# High Side NPN transistor load switching

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.

• That is not a low side switch. A low side switch is when the switch is on the "low side" of the load. Jul 28, 2017 at 17:08
• The actual resistor value doesn't matter, only the current through the base. 500uA is completely fine, but obviously as photon points out this is a high side switch, not a low side. Jul 28, 2017 at 17:25
• Put R1 between the collector and the ammeter. Tie the emitter to ground and you have a low side switch as Photon says. At present you have an emitter-follower Jul 28, 2017 at 17:36
• Corrected it to high side, Jul 28, 2017 at 18:02
• Also, notice the in this circuit we have a quite large voltage drop across the transistor. Vce = Vcc - (Vin - Vbe) and P = Vce*Ie. This is why in most of the time we don't connect the load into the emitter. Instead, we are using the collector as a "output" and PNP transistor for high side switch.
– G36
Jul 28, 2017 at 18:17