Path: utzoo!utgpu!news-server.csri.toronto.edu!cs.utexas.edu!usc!elroy.jpl.nasa.gov!sdd.hp.com!hplabs!hpfcso!hpfcdj!myers From: myers@hpfcdj.HP.COM (Bob Myers) Newsgroups: sci.electronics Subject: Re: Re: 12 to 9 volts dc (kinda long) Message-ID: <17660154@hpfcdj.HP.COM> Date: 26 Feb 91 19:52:48 GMT References: Organization: Hewlett Packard -- Fort Collins, CO Lines: 81 >The circuit that you're thinking of is this (Very) basic regulator, >An Emitter follower or Current amplifier... This one uses a PNP, other >variations can be used. > > Before anyone builds the circuit shown in the referenced article, please note that it is incorrect - it WILL NOT work if built as shown. There are two problems (or one, depending on how you look at it); the transistor is backwards (C for E) and the resistor is in the wrong place. A corrected version is as follows: + 13V in -------------\ /------------------> +9V out | C \ / E \ --------- / | B NPN power transistor \ R | / | | | ---------- | ----/ / /\ / \ 9.7V (or 9.6V) Zener diode ---- | | GND -----------------------------------------> GND Here's how it works: The regulation of the output voltage depends on the fact that the B-E drop of a silicon transistor is a *fairly* constant 0.6-0.7V, depending on the current through this junction. Thus, if the base is held to a constant voltage (which it is here by the Zener), the emitter will be this voltage minus the B-E drop. So, with a 9.6-9.7V Zener, we should get about 9V out at the emitter. The "lost" voltage (the difference between the input and output voltages) is not "dumped by the Zener, but appears across the transistor C-E. This is one drawback of this type of regulator - the voltage across the transistor (which will increase as the input voltage increases), multiplied by the output current, represents power which is doing nothing more than heating up the transistor. In other words, a linear regulator (which this is) isn't particularly efficient. (And note that you MUST have some minimum drop across the transistor - at a bare minimum, about 1.4V, and more reasonably 2V - to make sure that the transistor is biased properly.) The resistor R supplies both the Zener current and the base current of the transistor. The Zener current needs to be enough such that the Zener diode is sufficiently past the "knee" in its characteristic curve such that the voltage across the diode is reasonably stable; the base current required is determined by the maximum required output current and the minimum expected beta of the transistor. A Zener current of a few tens of mA at most is usually sufficient; if we say we want 20 mA through the Zener, and expect 1A load current with a transistor beta of 20 (meaning 50 mA base current), then the resistor must supply 70 mA with a drop of (13.6 - 9.7), or 3.9V. This calls for a resistor of 3.9V/70 mA = 55.7 ohms *maximum*; we'd probably wind up with a 51 ohm part here (nearest standard value). (Note that these numbers are examples only - use you own Zener requirements and transisor beta, etc., to get the right resistor for your application.) Also, please note that using a value slightly under the calculated maximum simply means that more current will be available to the Zener/base junction - and what isn't taken by the base will go through the Zener. Check to make sure that you won't have *too much* Zener current under any possible load/transistor combination - if you're in trouble here, go to a heftier Zener. The power rating of the resistor should be selected based on its expected power dissipation under the worst-case conditions (use at least the next larger standard rating). It's also a good idea to add some additional filtering to a regulator like this, in the form of capacitance at the input and output and across the Zener. If a capacitor across the output is used, a diode connected backwards across the transistor may be a good idea as well - to prevent any possible damage from trying to "run the regulator backwards" when the input power is shut down (or any other situation arises where the "output" voltage is higher than the "input." Bob Myers KC0EW HP Graphics Tech. Div.| Opinions expressed here are not Ft. Collins, Colorado | those of my employer or any other myers@fc.hp.com | sentient life-form on this planet.