I have been trying to run a W1209 module with 5V instead of 12 volts. My application is heating a small component to 70 degrees of Celsius, mostly using a power bank or a mobile charger (Hence the 5V). The original w1209 runs at 12V DC. After posting my initial question about running the module at 5V here, I did some experiments and found that module runs happily at 5V, except for turning on the 12V relay. Now, I am thinking to just decrease the resistor R7 value, in the schematic to lower value, so that it can turn on 12V replay with 5V input. I am not sure of this idea, please suggest me if this is possible.

Can W1209 be run with 5V DC only

enter image description here


Thanks for your answers, I understood that only option is changing the realy to 5V relay. So, I am going to buy the following 5V relay. Please let me know if this one looks alright.


Original relay is 12V, T78-S-HS enter image description here


I ordered the 5V relay, but it hasn't come yet. In the meantime, I tried to use a N-channel MOSFET. But as it turns out my MOSFET is not turning on at 5V operating voltage as the voltage after Q1 is below 2V. I had to increase my input voltage to atleast 6.2V, so that after Q1 it will be 2.8V. Moreover, this voltage at Q1 is high when the relay was usually in open position and becomes low when the relay was closed. So, when I replaced the relay with N-channel MOSFET, MOSFET is switching on when it is supposed to switch off and vice versa. I think I should have replaced the relay with a P-channel MOSFET instead of N-channel MOSFET. I am updating here my progress, so that it might be useful for others. I will update again after trying with a 5V relay. I added my new schematic figure to elaborate on how I replaced relay with a N-Channel MOSFET. R value is 10 kilo ohms. enter image description here


Following @Bruce's suggestion in the comments below, I connected the 'pull up' resistor to 5V and then it start to work without any issue with MOSFET.enter image description here

  • \$\begingroup\$ In the other question the answer tells you to replace the relay with a 5V relay \$\endgroup\$
    – r_ahlskog
    Feb 2, 2021 at 17:38
  • \$\begingroup\$ No. The only real solution is to change the relay. I can see it working at 5V but not reliably. You have a voltage drop across D2 and the AMS1117 + a load dependent voltage drop across R3. When you add the 5V relay jump out R3 and bypass the AMS1117 and use a good stable 5V supply. \$\endgroup\$
    – Gil
    Feb 2, 2021 at 17:46
  • \$\begingroup\$ Hmmm... 12V DC heater... Is this a job for a 5V MOSFET? Maybe I'm crazy. \$\endgroup\$
    – K H
    Feb 3, 2021 at 0:52
  • \$\begingroup\$ Thanks for your answers. I am going with a 5V relay, I edited my question with this. I will keep you posted. \$\endgroup\$
    – raspiguy
    Feb 3, 2021 at 12:09
  • \$\begingroup\$ "In the meantime, I tried to use a N-channel MOSFET, but my MOSFET is not turning on at 5V as the voltage after Q1 is below 2V" - you need a 'pullup' resistor (value ~1-10k) in place of the relay coil. \$\endgroup\$ Feb 9, 2021 at 19:35

4 Answers 4


Here's how to run a 12V relay from 5V.

  1. Remove the relay from the PCB.
  2. Disassemble the relay, removing the coil from the rest of the mechanism.
  3. Unwind the relay, counting the turns as you do so.
  4. Divide the number of turns by 2.4, measure the wire thickness, and buy enamelled wire with approximately twice the cross sectional area (rounding up if possible).
  5. Rewind the coil with the new wire, and the reduced turns count calculated at step 4
  6. Reassemble the relay and re-fit to the PCB.

In practice, just buy the matching 5V relay suggested in the duplicate question.

  • \$\begingroup\$ This is a good practical way to handle it. \$\endgroup\$
    – r_ahlskog
    Feb 2, 2021 at 20:03
  • 2
    \$\begingroup\$ just find the middle and fold the wire in half, that will give exactly twice the cross-section, then wind it back on, connect the middle (the fold end) to one terminal and the (original) ends to the other. but on such a small relay this will be difficult to do without accidentally breaking the wire. \$\endgroup\$ Feb 3, 2021 at 3:00
  • \$\begingroup\$ @Jasen A good desert island fix! \$\endgroup\$
    – user16324
    Feb 3, 2021 at 11:30
  • \$\begingroup\$ Thanks for your answers. I can't modify the reply, but thanks for the suggestions. \$\endgroup\$
    – raspiguy
    Feb 3, 2021 at 12:10

R7 is used to control the current into the base of transistor Q1. It will have been chosen to turn Q1 "hard on" so that it is almost a short circuit between its collector and emitter. This will effectively connect the bottom of the 12 V relay between the +12V supply and GND. You can't turn in on any more than fully on no matter value you choose for R7. Even if you could you still only have 5 V available.


No. If the 12V relay does not turn on with 5V supply, then it doesn't. Changing the series resiator for the transistor will not help. You need to change the 12V relay to 5V relay.

Edit: Oh this is a repost. The original problem was driving a load. Disconnect the relay and see if you can simply use the load in place of the relay. It depends on the transistor how much current it can drive.

  • \$\begingroup\$ Thanks for your answer. I don't think transistor can withstand 1A current I want to use. \$\endgroup\$
    – raspiguy
    Feb 3, 2021 at 12:08

You have a bunch of good answers about the relay- the suggested substitute should work. It draws around 2.5x the current so it would be prudent to parallel R7 with 1K.

...the operation from 5V is not great though.

Because of the series 1N4007, R3 and semi-LDO regulator your circuit will not be getting the regulated 5V it craves.

Consider bypassing (shorting) the regulator, R3 and 1N4007 diode (M7) .. removing and shorting input to output of the AMS1117 and shorting the other two components.

At a minimum the display will be a bit brighter and it may make it more stable, less prone to chattering.

  • \$\begingroup\$ Thanks for the suggestions. For now I will just go with replacing with 5V relay. I will see the current draw and then will see if I can improve my shorting the components you are saying. \$\endgroup\$
    – raspiguy
    Feb 4, 2021 at 10:18
  • \$\begingroup\$ You mean shorting the diode D2. I am asking this because both D1 and D2 are labelled as M7 \$\endgroup\$
    – raspiguy
    Feb 10, 2021 at 11:10

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