Why NPN BJT Pass Element Drops More Than PNP

[likephysics]

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.

 

[berkeman]

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?

 

[likephysics]

Kinda makes sense. But not really. Do you have an alternate explanation?

 

[berkeman]

Kinda makes sense. But not really. Do you have an alternate explanation?

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...

 

[cabraham]

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