mirrors using multiple transistors, such as the Widlar current source and the Wilson current source. Constant current source with thermal compensation One limitation with the circuits in Figures 5 and 6 is that the thermal compensation is imperfect. In bipolar transistors, as the junction temperature increases the drop (voltage drop from base to emitter) decreases. In the two previous circuits, a decrease in will cause an increase in voltage across the emitter resistor, which in turn will cause an increase in collector current drawn through the load. The end result is that the amount of 'constant' current supplied is at least somewhat dependent on temperature. This effect is mitigated to a large extent, but not completely, by corresponding voltage drops for the diode, D1, in Figure 6, and the LED, LED1, in Figure 5. If the power dissipation in the active device of the CCS is not small and/or insufficient emitter degeneration is used, this can become a non-trivial issue. Imagine in Figure 5, at power up, that the LED has 1 V across it driving the base of the transistor. At room temperature there is about 0.6 V drop across the junction and hence 0.4 V across the emitter resistor, giving an approximate collector (load) current of amps. Now imagine that the power dissipation in the transistor causes it to heat up. This causes the drop (which was 0.6 V at room temperature) to drop to, say, 0.2 V. Now the voltage across the emitter resistor is 0.8 V, twice what it was before the warmup. This means that the collector (load) current is now twice the design value! This is an extreme example of course, but serves to illustrate the issue. The circuit to the left overcomes the thermal problem (see also, current limiting). To see how the circuit works, assume the voltage has just been applied at V+. Current runs through R1 to the base of Q1, turning it on and causing current to begin to flow through the load into the collector of Q1. This same load current then flows out of Q1's emitter and consequently through to ground. When this current through to ground is sufficient to cause a voltage drop that is equal to the drop of Q2, Q2 begins to turn on. As Q2 turns on it pulls more current through its collector resistor, R1, which diverts some of the injected current in