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Zap a volt regulator w/ no cap??

Hi guys:

First of all, forgive me for being a dumb ME meddling in electrical work.

I have an application in a machine that has a 24-volt power supply. I have a tilt sensor (analog output) that needs 12 Vdc as its input supply. Until now this part of the machine used an embedded circuit that the sensor mounted directly to and that provided its own 12 V.

I converted this part of the machine's control to be read from a PLC analog input module, so I had to come up with my own 12 V source. What I did was add an LM2940CT-12 voltage regulator (TO-220 package) to the circuit right before the analog sensor. This provided my 12V and worked great for about a day and a half, then my analog sensor fried. When I measured the voltage from the regulator I found that it was now the full 24V input...so apparently the regulator failed first and then cooked the sensor with 24 V.

When I took a look at the spec. sheet on the LM2940, I see that they recommend putting capacitors from the input and the output to ground...I didn't do this. Other than this mistake, I can't see anything else wrong with the circuit. Could the absence of these caps have caused the regulator to fry? If not, what else could have caused it? I'd like to know exactly what my problem is before I toast another sensor (they're $175 a pop).

Thanks for any advice (or chastisement) that you can provide.

Don Kansas City

Reply to
eromlignod
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eromlignod wrote: (snip)

Low drop out regulators (of which the LM2940 is one) are only stable with certain values of capacitance (and sometimes only with certain values of series resistance in that capacitance). The graph on page 11 shows how tricky this is.

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Since you have lots of excess voltage, a follower output type linear regulator like an LM7812 would be less problematical. It still works best with a small capacitor from both input and output to ground, close to the regulator (say, .1 uF) but the values and series resistance are not at all critical, compared to the LM2940. See page 22 of:

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Reply to
John Popelish

"Works best" is a mild way of saying "leave them off at your peril". I stopped in the middle of build a circuit with an embedded 78L05 regulator; when powered up without an input cap it oscillated at 80MHz.

You should also check on power dissipation. These regulators are supposed to have thermal shutdowns, but that doesn't mean they work right, and you don't want a shutdown event anyway. Check the input current to your 12V module; you'll be dissipating that times 12V (24V -

12V), and more if your 24V supply goes higher.

Make sure you have enough heat sinking to keep the regulator cool. The best rule of thumb I know is to run it for a while then put your thumb on it. If you pull back and say "ouch" then the heat sink isn't dissipating enough.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

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"Applied Control Theory for Embedded Systems" came out in April.
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Reply to
Tim Wescott

Tim's caution about heat sinking should be heeded; I have worked with some industrial sensor modules (OPTO-22) and they can draw significant amounts of current. In your case, that would give a regulator power dissipation equal to the sensor (12V x load current) and even a TO-220 without a heat sink can't handle too much if it's within a box (so ambient is pretty high).

Do you know your load current?

Cheers

PeteS

Reply to
PeteS

I've always used the caps, and never had a regulator problem. But, being obsessive/compulsive about capacitation, I soldered the caps right to the regulator leads right where the leads neck down.

Cheers! Rich

Reply to
Rich Grise

Yes. All regulators should have input and output caps, but "Low dropout" regulators like the LM2940 in particular are very touchy and require an output capacitance of a certain type and value to stop it oscillating, and hence stop the vital smoke from escaping the case.

Unless you have a very low input voltage (

Reply to
David L. Jones

I can't speak for all thermal shutdown schemes, but on the parts I designed, the shutdown was not very intelligent. That is, it would sense the die temperature, shut down, then cool due to the shutdown, power up, and of course get hot again enough to trip the shutdown circuit. You could make an oscillator using the shutdown. As part of the characterization of the part, a diode on the chip would be characterized over temperature, i.e. with the part in an oven and no load so the die temp equals ambient. Then if you short the output, you can monitor this diode to see the temperature trip points.

There may be smarter regulators that stay shut down once tripped. It is a marketing issue since the smart design would need a power on reset from the input voltage. Not impossible, but POR circuits can be fooled.

Reply to
miso

Guess it's the 00882206 diagram "Output Capacitor ESR".

Makes me wonder if these LDO really are good. Seems more like liability :)

Reply to
pbdelete

(snip)

I have used them successfully, but I always use a low esr cap and an external series resistor.

Reply to
John Popelish

They are essential when you actually need real LDO operation! e.g. dropping say 5V to 4.5V with low noise.

Dave :)

Reply to
David L. Jones

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