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BJT current mirrors, have a 2 or more bipolar junction transistors (BJTs) wired up so that the current set through the reference transistor, will also flow through the load/output transistor(s).
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The reference current (Iref) BJT is the one that has it’s base and collector tied together. The resistance from collector to ground will set the current based on Ohms law. About 0.6V is dropped from the supply voltage due to the Emitter to Base forward voltage.
- I estimated 4.4V being across the resistors, as 5V supply minus 0.6V emitter to base (diode) forward voltage drop equals 4.4V.
- To get 1mA of current through the resistors, I put a 3,900Ω and 510Ω resistor in series. That gives a total of 4,410Ω of resistance. 4,400Ω/4.4V = 0.001A (or 1mA) of current. So, 4,410Ω will be a small amount less than that.
- The other transistor will pass that same amount of current through a light load (such as an LED). Of course, less current will flow through the load if the load limits it even more though or drops too much voltage.
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