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VBE voltage drop formula?

Appears long-gone... anyway - it's not difficult to guesstimate close-enough.

Assuming you are using a BJT not a mosfet as the pass device, and picking round numbers for ease of ilustration/ close-enough approximation:

  • There is the forward base-emitter diode drop, about 0.7v
  • There is the base current necessary to drive the pass transistor hard enough to pass the required current and that depends on having a useful estimate of hfe, the forward current gain at your desired current output. Check the part datasheet for a ballpark estimate.
  • Say you want 100mA output, and to use a regular TO220- style pass transistor, that has (for round numbers) forward current gain, hfe, = 100 at 100mA (check the graphs for the device you want to use)
  • Say you've also picked an R-C filter value of maybe 1Kohm /100uF, for a nice low frequency filter 'knee' (1.6Hz in this case - will take a c.36dB bite out of 100Hz ripple, and very much more out of HF noise.)
Then
  • So - to get the transistor turned-on enough to pass 100mA, it will need (100mA/hfe) = 1mA of base current.
  • That in turn means 1mA x 1Kohm = 1v dropped across the RC filter.
  • Add 0.7v for the base-emitter drop, and you can see - there will be about 1.7v drop from your reference voltage to the emitter on the pass transistor. Your reference voltage therefore needs to be 1.7v or more higher that what you want out of the transistor to feed whatever follows.
This is a close approximation - at least, close enough for most purposes.


[ETA: If you are using a mosfet instead as the pas device, you can ignore the B-E 0.7v drop, also the idea of voltage drop across the R portion of the reference filter! Instead you will just have to find -out/ measure whatever the Gate-Source voltage drop is at the desired current out of the mosfet you have chosen - could be 2-4V, and likely, rather-poorly characterised esp at such low currents in the mosfet datasheet.]
 


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