Simple NPN Switch

Thread Starter

philipm

Joined Jun 27, 2012
47
I'd like to use an NPN to switch a small load, either saturated or off.

In the attached circuit, +VBAT varies between 3V and 4.2V and the base logic operates at 2.8V.

I'm planning to use a BC817-40 (datasheet here).

My calculations are:
Ic = 4.2 / 10000 = 0.42mA
Ib = 2.8 / 4700 = 0.50 mA
I know it's not usual to have Ib>Ic, but with Hfe of approx 400, it'll be saturated!

If I'm reading the datasheet correctly, this condition should give:
Vce(sat) of 0.01v
Vbe(sat) of 0.5v

Does that all seem correct? Anything to watch out for with this circuit?
 

Attachments

crutschow

Joined Mar 14, 2008
34,464
It will work, but normally you don't need a base current higher that 1/10th of the collector current for full saturation. Under those conditions the base-emitter voltage would be closer to about 0.6V. For that R2 = (2.8v - 0.6V) / .0042ma = 520k ohms.
 

Thread Starter

philipm

Joined Jun 27, 2012
47
Many thanks. I have spare resistors in an array and another BC817 on the board, hence keeping the design simple with fewer lines on the BOM. =)
 
Top