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I am studying the Art of Electronics and trying to apply what I have learned to a motor driver circuit from an MITx intro to control systems course (https://www.edx.org/course/introduction-control-system-design-first-mitx-6-302-0x). I have a good understanding of the behavior when the driving NPN transistor is ON (pwmIn is 5V) but I am having trouble understanding the reason why the voltage at the base of the Power PNP is 5V (demonstrated on the scope) when the NPN transistor is cut off (pwmIn = 0). Most of the circuits in AoE had well defined DC bias voltages at the transistor base and in this case it seems the base is "floating."

Can someone please explain how best to understand the base voltage of the PNP when there is no base current.

enter image description here

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  • \$\begingroup\$ The voltage won't be 5 volts when the NPN turns off - maybe 4.6 volts but that makes a big difference. \$\endgroup\$
    – Andy aka
    May 26, 2020 at 13:59
  • \$\begingroup\$ Good catch. That schematic is broken. It may by happenstance work thanks to leakage in the PNP but its fragile. \$\endgroup\$
    – user16324
    May 26, 2020 at 13:59

1 Answer 1

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If the NPN transistor is cut off, then it doesn't allow any current to flow into its collector.

Therefore there is no current flowing through the base-emitter junction of the PNP transistor.

But the I-V relationship of the b-e junction is similar to that of a diode. The only way current is zero is when the voltage is also zero.

So the b-e voltage must also be zero in this case.

Another way to look at it is the b-e junction acts as a non-linear resistor, which is able to pull up the base voltage to equal the emitter voltage if no current is flowing through the junction.

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