Regulator 5v

194

04:18:29

published 12 years ago

thebugger

12 years ago

The zener should be at the base of the transistor. And you should replace the 100k resistor with a 22uf cap for better transient response. While you're at it replace the 9v source with a rectifier to better represent the given circuit. You might make it more load responsive by adding another transistor in a darlington configuration. Also a good operating point for the zener is 10-20mA so you might need to decrese the base resistor from 2,2k to 200-500R

Bintoro

12 years ago

Ok,thanks for the advise.

Voltransistor

12 years ago

Nice, to be efficient, replace the 500 ohm resistor with a 10k ohm resistor, put a PNP transistor from source to load. The NPN transistor no longer connects to the source, the base, instead. Base connects to the emitter of the NPN transistor, Emitter connects to the output, and Collector connects to the source. Darlington's amplification is more efficient

thebugger

12 years ago

Yeah I forgot to mention. An npn transistor reguires a 0.7v to enable itself. So a darlington npn configuration means 1.4v drop. This effect does not display in the pnp transistors so a little more efficient-maybe. But everything else remains the same ;)

rbrtkurtz

12 years ago

Agreed. Though to be clear, using either both PNPs or both NPNs is Darlington. A PNP and an NPN is a Complimentary Darlington, or Sziklai Configuration. I don't think a standard Darlington would help out to much here. Anyhow, setup a Sziklai Configuration, and you should be good to go. Also, personally, I would use a 5.6v zener to compensate for the voltage drop at the BJTs, maybe even a 6.2 for higher power applications.

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