I am by no means an electrical engineer, so I'm guessing that there is a pretty simple solution to this problem and I'm just being thick about it.

In general, I want to read the voltage of 4 different relays into cDAQ module NI 9402. This particular DI/O module has a voltage read limit of 5V, but since all relays will be operating at 12V, I'm wanting to step down to 3V3. Essentially, the NI module will read the voltage as either 3V3 or 0V thus providing info to the state (open or closed) of the power relay(s).

After some basic googling and looking through parts on digikey and mouser, I'm having a difficult time finding power converters that can handle 12V --> 3V3. Is this too large of a step-down for a single unit to achieve? Wondering if I'll need to go from 12V --> 5V --> 3V3? Help is much appreciated, Thanks!!

Don't think it's required information, but here is the particular relay that I'm working with: https://www.digikey.com/product-detail/en/phoenix-contact/2906224/277-14896-ND/5699611

  • 2
    \$\begingroup\$ Um... what's wrong with a simple voltage divider resistor network? Or are you not telling us something.. \$\endgroup\$
    – Trevor_G
    Aug 28, 2017 at 17:16
  • \$\begingroup\$ +1 Trevor's comment. Unless you intend to power something off of the 3.3V step-down, sensing a voltage divider resistor ladder is fine. \$\endgroup\$
    – Daniel
    Aug 28, 2017 at 17:18

2 Answers 2


If you just want to measure the coil voltage to sense if the relay is on then a simple voltage divider will do the job.


simulate this circuit – Schematic created using CircuitLab

Figure 1. A simple voltage divider will reduce the voltage to a level suitable for the cDAC input.

I'm not familiar with the cDAC module so this answer gives you the means of working out the voltage divider.

Assuming that R3, the input impedance of the cDAC input, is high (x 10) relative to R2 the voltage at INPUT will be given as the simple ratio \$ V_{IN} = V_S \frac {R2}{R1 + R2} \$, where \$ V_S \$ is the supply voltage. In the example of Figure 1 we get \$ V_{IN} = 12 \frac {12}{33 + 12} = 3.2 \; V\$. I think this is a bit low for a 5 V input but check this from the datasheets.


Using @Transistors solution, I believe I can use a 9100 ohm and 3600 ohm resistor to step down from 12V to 3V3


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