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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.

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    \$\begingroup\$ That is not a low side switch. A low side switch is when the switch is on the "low side" of the load. \$\endgroup\$
    – The Photon
    Commented Jul 28, 2017 at 17:08
  • \$\begingroup\$ 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. \$\endgroup\$
    – BeB00
    Commented Jul 28, 2017 at 17:25
  • \$\begingroup\$ 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 \$\endgroup\$
    – sstobbe
    Commented Jul 28, 2017 at 17:36
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    \$\begingroup\$ Corrected it to high side, \$\endgroup\$ Commented Jul 28, 2017 at 18:02
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    \$\begingroup\$ 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. \$\endgroup\$
    – G36
    Commented Jul 28, 2017 at 18:17

1 Answer 1


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.

  • \$\begingroup\$ Direct short 24V to base , couldn't damage the transistor ? Since datasheet say max Vbe is 5V only. \$\endgroup\$ Commented Jul 28, 2017 at 17:55
  • \$\begingroup\$ The emitter voltage will rise to stay with the base. \$\endgroup\$ Commented Jul 28, 2017 at 17:56
  • \$\begingroup\$ Wondering why to use high side switch with NPN( as i used) , since only advantage i can see it that output signal will be in phase with input. where is Low side NPN switch (popularly used) , where we can get full voltage but out of phase with respect to input. \$\endgroup\$ Commented Jul 28, 2017 at 18:06

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