🚀 go-pugleaf

In article <1991Feb26.235009.1389@cbnewsj.att.com>, asd@cbnewsj.att.com (Adam S. Denton) writes:
> DON'T use the posted circuit -- you will cause Discman Frying.

I knew someone would follow up on this -- good treatment!

> Another solution is to use a 7809 if you can find one, or a
> 7805 with a 3V Zener (or 3-4 series-connected diodes) in series
> with the ground lead.  If you go the latter route you will need
> capacitors at the input, output, and ground terminals of the 7805
> to GROUND in order to ensure stability.

I understand that it is *not* good practice to boost the output of
the 78XX-series regulators by putting diodes (Zener or otherwise)
in the ground lead.  Something to do with the short-circuit protection
(I haven't analysed it myself).

I prefer the LM317-series of regulators.  These let you set the
output Voltage with two resistors and are more stable than the 78XX
regulators.  If you *must* use the 78XX regulators in boosted-
output Voltage applications, here's what I do:

You can use a series resistor from the ground pin to circuit-ground, 
or better a resistor from output to the ground pin, and another from 
the ground pin to ground:

           7805
        ------------
in o----|in     out|-----+-----o out
        |   gnd    |     |
        -----+------     R1
             |           |
             +-----------+
               Iq ->     |
                         R2
                         |
                       -----
                        ---
                         -

The regulator quiescent current (Iq) flows from the gnd pin, through R2
to ground, and is typically 4.75 mA in the 7805. 

The Voltage across R1 will be 5V (for a 5V regulator), and the current 
through it will be 5V/R1.

Set the value of R1 so the current through it is about 5 * Iq, to provide
for some swamping of Iq variations.  This gives a R1 value of:
R = E / I, R = 5V / 23.75mA, R1 = 210 Ohms 
(approx, use 220 for I(R1) = 22.7mA

Choose R2 so that the current through it (Iq + I(R1)) gives the Voltage
drop required to boost the regulator output to the desired value:
Vout = 9V (in this example)
Vreg = 5V (for the 7805)
Vboost = 4V
I(R2) = Iq + I(R1), I(R2) = 4.75mA + 22.7mA, I(R2) = 27.5mA (approx)

R2 = Vboost / I(R2), R2 = 4V / 27.5mA, R2 = 145 Ohm (approx)

If you use a 120 Ohm resistor for R2, you get:
Vout = Vreg + (R1 * I(R2)), Vout = 5V + 3.3V, Vout = 8.3V

Power dissipation in R1 is about 0.114W, in R2 is about 0.133W

All in all, the LM317 is a better bet, though.

-- 
      Paul M. Elliott      Optilink Corporation     (707) 795-9444
            {uunet, pyramid, pixar, tekbspa}!optilink!elliott
 "If I had known it was harmless I would have killed it myself." - P.K. Dick