Zap a volt regulator w/ no cap?? 3

[eromlignod]

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

 

[geekEE]

Well, without the caps on the regulator, I've seen them work fine, output a voltage that's too high, or oscillate wildly. You definitely should put them on. And if the data sheet has any specific recommendations on the value or type, you should follow it.

Also, I don't know how much current your circuit draws, but make sure that you calculate the power dissipated by the regulator and make sure that you have adequate heatsinking if needed.

 

[OperaHouse]

It is often a good idea to put a resistor between the supply voltage and the input pin of the regulator. This will provide a little electrical isolation that will make the input cap more effective on voltage spikes. It will also lower the voltage to the regulator and let it run cooler. Select a value that gives you at least 5V more than the 12V needed for the sensor. Often a 24V supply is mich higher than the stated voltage.

 

[Skogsgurra]

There are so many small switched power supplies available and at a very low cost that it is hardly worth the time doing anything yourself nowadays. Especially if you are not used to it.

Look at so called wall warts with a regulated 12 V output. They won't cost you more than 20 - 30 dollars. And they are quite good and very reliable.

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[itsmoked]

You are using a low head regulator that you do not need.

They absolutely require the caps - no if or maybe.

It did oscillate.

It will fry.

Non-'low head' regulators are less touchy and are less expensive.

I do not see how you add a voltage regulator piece-meal to a PLC in any kind of a clean and robust manner.

I would look for small DIN RAIL power supplies with all the Certs and hence some pedigree.

Something like these: (Note: there are a huge number of different makers and styles)

http://www.sola-hevi-duty.com/products/powersupplies/pdfs/scl.pdf

They will all cost far far less than $175!!!





Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[ScottyUK]

Just a thought: you did heatsink the regulator didn't you? If your circuit draws more than a few tens of mA it is going to get very hot very quickly dropping 12V across it.


----------------------------------

Sometimes I only open my mouth to swap feet...

 

[eromlignod]

The voltage regulator in the old embedded circuit never had a heat sink, so I didn't think it was necessary.

Don
Kansas City

 

[Skogsgurra]

Dropping 12 V? Or some lower value? Heat produced is proportional to voltage dropped.

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[ScottyUK]

Hi Don,

Linear regulators dissipate heat according to P = V.I where I is the load current and V is the drop across the regulator, i.e. the differential between input voltage and output voltage. With a high input-output differential you'll see from P = V.I that you can quickly get to a fairly high dissipation even with light load currents far below rated current of the regulator. A TO-220 package can't dissipate much power without a heatsink, probably less than 1W before it reaches a damaging temperature.

----------------------------------

Sometimes I only open my mouth to swap feet...