# Relay not switching

This is a relay circuit I am using. I got the PCBs and the relay is not switching. After debugging, I noticed that one terminal of the relay is not going to 0V while the other is 5V. It stops at 2.5V or near that. Any ideas on how I might be able to fix this issue?

Note: I have not populated R48. It is left open. You can assume that it is not there.

• What is VCC? +5 V? Oct 17, 2023 at 12:06
• And why are you using an opto, if your have the same VCC on both sides of the opto? Oct 17, 2023 at 12:07
• @SteveSh VCC is +5V opto was added to provide isolation but it is redundant as it has the same VCC on both sides Oct 17, 2023 at 12:08
• Finally, with 2 LEDs in series on the input, you probably can't drive enough current through the opto's diode to turn it on hard. Oct 17, 2023 at 12:09
• @SteveSh VCC is a +5V,2A supply so I dont think its a current issue. Oct 17, 2023 at 12:12

You have a number of problems.

Your relay coil is rated at 180 mA, and the transistor collector current is rated at 100 mA max, so you need a beefier transistor. Something like a 2N4401 should handle the relay current.

The next problem is driving the transistor into saturation, your transistor has a minimum current gain of 110. So to get 180 mA collector current you would need 1.63 mA. Starting with a 5 V supply and subtracting the transitor's $$\V_{BE}\$$ and the opto's $$\V_{CE}\$$ you have maybe 4.2 V, with a 10K resistor that's 0.42 mA, nowhere neer what you would need. And that's not even forcing saturation, you need more base current than just $$\I_C/Beta\$$.

Let's look at the 2N4401 data sheet, it shows a $$\V_{CE}(sat)\$$ rating of 0.4 V for $$\I_C\$$ = 150 mA, $$\I_C\$$ = 15 mA, so let's figure on needing 18 mA base current.

$$I_B = \frac{4.2V}{18mA} = 233\Omega$$ So your 10K resistor is not going to do.

Now, I'm not sure if the opto can handle 18 mA, so what you could do is use a darlington transistor, or two transistors wired as a darlington. This would reduce the needed base current to something like 180uA and allow you to keep the 10K resistor, and your opto shouldn't have any problem with that.

• Whether the opto can handle it is less of a problem than whether the thing driving the opto can handle it, because that opto only guarantees a CTR of 20%! So to get 18 mA out, you'd need at least 90 mA in, and that's only guaranteed when Vce is kept at 10 V. Oct 17, 2023 at 15:49