EH bi-polar from wallwart power supply. Any good?

[Dan N]

Seems I have seen other means of getting +/- from wallwarts here, but "search" has failed me. How about this one from my EH Frequency Anal-yzer? Could I just change the zenor values for like +/- 10vdc? Any downsides in similar low current circuits?

Thanks!

[R.G.]

Looks fine for most cases. There are some limitations.

Zeners are shunt regulators. They only regulate in concert with a limiting resistance, that being the two 150 ohm resistors.

The first 150 ohm limits the voltage on the 100uF filter cap. This produces some pretty significant ripple, which presumably is filtered better by the second 150 and the two 470s. For the zeners to regulate at all, you have to have the voltage on the 100uF cap being more than 30V. The zeners eat current until the voltage across the second 150 increases to drop any excess voltage above 30V.

The 150s limit the total current that can go to the load at +/- 15V. If you have 40Vdc from the wart, then you lose 0.7V in the 1N4002 and 9.3V in the two 150s. That means the current that flows in the 150s is 9.3/300 = 0.031, or 31ma. That is what current flows through the combination of the load and the zeners. Under no load, the zeners eat it all. As load current comes up, the load steals current from the zeners until when load current is 31ma, the zener current is zero. Above that load current, the load voltage falls as the current through the 150s reduces the available voltage; regulation is lost.

What you can do to modify this supply:
Change the zener voltage: if you change the zeners to 10V units and make no other changes, then the 150s drop 20V, not 10V. The total current in the zeners under no load goes up to 62ma. The power in the zeners is the product of their voltage and current, so each zener dissipates 62ma * 10V = 620mW. You will have to have a 1W zener to do this. The next lower value, a 500mW zener, will burn out if you unload the power supply. The power in the two 150 ohm resistors is also 620mW, and so you need 1W resistors for these to live a long and happy life. You could restrict the curent by changing the value of the two dropping resistors so that only 500mW of power is dissipated in zeners and resistors. This would require the total current to be reduced to I = 0.5W/10V = 0.05A, and the resistor values become R = 10V/0.05A = 200 ohms. But now the total load current maximum is 50ma, not 62. That may be OK, depending on what you want to power with it.
Change the available current: ...oops, I just told you how to do that, out of sequence. Calculate the power dissipations carefully or you will burn parts up.

I don't like zener power regulators. IMHO they are useful only for tiny currents, not suitable for powering whole effects. It can be done, but it's not all that reliable.

I personally would stack two 15V three-terminal regulators. The peewee 100ma kind work OK. Hook some minimal load, like 2.2K to 4.7K across each one to keep ground currents from confusing the regulators. Now change the value of the "150ohm" resistors to allow 100ma to flow. For 30V, that's (40V - 30V)/0.1A = 100, so each resistor could be 51 ohms. For 20V (two 10s) it's (40V - 20V)/0.1A = 200 ohms, and each one is 100 ohms. The resistors now have to carry the 100ma current, so in the first case (30v) they have to be 51 ohms and 0.1A so P= 51*(0.1A*0.1A) = 0.51W - use a 1W. In the second case, they can still carry 100ma, but now they are twice as much resistance, so the power is 1.02W and you need 2W resistors.

Ooops. The regulators need 2V to regulate. So the resistors need to supply two more volts. The 30V case is (40-32V)/0.1A = 80 ohms, so use 43 ohm resistors, and the 20V case is (40-22)/0.1 = 180 ohms, each one should be 91 ohms. Use 1W and 2W still.

You can modify all of the parts. The resistors set the maximum current that can flow and still keep the voltage in regulation. The regulators set how much of the current the load gets. With the zeners, maximum dissipation for the zeners is at no load and the resistors burn constant power. For the linear regulators, there is essentially NO dissipation with no load, and the resistors are at maximum dissipation at max load. The linear regs are loafing all the time.

[Dan N]

Thank you very much for the detailed answer!