\/ckt_orig.png\")
<\/a>Original circuit of the active load.<\/p><\/div>\n
The opamp power supply is not shown, Vext and Zl are external to the circuit. Assuming for a while that the opamp is a bit more ideal and its output can swing to at least Vext the circuit can be reorganized as follows without changing the behaviour:<\/p>\n
<\/a>Active load circuit, reorganized.<\/p><\/div>\n
And finally, we can dispense of the transistor, arriving at the bare minimum:<\/p>\n
<\/a>Active load, simplificated.<\/p><\/div>\n
If Zl is an inductor we can model it as a ‘pure’ inductor L in series with a resistance Rl. Then, the voltage at Iout will be:<\/p>\n
We can build a block diagram of the whole system:<\/p>\n
<\/a>Uncompensated system.<\/p><\/div>\n
So we have a three pole system (p1 and p2 are the opamp poles). This is bound to oscillate sooner or later unless gain is reduced or the response modified with a compensation network.<\/p>\n
The traditional recipe for this cases is to put a small capacitor from the output to the inverting input of the opamp. Given the very low impedance present at V- we isolate it with a moderate resistor, thus the new circuit is:<\/p>\n
<\/a>Compensated active load.<\/p><\/div>\n
To obtain the voltage at Vfb we use the superposition theorem. Neglecting the loading effect that Rcc and Cc can present at the Vout and Iout nodes we consider the two cases:<\/p>\n
<\/p>\n
<\/a>Models of the compensation network for Iout and Vout to Vfb<\/p><\/div>\n
For Vout it acts as a high pass filter and as a low pass for Iout. In the s domain that is:<\/p>\n
And in a block form:<\/p>\n