LM320T-12 oddity - Page 1
Forum section: electronics / repair

author: clay1905
author_rank: Regular Contributor
timestamp: May 10, 2020, 12:08:33 pm
content: Hi,A while ago I began to look at a bench power supply. It has a selection of faults, and the first I think I should tackle is the voltage regulator in the power supply for the internal amplifiers etc of the unit.The supply in question consists of a current source (transformer secondary) a full wave rectifier (4 silicon diodes), and thence two regulators in a kind of mirror image arrangement. One regulator is a LM317, the other a LM320. The output should then be +15 volts from the LM317, and -15V from the LM320 about a common 0V ground.The usual divider networks are used to set the output voltages of the regulators, and in this case for the LM320 includes a 820R fixed value resistor and a 1K trimpot. This network *should* be able to be set so -15VDC is output from the regulator, but the 1K trimpot can't seem to quite make it happen. The highest it can achieve is -17.97V.My arithmetic tells me that if the trimpot is set fairly close to centre, then the regulator ought to be outputting close to -15 Volts so something drifted significantly. The thing is, I can't seem to be able to determine what. The resistors in the circuit all measure well within tolerance. I have tried swapping the regulator for another. Input voltages are within acceptable ranges.Attached is a schematic of the circuit. Appended are expected voltages, and actual in (brackets).Suggestions for why the voltage cannot be brought up to -15 Volts would be welcome.

author: PKTKS
author_rank: Super Contributor
timestamp: May 10, 2020, 01:21:57 pm
content: Weird combination.IMHO? Get rid fo this 320 and use a regular 337 instead.I would also get totally rid of those bypass 330 resistors.Bad design. I would not even try it, just replace the 320for a good 337Paul

author: David Hess
author_rank: Super Contributor
timestamp: May 10, 2020, 01:31:37 pm
content: The 330 ohm resistor across the LM320 is pulling the output more negative when insufficient load is present.As PKTKS points out, the LM320 is not suited for floating regulator applications although it can be made to work. Its common pin current is relatively high so the divider needs to be lower impedance for good regulation. If you insist on using the LM320, then replacing the divider with a 10 volt reference (zener, TL431, whatever) or driving it with an operational amplifier as a tracking regulator would be be better.

author: clay1905
author_rank: Regular Contributor
timestamp: May 12, 2020, 01:14:29 am
content: Thanks guys,I think I should have said that the bench supply is over 40 years old now. I should also add that I'm not the most knowledgeable with solid state devices.It's interesting that the 320 was originally selected rather than the device that was designed to compliment the 317. And I wonder a bit what was in the designing engineer's mind.But first off, I think I'll see if I can find a bit more of what's going on. David's comment about there being insufficient load to get the regulator working as intended looks like a big hint.As I mentioned, there are a number of faults in this unit. It would seem that one of these faults is a load that's disappeared, allowing the LM320 with the 320R to go over voltage. Perhaps this is a better place to be starting, figuring out what circuit(s) in the supply unit should be constantly drawing some power.Part of me is very unwilling to make changes to circuits that worked well for a long time and then started doing something new. I'd prefer to understand what's gone awry before doctoring the circuit to address symptoms rather than causes.Thanks again,Clay.

author: TheMG
author_rank: Frequent Contributor
timestamp: May 12, 2020, 01:38:57 am
content: Unless I'm missing something, the only reason I can really think of for the resistor between the input and output of the regulators would be to reduce the power dissipated in the regulator. As long as the regulators have sufficient cooling and are rated to dissipate the power, the resistors wouldn't be necessary.

author: clay1905
author_rank: Regular Contributor
timestamp: May 12, 2020, 02:29:16 am
content: I don't know what they're for either.In some cases I have seen a diode in the same place. My understanding is that when a diode is in this place, it's there to bypass EMF that may be generated by switching off an inductive load.That doesn't go far in explaining the purpose of these resistors. Theymightbe for discharging the two 4700uF capacitors though. That's about all I have on them.Clay.