Re: peak current source
I just stumbled across this, and couldn't resist an answer, here 2 years after it was posted.
Here's a quickie approach.
The odd resistor in the collector of the left hand transistor, should be set to 1/gm1. Then a cool voltage division occurs for smal variations in collector current for the left hand side. A slight increase in Collector current will cause VBE1 to rise by deltaI/gm1. But the resistor pushes it back down by deltaI*gm1. Thus they cancel, and you gain some slight protection against small vaiations in Current on the left side. For example, the setup resisistor may have a tolerance or TC drift. To a first order, this cancels out much of it.
Of course, gm1 = Ie/Vt ~ Ic/Vt
The transistor on the right has 4x the area of the one on the left. Since VBE changes by about 18mV for every doubling in current desity, you can anticipate the current of the right hand side if you simply assume you want the input current to match the output current. Then the device on the right has 2*18mV less of Vbe (two doublings). That extra 36mV is soaked up by the emitter resistor on the right. if you want 100uA coming out of this thing, set the input to 100uA (using the bias resistor R1) and select the emitter reesistor at 36mV/100uA = 360 Ohm. If the TCR of the emitter resistor is positive, you can expect some happy temperature compensation, as the delta VBE is also postive.
If you do not want the left and right currents to match, the equations already posted will stand.
- Michael