I am working on a project which require to activate a 24V Rated relay using a 3.3V rated IO. To provide adequate isolation I have added a Opto-isolater stage.

In order to increase the current output capacity I have added a NPN Transistor Switch and controlling it using the Opto-Isolator output.

The main issue I have encountered in past (Using only Opto-isolator to drive relay) is the output is accidentally connected to the 120V AC supply which potentially blow up the Opto-isolator.

I have following questions related to the over-voltage protection concept:

  1. The transistor's reverse breakdown voltage (VEBO) is 6V, even if I use a over-voltage protection diode the Emitter will observer at least -25V upon AC application (if Zener is 25V rated), In this case how can I ensure that the transistor switch will tolerate the reverse bias (By adding addition components?)

  2. Is there any other method can be used to provide the isolation with 120V AC protection?

Here is the circuit-

Transistor Datasheet:https://www.fairchildsemi.com/datasheets/MM/MMBT3904.pdf

Option A and B can be used to activate the Transistor switch

Relay Specification:

The relay I am using is Omron -G7SA-2A2B, rated 24V, Coil resistance 1600 ohm, Min vol = 75% of rated 24V, Max Vol =110% of rated 24V

The circuit is for the failure case when the 24 V/80mA rated output get connected to the 120V supply. In this case the transistor output gets +/- 120V.

Failure case: 120V applied on 24V rated output

  • \$\begingroup\$ Vf of the Opto-isolater emitter is max - 1.4V \$\endgroup\$ Commented Oct 15, 2018 at 20:19
  • 1
    \$\begingroup\$ Where is the 24 volt relay in your circuit? And why are you connecting 120 VAC to the right-most transistor's emitter? \$\endgroup\$ Commented Oct 15, 2018 at 23:02
  • \$\begingroup\$ A simple \$24\:\text{V}\$ relay will provide a coil designed to be driven with \$24\:\text{V}\$ and contactors (switch) designed for some other purpose (such as 120 or 240 VAC at some maximum current compliance.) You have not specified what your AC load or relay is, so we cannot verify that side of things. But most \$24\:\text{V}\$ relays can be driven with as little as 70%, so your DC voltage range seems okay, at least. You will drive the coil with your MCU and if you want, you can opto-isolate that from the MCU. But it may not be necessary. A relay AC contactor switch is already isolated. \$\endgroup\$
    – jonk
    Commented Oct 16, 2018 at 6:07
  • \$\begingroup\$ Of course, you might be using a \$24\:\text{VAC}\$ relay coil? (I've not seen these, though I have and do use \$120\:\text{VAC}\$ relay coils.) For those, you would want to use a opto-triac, such as the MOC302x and MOC306x series. \$\endgroup\$
    – jonk
    Commented Oct 16, 2018 at 6:10

1 Answer 1


You have a bunch of problems with your design.

enter image description here

Figure 1.

  1. You are feeding 21 V into a 3.3 V micro-controller output. Unless it has high voltage open-collector outputs this won't end well.

enter image description here

Figure 2. The problem is similar to this High-side fail. The internal protection diodes will conduct and the LED will light continuously. The internal protection diodes may burn out and then the output stage will be over-stressed and fail.

  1. Connecting Option A will feed a permanent 'high' into the base of the transistor. All the circuitry to the left then becomes irrelevant.

The main issue I have encountered in past (Using only Opto-isolator to drive relay) is the output is accidentally connected to the 120V AC supply ...

  1. You seem to have solved the 'accidentally" problem by permanently connecting it to the 120 V, 60 Hz supply! As shown you are feeding 120 V AC into your circuit, the transistor, the 100k resistor and possibly back into the supply. This is not correct and needs to be redesigned.

  2. You are showing your 120 V supply referenced to earth. You should be showing it referenced to neutral. You can call this ground or GND and use the GND symbol (a triangle as on my Figure 2) and reserve the 'earth' symbol for mains earth.


  1. Power the opto-LED from the 3.3 V power rail. The forward-voltage, Vf is only 1.4 V so you have plenty of headroom.
  2. A solid-state relay (SSR) is a readily available solution to replace the rest of the circuit and will save you much trouble. You need to select one that will work with a 3 V input and handle 120 V AC at the current you require. (Over-spec the current rating for safety margin.)
  3. If you must roll your own circuit then replace your opto-isolator with an opto-triac. You will find plenty of examples in a web search.

From the comments:

I want to switch 24 V relay, but to make the circuit fail-safe I want to add protection of some kind which can prevent damage to internal circuit due to 120 V supply connection. In other words -24 V rated output with, 120 V protection.

This is a very strange requirement. As you have drawn it the 120 V supply will probably have enough energy to destroy your protection devices and then destroy the rest of the circuit. We protect against this type of fault by choosing good quality, well designed components and wiring correctly.


simulate this circuit – Schematic created using CircuitLab

Figure 3. A relay or SSR adequately isolates the micro-controller from the mains.

  • \$\begingroup\$ 3. I have missed to mention the circuit I have added in question is for the failure case, when 120V get connected to the 24Volt rated output. \$\endgroup\$ Commented Oct 16, 2018 at 12:23
  • \$\begingroup\$ What could happen if the npn transistor's Emitter is connected to 120V supply? The overprotection diode will short the 120 V to GND in that case the fuse will blow, But as the fuse take at least a couple of ms or more before blowing, It might be the time where Transistor can be damaged? Most of the transistors I have searched are having a reverse breakdown voltage of 6-7 Volt maximum for VEBO(npn) transistor. In that case if I want to protect the low voltage circuitry, what are my options? \$\endgroup\$ Commented Oct 16, 2018 at 12:51
  • \$\begingroup\$ What is the real problem you are trying to solve? I think you want to switch a 24 V DC relay from a 3.3 V GPIO and use the relay to switch a 120 V AC load. Is this correct? \$\endgroup\$
    – Transistor
    Commented Oct 16, 2018 at 13:17
  • \$\begingroup\$ I want to switch 24V relay, but to make the circuit fail-safe I want to add protection of some kind which can prevent damage to internal circuit due to 120V supply connection. In other words -24V rated output with, 120V protection. \$\endgroup\$ Commented Oct 16, 2018 at 14:02
  • \$\begingroup\$ See the update. The real solution is to not apply 120 V to the low-voltage circuit. It sounds as though someone was being careless. This could cost more than some electronics. Use proper isolation techniques when carrying out work on mains equipment. \$\endgroup\$
    – Transistor
    Commented Oct 16, 2018 at 20:26

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