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I'm using a PNP transistor with V_CEO= 30V, V_CBO=40V, and V_EBO=10V as a signal inverter for the output from a control IC (TL5001). So, the power source for the TL5001 will be also applied to the emitter terminal which is around 6V, the output voltage from the TL5001 will be applied to the base terminal and the output from the collector terminal will go to the gate driver in my buck converter circuit.

Originally, I was thinking of using the same power source for both my buck converter and TL5001 which is 24V but I used another power source of 6V for TL5001 because of the emitter-base breakdown voltage of the PNP transistor.

My circuit is not working properly so I am not sure I did the right consideration here.

If V_EBO of transistor is 10V, then the voltage which applies to the base terminal should also not exceed 10V?

enter image description here

This is my rough circuit schematic. Vcc2 is 6V of power supply.Photocoupler is used for the gate driver. So, PWM voltage will be applied to the base terminal of PNP transistor.

With those breakdown voltage ratings, using 24V which is a input voltage supply for the buck converter will not be able to be used as Vcc2?

https://html.alldatasheet.com/html-pdf/50728/FAIRCHILD/BC516/404/1/BC516.html This is the datasheet for the used PNP transistor.

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    \$\begingroup\$ I suspect that if you look at the data sheet again you'll see that V_EBO is a negative value.. It's the breakdown voltage when the junction is reverse biased. \$\endgroup\$
    – Finbarr
    Commented May 22, 2022 at 9:48
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    \$\begingroup\$ Welcome to EE.SE. A schematic is very important to help us understand the circuit. \$\endgroup\$
    – devnull
    Commented May 22, 2022 at 9:49
  • \$\begingroup\$ If the applied voltage exceeds the breakdown voltage, it will break down! However, it may break down destructively, or may break down safely (zener diodes break down safely). Be careful about reading the specification for PNP transistors, make sure you have the voltage polarity correct (a schematic in your question would be a great benefit). Make sure you are reading between the correct terminals. You can safely apply a forward current to a BE junction which will then limit it to 0.7ish volts, and a reverse voltage of maybe 5 V without worries for 'most' transistors. \$\endgroup\$
    – Neil_UK
    Commented May 22, 2022 at 11:04
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    \$\begingroup\$ Thank you for your comments. I have added the circuit schematic diagram above. I can't find the V_EBO negative value on the data sheet... \$\endgroup\$
    – christine
    Commented May 22, 2022 at 12:32
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    \$\begingroup\$ Where's the datasheet for your transistor? \$\endgroup\$
    – brhans
    Commented May 22, 2022 at 14:10

1 Answer 1

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If V_EBO of transistor is 10V, then the voltage which applies to the base terminal should also not exceed 10V?

VEB0 is the reverse breakdown voltage.

Your circuit diagram is a little confusing since it shows the transistor 'upside down' compared to the recommended orientation (lower voltages towards the bottom) and has no ground connection to the TL5001. After drawing the circuit more conventionally it becomes obvious that the transistor never receives a reverse voltage between the Base and Emitter, so you have nothing to worry about.

schematic

simulate this circuit – Schematic created using CircuitLab

In circuits that do create a reverse voltage between Base and Emitter, it must be limited to less than the absolute maximum rating, which is less than or equal to the minimum reverse breakdown voltage (if specified). If the actual breakdown voltage is reached the junction will act like a Zener diode. Operating in this region is not tested or guaranteed. Even if current is limited by external Base resistance it could cause degradation of the transistor's characteristics or damage due to overheating etc.

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