Path: utzoo!utgpu!news-server.csri.toronto.edu!cs.utexas.edu!sun-barr!olivea!uunet!optilink!elliott From: elliott@optilink.UUCP (Paul Elliott x225) Newsgroups: sci.electronics Subject: Re: WAIT! - Re: 12 to 9 volts dc (kinda long) Summary: non-standard output voltage regulators Message-ID: <5825@optilink.UUCP> Date: 28 Feb 91 20:33:11 GMT References: <1991Feb10.220719.17078@cbnewsl.att.com> <1991Feb26.235009.1389@cbnewsj.att.com> Organization: Optilink Corporation, Petaluma, CA Lines: 70 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