Transistor Reverse Biasing - AC

Question

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.

Answer 1

You have a bunch of problems with your design.

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.

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.

2. 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 ...

3. 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.

4. 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.

Fixes

1. Power the opto-LED from the 3.3 V power rail. The forward-voltage, V_f 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.

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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.