Why NPN BJT Pass Element Drops More Than PNP

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Discussion Overview

The discussion centers on the differences in voltage drop when using NPN and PNP BJTs as pass elements in series pass regulation. Participants explore the reasons behind the observed voltage drops, specifically comparing the VCE saturation and VBE characteristics of both transistor types.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants note that the NPN drop includes both VCEsat and VBE, while the PNP drop is primarily VCEsat.
  • One participant suggests that the biasing requirements for NPN and PNP transistors affect the voltage drop, with NPN requiring control circuitry to be above the base voltage, increasing the C-E voltage drop.
  • Another participant expresses confusion about the explanation and requests an alternate perspective.
  • A further contribution explains that in an NPN emitter follower configuration, the base drive must exceed the output voltage by one forward voltage drop, while a PNP common emitter stage can operate with a much lower VCE.

Areas of Agreement / Disagreement

Participants do not reach a consensus, as there are differing interpretations of the voltage drop mechanisms and the explanations provided do not fully satisfy all participants.

Contextual Notes

There are unresolved assumptions regarding the specific circuit configurations and biasing conditions that may influence the voltage drop characteristics of NPN and PNP BJTs.

likephysics
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When using BJT as a pass element in series pass regulation, why is the NPN drop more than the PNP.
Both have BE diodes. Why is the drop (VCEsat+VBE) in case of NPN and drop is equal to VCE sat in case of PNP.
Can someone also explain VCE sat using diode equivalent of BJTs.
 
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likephysics said:
When using BJT as a pass element in series pass regulation, why is the NPN drop more than the PNP.
Both have BE diodes. Why is the drop (VCEsat+VBE) in case of NPN and drop is equal to VCE sat in case of PNP.
Can someone also explain VCE sat using diode equivalent of BJTs.

It's not so much the PN drop in the BJT, it's where you have to bias the transistor in order to be able to control its current. For the PNP highside regulator topology (used in low-dropout applications), you can put the control circuitry "below" the base voltage. But for the traditional NPN highside regulator, you need your control and bias circuitry "above" the base, which adds to the C-E voltage drop.

Make sense?
 
Kinda makes sense. But not really. Do you have an alternate explanation?
 
likephysics said:
Kinda makes sense. But not really. Do you have an alternate explanation?

:smile:

Nope, that's the one and only. Have you looked at the schematic for typical NPN and low-dropout PNP regulator ICs? You should be able to see how the control circuitry dictates the minimum C-E bias voltage required. Maxim makes low-dropout regulators -- you could check their website for some datasheets...
 
With a positive input & output voltage, an npn pass element is an emitter follower. A pnp is a common emitter. In the ef, the base drive must be one b-e forward voltage drop above the output. Not the case w/ the ce stage. A ce stage, built w/ a pnp, can have a Vce of just a few tenths, or even a few hundredths of a volt. With non, at least 0.65 volts is needed due to Vbe forward drop.

Did I help?

Claude
 

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