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7805 / 7905 Power supply problem 1

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walker1

walker1

Industrial
Joined
Dec 27, 2001
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117
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DK
We have eight IF boards in our equipment. Most of the IF part run on +15V.
The last demodulator (controlled by a shift register running at 40 MHz) runs on +5V, and the ADCs run on +/- 5V.
The analog filter on the ADC inputs runs on +/- 15V.

The principle in the power supply is shown in the attached schematic.

Now for the problem part.
Every now and then, when we turn the +/- 15V on, the 18 Ohm resistor (R1, surface mount) burns! (usually jumps to 3-6 kOhm or so)
At normal operation the voltage drop over it is about 2V, ie. 110 mA or so, which is as expected.

At 'burn time', that current has been measured in the 600 mA area! A quick turn off usually saves the resistor. (max. 1 sec!)

We have tried separate + and - 15V supplies, and turning them up individually, in different order and slowly.
Something inside one or both the 7805s or in the load still pulls a large amount of current from time to time.

The questions are what and why?

All regulators are from Motorola in standard TO-220 housing.
There are no protecting diodes. Not across the output and not backwards across the regulator.

The designer of the board has long retired, unfortunately.
 
 https://files.engineering.com/getfile.aspx?folder=cf99f394-acf9-4219-8da5-9263b7102104&file=Kanal-PS.gif
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Why do you have resistors inline with the power supplies in the first place?

I would suspect the equipment attached to the 7805's pulling more current before I suspected the 7805's themselves... slow slew rates can do bad things for a lot of logic, including power shoot-through. You have a resistor followed by a cap, which will provide a longer-than-necessary slew rate.

Dan - Owner
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Probably too easy but have you checked that the polarity of the capacitors are correct?
 
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I don't understand why you need two 7805's? I would think a larger heatsink would be in order, but one would do. Also the resistor likely is not needed, and that maybe the issue, is on start up the resistor is limiting the current to charge the capacitor, and power the circuit at the same time.

You can always try a zener across the resistor to limit the voltage drop, if that is the problem.
 
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"There are no protecting diodes. Not across the output and not backwards across the regulator. The designer of the board has long retired, unfortunately. "

If the output of a positive voltage regulator gets pulled below ground because a negative supply (the LM7905) starts up first, it can cause the positive regulator to latch up and basically look like a short circuit. I quick look at an LM78MXX regulator (dated 2005) shows no mention of this, so it may be a concern. They typical fix is to put a Schottky diode between the 7805 output (cathode) and ground (anode), so the output cannot get dragged more than 0.3V below ground.

If your purchasing department recently changed vendors for the 7805 or 7905, that can easily cause this type of issue.


In your schematic, the LM7905 capacitors are shown with improper polarity (LM7905 output is -5V, not +5V).

Z
 
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To add to zappedagain suggestions (he addressed the most likely issues). You have one input cap shown for both LM7805 regulators. I would make sure you have two. And follow the directions in making sure each input cap is close to the LM7805 it is intended for. LM7805 do need that cap for stability. Since you indicated TO-220 packages, I would just solder a 0805 or 1206 SMT package size between the Input and Ground leads 1 & 2. Spacing should be just about right.
 
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