Electrical – High Side NPN transistor load switching


enter image description here

I have one circuit ( Circuit A) , transistor used is BCP56-16T1G, R1 is fix load of 500 ohm. I am measuring output at VF1 , that comes to around 19 V, I simulate the same circuit , i found the Vf1 is actually come 19 V in simulation too.

And as per calculation too it should be come
24= 10KIb+Vbe+R1(Ic+Ib)
24= 10KIb+ 0.7+500(100*Ib+Ib) ( since Ic=Hfe*Ib)
so finally Ib=385uA
so Ic= 100*385 = 38.5mA
Drop across R1 = (38.5+0.385)*500 ohm =19.442 Volt.

problem is I need VF1 to be near 24V. off course it will go above 23.3V. but i am getting only 19.4V, and r1 is fixed , so only thing i can change is base resistor. Transistor data sheet does not give max base current limit. but they give max collect current. so i can see max base current will be = 1A/Hfe= 1000mA/100= 10mA. So i try to reduce the base resistor to 10ohm , and I got VF1 near to 23.2V( which I desire circuit B).

So will this Ok to reduce base resistor to this much lower value , is it safe for transistor also.

Best Answer

You can take out the resistor completely and leave a short. As you have already pointed out, the emitter voltage is ~23.3 (one diode drop down), so the emitter current is 23.3/500 or 46.6 ma. Base current will be 46.6 ma/Hfe. This would be an "emitter follower" rather than a low-side switch.