All is well, except I'm getting -21v from the output of the 7912 voltage regulator instead of -12v. The +12, and +/-9 outputs are all fine. I tried removing the 7909, then the 7809. I was still getting -21v from the 7912. I then tried the 7909 in the 7912 slot, and I got -9v as expected, so the circuit seems sound. I have tried substituting the 7912 twice. Same result. The AC power supply is 9V, 500mA. I'm measuring the voltage going into the positive voltage regulators as 30v, and -30 into the negative voltage regulators. Could I have a bad batch of 7912s, or can anyone suggest something else I might be doing wrong?
Thanks!
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P
Phil Allison
"beatbox"
** Suspect your 7912s are mislabelled or fakes.
May really be 7812s or just some cheap BJT.
....... Phil
T
Tom Biasi
Where are you getting the 30 volts with 9 volts in? Check the orientation of the diodes.
Tom
F
Falko Rudolph
Hi,
The 7912 needs a minimum load, then it works fine. If you understand german you can read the thread
"7912 braucht Mindestlast?"
in d.s.e. There the same problem is discussed.
HTH Falko Rudolph
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Phil Allison
"Falko Rudolph" The 7912 needs a minimum load, then it works fine.
** TOTAL BOLLOCKS !!!
** Mein Kampf is damn good page turner too.
....... Phil
M
mike
B
beatbox
I'm posting from a couple of accounts here, just to confuse things ;) My (very basic) question, after reading that thread, is how might I change the load in my circuit to see if that has any effect? Thanks.
H
Hammy
If you're saying you want to test the PSU's response to a fast load transient? Use a Mosfet to switch from no load to full load. If you don't have a function generator to generate your gate pulses just use a pwm like a 555 timer and a driver IC.
H
Hammy
Sorry here is a schematic.
formatting link
I just did this a couple of days ago so you can see a scope image here.
formatting link
A 7808 going from no load to 1 amp (2.3uS pulse per 26 uS).Channel 1 (yellow) is the 7808 output voltage dropping by 1.5V. Channel two is the gate signal to a logic FET the driver gets its Vcc from the output of the 7808.Thats why the peak gate voltage drops.
P
Phil Allison
"beatbox"
** No load is needed on a 7912 to get -12 volts out.
........ Phil
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Falko Rudolph
schrieb im Newsbeitrag news: snipped-for-privacy@4ax.com...
Just add a Z-Diode >12V from Output to ground. When they are right in the german thread this will create a load current whenever the output voltage is too high. If it works after that you have found the problem. IIRC the input/output capacitors had an influence, too.
Gern geschehen.
Falko
D
Donald Watland
Hi,
If you were to start with an input of 9Vac rms, in theory you would have an AC input signal that was about 12.7 Vac, peak-to-peak, which you want to rectify to a useable source for your +/- DC inputs to the voltage regulator chips. By the time you subtract off the forward bias drop across the diodes, and the voltage lag from the numerous capacitors in the circuit, you may end up with an input voltage 10 or 11 Vdc (+/-) , maybe .... For the 7912 regulator, you are supposed to have an input voltage is between (-14.5Vdc) to (-30Vdc), which may even change to (-16Vdc) to (-22Vdc), depending on the how the load effects the operating temperature. Now the dropout voltage is
2V, which means that you need keep the input voltage at least 2V greater than the expected -12Vdc output, or the chip will not conduct. You should check to see that you really are supplying -13Vdc (or a value more negative) to pin 2 of the 7912, and the ground (0V) on pin 1. If you don't have at least a negative 14 volt differential between these two pins, then the regulator will stop conducting, and I'm not sure what you would then read as your output voltage, but if the input voltage should read something close to -9Vdc, then the -21Vdc you have read on the output seems suspiciously close to the sum o (-9Vdc actual input) + (-12Vdc expected output), with no output current to available... I got cross-eyed trying to follow the leads on the PC drawing. An equivalent schematic would have been easier, or quicker to digest. Plus, I'm not sure what you meant by the statement: "I'm measuring the voltage going into the positive voltage regulators as 30v, and -30 into the negative voltage regulators." If all you had for your input voltages were the rectified DC voltages you got from a 9Vac source, you'd have a pretty hard time coming up with either a plus or minus 30Vdc as an input to anything in that circuit. If you were lucky enought to squeeze out 11-12Vdc (plus and minus) after rectifying the AC input, that would probably be enough over the 2Vdc cutout voltage for everything up through the 9Vdc outputs (plus and minus), but may not have been enough to stay above the extra 2Vdc needed to allow the 12Vdc output to stay active. The absence (or presence) of a load can also effect how much "extra" input voltage is needed to maintain the expected output voltage.
I hope this helps. Maybe it's all smoke.
-DW
"beatbox" wrote in message news: snipped-for-privacy@g4g2000hsf.googlegroups.com...
H
Hammy
Agreed.
The circuit is a voltage multiplier thats how he's getting the 30V. ;)
>
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beatbox
Hi, Thanks for that useful explanation. I'm learning as I go along here. There's a schematic here:
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The voltage differential I'm getting between the input pin and the ground pin of the 7912 is -30v. The voltage between the input pin of the 7812 and its ground pin is
30v. The 7812 outputs 12v quite happily. Would that fit with your theory? (The same voltages are measured between the inputs and grounds of the
7809 and 7909 by the way, both of which are working OK.) If you could explain what a "load" means in this context that would be really useful. (Is that basic enough for a "basics" group?) I'm not sure where to start looking for a fix, but I have started by ordering some new 7912s by a different manufacturer. Will report back on that...
B
beatbox
Newsbeitragnews: snipped-for-privacy@4ax.com...
Thanks, I'll try it. Mike
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Brendan Gillatt
I don't want to insult your intelligence but are you positive you're measuring from the ground to the 1912s? If you happen to measure from
+12 to -9 (or +9 to -12) then you'd get your odd 21volts. check for shorts between the regulators and the ground rail.
Hi, Yes, sure as can be. I'm testing from the designated ground and output points in the circuit, as well as from the regulator pins to double check. Mike
B
beatbox
I replaced the 7912 with one from a different manufacturer. It worked. So I got there in the end, though my nice PCB is now in a mess from all that soldering and desoldering. Thanks all.
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