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Hello guys this question is regarding a gap in my knowledge regarding transistors specifically the driver of the totem pole configuration. Based on the following figureTotal pole driver

And these results here: enter image description here When the control voltage goes high why does the base of Q2/Collector of Q1 rise to 0.8V, and why does the base of q1 also rise by that same value?

schematic

simulate this circuit – Schematic created using CircuitLab

Partial Answer: The answer to this question I believe has to do with the fact that Q1 is in reverse bias configuration.

After looking at this resource: https://learn.sparkfun.com/tutorials/transistors/operation-modes.

The Emitter voltage > Base voltage > collector voltage. I will have to look into this further but, I didn't realise the value of this configuration as I barely covered it in my engineering classes. If I find out more I will add further to this answer, but please give a more in-depth explanation as it still isn't intuitive to me why the collector voltage increases.

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  • \$\begingroup\$ 'driver' for the totem pole? More like the logic gate section. \$\endgroup\$
    – Kartman
    Jul 27, 2021 at 11:49
  • \$\begingroup\$ Edited the question, and that is technically a driver as it is selecting what the state of the bjt should be. \$\endgroup\$
    – Raj Nar
    Jul 27, 2021 at 11:53
  • \$\begingroup\$ what about the 1K resistor on the emitter of Q2? \$\endgroup\$
    – Kartman
    Jul 27, 2021 at 12:04
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    \$\begingroup\$ It doesn't change much regarding the operation of the circuit, it raises the voltage at the gate proportional to the amount of current flowing through it, but it doesn't change the question as to why the collector of q1 goes to the Vbe saturation voltage of q2 when the control voltage is at 5v. \$\endgroup\$
    – Raj Nar
    Jul 27, 2021 at 12:08
  • \$\begingroup\$ Where's the totem pole driver? \$\endgroup\$
    – Andy aka
    Jul 27, 2021 at 12:08

1 Answer 1

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When the control voltage goes high why does the base of Q2/Collector of Q1 rise to 0.8V, and why does the base of q1 also rise by that same value?

Because the is a pn junction between the base of Q1 and the input (essentially a diode), the voltage at the base of Q1 will never be more than somewhere around 0.6-0.7V above the input. For a similar reason, the collector of Q1 and the base of Q2 will never be more than 0.6-0.7V below the base of Q1.

This means that when the emitter of Q1 is low, the base of Q2 will also be low.

When the emitter of Q1 is high, the base of Q1 will rise to about 1.2-1.4V. There will be one diode drop between the base and collector of Q1, and one diode drop between the base and emitter of Q2.

If the circuit seems strange at first, it is probably because when the input is high, the base-emitter junction of Q1 is reverse biased. That is not normal for CE, CB or CC amplifiers. But this transistor in not being used in as a normal CE, CB or CC amplifier. It's function is basically that of two diodes. However it has an advantage over two discrete diodes. In discrete diodes, when current switches from one diode to the other, there is a longer reverse recovery time. Minority carriers in the diode need to be cleared. With the transistor used in place of two diodes, carriers in the base do not need to be cleared. They simply change direction regarding whether they are flowing to/from the emitter to flowing to/from the collector. Hence using a transistor this way allows faster switching than using two discrete diodes. (Hence TTL or transistors transistor logic, replaced DTL, or diodes transistor logic.)

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  • \$\begingroup\$ Thanks man, I forgot that the voltage could act in both directions... Pretty embarrassing since I am about to graduate with a degree in electrical engineering and I remember doing a course on this distinctly. I guess that's what happens when you don't give your full attention in class... \$\endgroup\$
    – Raj Nar
    Jul 27, 2021 at 12:37

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