Currently I’m working on a small portable amp which will be run either with 17 to 24 volts, depending if I drive it with a battery or a wall wart. It’s my first electronics project, so I have some fairly basic questions...

To power two single coil non latching relays (G6K-2P DC12 by Omron Electronics, coil current: 9.1 mA, coil voltage: 12 V, used as input selector, and stereo-to-mono switch) and an LED, I plan to use a voltage regulator. After reading an input by Russell McMahon on the Electrical Engineering Stack Exchange, it seems to me that in my situation (voltage drop of up to 12 volts) it would be the best to use a switching regulator, like the OKI-78SR-12/1.0-W36-C by Murata (input: 15 - 36 V, output: 12 V).

I want to design a little PCB with the regulator, the relays, a power output for the LED, and all the circuits and resistors needed for the input selector, and the stereo-to-mono switch). I got the audio part all figured out, but I’m not sure if I got the power part right.

Here is my schematic (just the power part), showing how I would connect the regulator the relays, and the LED:

Schematic with switching regulator, relays, and LED

I already tried to sketch out, how I would put this on a PCB in the size that would fit into my amp case. I will not solder the LED and the SPST switches to control the relays directly on the PCB therefore on the PCB you see the footprints of JST connectors at their place (SW1 = P2, SW2 = P3, LE1 = P7).

Sketch of my PCB

Now, I have the following questions:

  • Is is a good choice to use a switching regulator, as the one linked from Murata?
  • Is my schematic correct?
  • Is the routing in my PCB correct, and supposed to work?
  • 1
    \$\begingroup\$ At 9 mA, the Iq of the regulator is still not neglecable. You will get a bit better efficiency with a buck converter, but not by much. I would go with 7812 or 1117-12. Your flyback diodes are upside down due to your crossovers of wires. \$\endgroup\$
    – winny
    Jun 26 '17 at 21:06
  1. Don't use SMPS. Yes, you have 12V dropout voltage but by only ~20mA load power loss is approximately 250mW which won't be troublesome for many of the "off the shelf" linear regulators.

  2. Use ceramic capacitor for input and output decoupling of linear regulator. Seriously, your circuit will not work without them. About 470nF at input and 1uF output will most probably do the job. However, read the datasheet of regulator thoroughly!

  3. Both diodes are orientated wrongly! Replace cathode and anode of the diodes - you will blow them otherwise!

  4. Friendly advice - read some basic books about electronics before building anything. Predict what could possibly go wrong and deal with it. I mean seriously, otherwise it's just waste of money.

  • \$\begingroup\$ Thanks. I see that a linear regulator will do. A recapitulation for my own understanding: My relays draw 9.1mA each. Given the 12V regulator, and with a 3.3k resistor the LED will draw ≈ 3mA. So, the total load current will be ≈ 21mA (9.1mA + 9.1mA + 3mA). With a voltage drop of 12V, and the total load current of 21mA the power that the regulator dissipates as heat, will be ≈ 0.250W, as stated above (W = V * A, in my case: 12V * 0.021A = 0.0252W). If the thermal resistance junction-air of my regulator is 65°C/W, the maximum rise of temp will be a bearable 16°C (65°C * 0.250W = 16.25°C). \$\endgroup\$
    – Monrin
    Jul 1 '17 at 19:04

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