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#1
12-30-2013, 07:51 PM
 philipm Junior Member Join Date: Jun 2012 Posts: 47
Simple NPN Switch

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?
Attached Images
 npn.png (12.3 KB, 26 views)
#2
12-30-2013, 08:23 PM
 crutschow Senior Member Join Date: Mar 2008 Location: L.A. USA Posts: 6,317 Blog Entries: 1

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.
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#3
12-30-2013, 09:14 PM

2N7000 'Fetlington'
Max.
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#4
12-31-2013, 09:48 AM
 philipm Junior Member Join Date: Jun 2012 Posts: 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. =)

 Tags npn, simple, switch

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