Re: what's wrong with this circuit?


"Jeff Liebermann" <jeffl@xxxxxxxxxxxxxxxxxxxxxx> wrote in message
news:v3mp521a26ijqvrfdg54iivatkb61u52t3@xxxxxxxxxx
"Peter" <pmhughes@xxxxxxxxxxxxxxxxxxxx> hath wroth:

http://members.ozemail.com.au/~pebarhug/radios/2N3055schem.jpg
A fellow hobbyist told me it was not a good circuit arrangement as it had
poor regulation with the NPN pass transistors. Can someone explain to me
why?

Argh. That's an awful design. The lack of voltage ratings on the
output, xformer, and cazapitors should be a clue.

1. The output voltage isn't regulated. The feedback stops at the
base of the 2N3055 xsistors. The regulation is further mangled by
having 3ea 22 ohm resistors in series with the output. The voltage
drop across these resistors will vary with the load current, which is
not exactly my idea of voltage regulation. Ummm, doing the math:
E = I * R = 5A * 7 ohms = 35 volts
The 7 ohms is from 3ea 22 ohms effectively in parallel. In order for
this piece of junk to output any voltage at 5A, the regulator input
voltage will need to be 35 volts plus the regulators dropout voltage
and Veb. It won't at all work with the 22 ohm resistors.

2. In normal operation, the 22 ohm resistors will cook. For example,
if this power supply really does mangage to output 5A then the
dissipation in the resistors will be:
P = I^2 * R = 5^2 * 7 = 175 watts.
Not even close.

3. The output isn't current limited or protected. Short the output
to ground and you get 1A (output current limit of the 7805) through
the base-emitter junction of the 2N3055. The usual value for such
emitter follower load equalizing resistors is prehaps 0.22 ohms. Were
those used in this design, the current would smoke the EB junction on
of the 2n3055's.

4. The diodes listed are all plastic parts that will get rather hot
near their current limits. For anything over about 8 Amps, one really
should use diodes with heat sinks.

5. Using a full wave center tap xformer means that the xformer will
be twice as large as a smaller xformer with a diode bridge.

Start over.



There is a decimal point on the schematic. That's zero point two two ohms.
That makes the Rt 0.07R. That makes the voltage drop across them, at 5 amps,
0.35v , not 35v ... How do you arrive at a maximum output of 1A ? Whilst the
7805 can deliver a maximum output of 1 amp, this figure is multiplied by the
current gain of the 2N3055's. Assuming a ( poor ) gain of 20 on them, that
would result in an output current of some 20 amps with 1 amp of drive to
their bases. That is the whole point of having external series pass
transistors. The output current is drawn from the collector-emitter circuit,
not the base-emitter circuit, so there is no reason why the base-emitter
junctions should fry.

The output is current limited and protected by the fuse feeding the series
pass transistors' collectors.Admittedly, this is not very elegant, but it is
protection, no matter which way you look at it.

It doesn't make any difference to the size of the transformer, if you have
one 20v 20A winding, or two 20v 10A windings, series'd and grounded at the
junction. It's still 400vA either way

I agree that this is not a *good* design, and will suffer from poor dynamic
regulation due to the current-dependant drop across the current sharing
resistors, but it is at least functional, and a simple useable design to
produce an adjustable, reasonably high current output. Depending on what it
is needed for, it might be quite adequate, and its shortcomings, of little
or no consequence.

Arfa


.

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