# The operational amplifier implements a constant current source circuit

2023-10-26Archive

The figure above shows a constant current circuit built around an op-amp, same as in previous issue, implementing a 1 mA constant current source. Its working principle is mainly in op-amp voltage follower circuit and triode emitter follower circuit.

Suppose op-amp U1A used here has a magnification of 100,000 times and triode magnification is β=100. At start of power-up, assuming that Vin input is 2.5 V, op-amp has no negative feedback, and op-amp output pin 2 rises from 0 V until it reaches op-amp saturation voltage. ampere, that is, supply voltage of op-amp is + 5V. If negative feedback is added, Vin = 2.5V, assuming op-amp output pin 1 is currently rising to 1V, since load R4 needs a 1mA DC source, Ic Q1 = 1mA, β = 100. and Ib is about 1mA/100=0.01mA=10µA, from above data, current flowing through R2 at this time is µA level, very small, so voltage drop across R2 is about 0V and voltage is 1 V is added to B pole of Q1, Vb = 1V, E pole of emitter of Q1 Ve = 0.3V, current Ie of Q1 can be Ve / R3 = 0.3V / 5.1k = 58.8uA, and at that At same time Ve = 0.3V, it is fed back to pin 2 of op-amp, due to coupling, feedback occurs at inverting terminal of op-amp, so it is also called negative feedback. As shown in figure above, pin 2 receives a potential equal to Ve, potential Ve=0.3V, Vin is still 2.5V, Vin-Ve=2.5-0.3=2.2V, and gain operational amplifier is 100,000 times. no negative feedback output, it will be directly saturated. The feedback here can cause output of op-amp to not saturate, it will reach Vin 2.5V, since op-amp gain is 100,000 times, voltage difference between IN+ and IN- of op-amp can be ignored, that is, get VIN+=VIN-, which is what we often refer to as "false short" of op amp.

If Vin is currently 2.5V, why is VIN+ also 2.5V? The answer is obvious, input impedance of op-amp is close to infinity, input current is only pA, and a large one is uA. The voltage drop across R1 is about 0 V, so Vin=VIN+, i.e., "virtual break" of op-amp.

According to virtual break principle, Vin=VIN+; according to virtual break principle, VIN+=VIN-=Vin=2.5V, when op-amp is dynamically balanced, VIN- will be stable at 2.5V, for same reason Ve=2.5V, according to Q1N Next is emitter of triode, Vb \u003d Ve + 0.7V \u003d 2.5 + 0.7V \u003d 3.2V, that is, Vout OU \u003d 3.3V. Based on data above, Ve = 2.5V, R3 = 5.1K, I = 2.5V / 5.1K = 0.49mA = 490µA The target value I developed here should reach 1 mA, according to Vin=VIN+=VIN-=5.1V, get Ve=5.1V, R3=5.1K, Ie=Ve/R3=5.1/5.1K=1mA Q1, which meets my design requirements so that R4's load can reach 1mA DC.