0
\$\begingroup\$

I have a very simple board, two layers, two gnd polygons, 8 analog inputs incoming from 8 potentiometers and 8 analog outputs. In the PCB are 2 resistors (R1||R2) between input and ouput. So, output just will drive 8 voltage dividers.

Board source supply is 24V, but input signals needs to be fed under 20V, so I used a MC78M12G regulator (12V). As this board outputs will be connected to 8 ADC in a PLC block, 2 bypass capacitors were connected at regulator input and regulator output.

Capacitor components have not arrived yet and board were sold with the rest of components.

Design considerations talks about capacitors will not be needed if it's used in a low current application.

Well, taking this into account, in order to get some advantage, I thougt to do some little and simple tests:

  • only one active input (only one potentiometer is connected and its current is < 20mA),
  • without connecting the board to the next block
  • and then measuring its related output with a tester.

I thought the capacitor will be not needed -according to datasheet- and any signal freq will be passing through the circuit.

After switching on the board I can see how voltage output has unexpected low levels. I put the tester on the regulator output pin and I can see 4.6V.

Can the regulator be damaged due to Capacitors absence during switching on? Or can it be another factors?

This regulator it's able to drive until 500mA. I think the over current can not be the cause.

I let you a channel diagram in a pict. enter image description here

\$\endgroup\$
4
  • \$\begingroup\$ From the very Design considerations: "However, it is recommended that the regulator input be bypassed with a capacitor if the regulator is connected to the power supply filter with long wire lengths, or if the output load capacitance is large." You checked the last condition, did you also check the first condition? \$\endgroup\$
    – Huisman
    Commented Jan 31, 2019 at 10:19
  • \$\begingroup\$ Can you add a diagram for this board (or at least one of its channels)? From the description alone you just talk of it containing voltage dividers – no mention of any "active" components that would require a power-supply (as I read it, the ADCs are on the next board...). \$\endgroup\$
    – TripeHound
    Commented Jan 31, 2019 at 10:26
  • \$\begingroup\$ @TripeHound ADCs are the inputs in a PLC device, I have not connected the output board to the analog input (ADC) of the PLC. It only has resistors and one potenciometer sensor, but a green led as powering indicator. I will edit the post with one channel and the diode with some extra notes. I hope it can be understood. \$\endgroup\$
    – Suvi_Eu
    Commented Jan 31, 2019 at 10:38
  • \$\begingroup\$ @Huisman I have used two banana wires from the supply source. They may be a length = 50cm. After connector the Vin track may be less than one cm. \$\endgroup\$
    – Suvi_Eu
    Commented Jan 31, 2019 at 10:42

1 Answer 1

Reset to default
1
\$\begingroup\$

The regulator will certainly not be damaged by leaving capacitors out.

It will probably work in the situation you describe, but it would be better to have the capacitors on there.

I suspect you may have miswired the chip (swapped input and output, perhaps) or the chip has been damaged from some previous mishap such as reversed input voltage.

On a 12V 78xx regulator you should also consider having a diode across it if the input power can be shorted or have a heavy load applied.

\$\endgroup\$
2
  • 1
    \$\begingroup\$ maybe you are right, chip could be damaged before. The fact is that I have replaced regulator and Vout = 12V. I'm agree with you. Even it could work without capacitors my idea is to keep them to avoid any perturbation and unexpected transients. \$\endgroup\$
    – Suvi_Eu
    Commented Jan 31, 2019 at 13:31
  • \$\begingroup\$ That's a pretty good indication there was something awry with the chip. Good to close the loop so we know what happened. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Jan 31, 2019 at 13:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.