I have a rather simple circuit that acts as an overvoltage protection switch with an input control signal to enable/disable the output.On the final output I am driving a motor that consumes around 2A in the startup phase for 200-300ms and around 500mA constant. My input voltage is in the region of 12-30V

The issue is that I have a very large voltage drop across the transistor when the motor starts to spin up, but according to the datasheet this transistor should be able to handle large current. main pnp datasheet

My schematic

Does anybody know why the drop across my FZT956 PNP transistor is so high? I need to reduce it because the motor is used in a pulsed fashion so I need to minimise the heating in this period.

  • 1
    \$\begingroup\$ Calculate what base current you need for the 2 A collector current when that is flowing. Does R4 (4.7 k) deliver enough base current? \$\endgroup\$ Apr 22, 2020 at 7:56
  • \$\begingroup\$ I have been playing with R4 and this is helping to reduce the drop, I'm now at 1K and with 1A flowing i still see a drop of around 2V, does that sound right? \$\endgroup\$ Apr 22, 2020 at 8:20
  • 2
    \$\begingroup\$ I would also check that T2 isn't dropping too much voltage, there should be less than 0.5 V across the Collector - Emitter of T2 when it is conducting. Maybe R6 needs to be lowered as well. I would also consider using a P-MOSFET for T1 as then you can get much lower voltage drops without the need for a high base current. \$\endgroup\$ Apr 22, 2020 at 8:45
  • \$\begingroup\$ I initially had a PMOS, but unfortunately for the overvoltage case of 200V+ at the input it was difficult to turn it off quickly enough and my output was going too high \$\endgroup\$ Apr 22, 2020 at 8:55

1 Answer 1


I suppose the problem is that during start-up you are not providing enough base current to \$T_1\$ because of the capacitor \$C_3\$. Once \$V_{CTRL}\$ is pulled high, you are slowly charging the base capacitor \$C_3\$ of \$T_2\$ before turning it on. Reducing \$C_3\$ and \$R_6\$ should turn on the logic / power BJT faster, thus causing a smaller voltage drop across the power PNP transistor.

Edit #1

Considering the current gain for the case where the collector current \$I_C\approx -3A\$, you would need a base resistance smaller than:

$$R_4< \dfrac{12V - V_{BE, T1} - V_{CE, T2}}{I_B}$$

$$R_4< \dfrac{12V - 1.1V - 300mV}{\dfrac{3A}{75}} \approx 265\Omega$$

Where \$V_{CE, T2}\$ is assumed to be the approximate collector emitter voltage of a common logic BJT.


  • \$\begingroup\$ I have played with this and this it does impact the initial behaviour for a couple of milliseconds, but it is the steady state behaviour that is the problem, even with a constant 1 A at the output I still have a few Volts across the PNP \$\endgroup\$ Apr 22, 2020 at 8:17
  • \$\begingroup\$ Wow, 265ohm is a lot lower than my current resistor, I will decrease it further and see how much drop I get \$\endgroup\$ Apr 22, 2020 at 8:34
  • \$\begingroup\$ This considers an output current of \$3A\$, and you have to make sure that your \$T_2\$ is able to withstand this current (\$\approx 3A/75=40mA\$) in steady state. \$\endgroup\$
    – vtolentino
    Apr 22, 2020 at 8:39
  • \$\begingroup\$ T2 is eu.mouser.com/datasheet/2/115/ds30062-87055.pdf \$\endgroup\$ Apr 22, 2020 at 8:51
  • 1
    \$\begingroup\$ @EoinOConnell: Can you show the schematic of the MOSFET circuit you used? Because the current overvoltage protection circuit is suited for a bipolar transistor, but not for a MOSFET. With a slight modification, it can be made to work with a MOSFET. \$\endgroup\$ Apr 22, 2020 at 19:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.