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Why are there PNP transistors here?

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from: https://www.youtube.com/watch?v=5_uYE4EWSKY&t=884s&ab_channel=SparkFunElectronics

This shows a charge pump circuit that will be controlled via the digital pins of a microcontroller. Why does it use 3 transistors each? Wouldn't 2 NPN's each give the same output? like this:

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More specifically, can I replace the PNP transistor with an NPN transistor(as shown in the above schematic)?

(I am an amateur in electronics, so please forgive me if this is a dumb question -_-; )

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3 Answers 3

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The D1 signal (presumably 5 V logic level) is switching a 12 V circuit.

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Figure 1. The output supply is 12 V.

The PNP transistor will turn on when the base is pulled low and the emitter-collector voltage, VEC, will drop to about 0.2 V meaning that 11.8 V will be available on the output.

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Figure 2. Q3 is working in emitter-follower mode.

The NPN, Q3, of Figure 2 is working as an emitter-follower. The highest output voltage you can get is VD1 - 0.7 V. If D1 is driven by 5 V logic that means that you can only get 4.3 V out maximum.

... can I replace the PNP transistor with an NPN transistor?

Nope!

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Figure 3. The next problem is that there are no base resistors.

Without base resistors on each transistor Q4 and Q6's base-emitter junctions act as diodes connected to GND. These will clamp the D1 and D2 voltages to 0.7 V preventing Q3 and Q5 from ever turning on. D1 and D2 will only be able to switch between 0 V and 0.7 V and whatever is driving them could possibly be damaged due to the high current drawn through the base-emitter junctions.

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That's because NPN would not work. Try simulating the circuit in a simulator.

PNP is used in the high side drive because it can switch the output properly to VCC, just like NPN is used in the low side drive because it can switch the output properly to GND.

The Vbe is 0.7V, so if you try to set D1 high, Q6 will clamp Vb it to 0.7V, and because Q3 also has 0.7V Vbe, Phase1 is 0V.

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To activate Q3 and Q5 to turn Phase high (connect it with VCC), you need to send a current from base to emitter by pulling the base low. That won't work (at least not as intended) when the emitter is connected to Phase. The only option is to use a PNP transistor.

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