I recently purchased this signal relay board.

When Input 1 and Input 2 are at 3.3VDC, both D1 and D2 illuminate.

Here is my set up,

+3.3VDC for IN1 as well as a separate +3.3VDC for IN2. For VCC we are feeding with +5VDC. Each +3.3VDC input line has more than 20 mA of drive strength, but the relays do not trigger as expected. If we apply +5VDC to to IN1 or IN2 for the relays which do not work with +3.3VDC, the relay triggers as expected.

Can someone review on my set-up and confirm that both relays should indeed be working as it's set up like this?

  • 1
    \$\begingroup\$ Seems to be all good. It might be worth probing voltages at the transistor pins (Q1 and Q2), when IN1 and IN2 are set to 3.3 Volts. \$\endgroup\$ Feb 6, 2013 at 10:37

1 Answer 1


The control circuit on the relay board is poorly designed.

enter image description here

If you apply 3.3 V then there's 2.6 V across R1, so the current will be 260 µA. Across R7 we have 0.7 V, so that's 70 µA through R7, leaving 190 µA for Q1's base. If Q1 has an \$h_{FE}\$ of 100 (typical for a general purpose transistor), then you have 20 mA collector current. The schematic doesn't give a value for R2, but even at a few mA through the LED there's not enough current left for the relay.

The solution: replace R1 with a 1 kΩ type, which will give you 2.6 mA base current, and 260 mA collector current. That should be enough to drive both LED and relay.

The reason it works with 5 V is that at 5 V you have a higher base current, a bit the same effect as decreasing the value for R1. Apparently the design was not tested at 3.3 V.

BTW, remove R7, it serves no function. It only takes current away from Q1, which you don't want.

  • \$\begingroup\$ R7 keeps the transistor turned off when IN1 is disconnected. Instead of removing R7, change the value of R1 to 1k; that will provide sufficient drive to make this work reliably. If removing R1 is difficult, just solder a 1k resistor across it. 1k in parallel with 10k is just a mite under 1k (909 ohms) which is close enough. (I just noticed I repeated your solution; I started replying about R7 and kept going. :-) \$\endgroup\$
    – akohlsmith
    Feb 6, 2013 at 18:56
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    \$\begingroup\$ @Andrew - R7 would have a function if Q1 were a FET, but BJTs really don't need the pull-down. And even if they would it's unlikely that the input would be disconnected while the rest of the module would be operational. \$\endgroup\$
    – stevenvh
    Feb 9, 2013 at 10:09
  • \$\begingroup\$ @stevenh bipolar transistors are not as noise-sensitive as FETs but I can assure you that pulldown is very much needed if you don't want it switching on in noisy environments. As far as the unlikeliness of the input being connected... stuff happens. Loose connections, broken wires/solder joints... For the cost of a pulldown resistor it's hardly worth arguing about for a more robust design. \$\endgroup\$
    – akohlsmith
    Feb 9, 2013 at 21:19

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