# Explanation : Latching Relay Circuit

I need to understand the flow of this circuit well. The objective of this circuit is to switch the latching relay "OFF" or "ON" (in two states) depending on the state of the I/O's of the MCU. I need to understand how this works. The relay is a 9VDC, 60A, 250VAC relay, and the bridge gives an output of 12Vdc. Thanks!

• Is it me, or is the whole circuit backwards? – Majenko Dec 19 '11 at 11:26

This looks very similar to a half bridge.

When both the IOs are at the same level, both the left and right circuits are providing the same potential to the relay.

When one of the IOs changes, the potential provided at that side of the relay changes.

The difference in these two potentials is what switches the latching relay. It's just like driving a motor in forward/reverse.

Let's take a closer look at the right-hand side of the circuit.

When the IO pin is low, what is happening?

Well, Q4 and Q6 both have their bases pulled low. This switches them off. In doing so the pull-up resistor R2 pulls up the base of Q3, switching it on. Thus, the power from the +12V rail is allowed to flow to the relay.

If the IO is set to high, what then?

Well, Q4 and Q6 are both switched on. This connects their collectors to ground. Q4 links the relay to ground, and Q6 links the base of Q3 to ground, switching it off. This isolates the +12V from the relay so you don't get a nasty short. so that side of the relay is now at ground potential.

Combine that with the same thing happening to the left side of the relay, and you can see how different combinations of IO can give different potentials at the relay terminals.

A | B || L | R | Vdiff
======================
0 | 0 || + | + |   0v    Nothing happens
0 | 1 || + | G | +12V    Relay switches to one position
1 | 0 || G | + | -12V    Relay switches to other position
1 | 1 || G | G |   0v    Nothing happens

• Thanks for the explanation! But I want to know if there is a peculiar need for using 3 transistors each, because I want to believe that 2 transistors can do exactly the same. From the data sheet, Q5 and Q6, are switching transistor, and it seems like Q1 - Q4 are amplifying transistors. So is there any form of amplification in the circuit? – Paul A. Dec 19 '11 at 11:55
• All the transistors are switching. The purpose of Q5 and Q6 is to invert the signal, so either Q1 or Q2 is on but not both - the same for Q3 and Q4. This routes either +12V or GND to the relay pin, but never both. You could replace Q5 and Q6 with an inverter logic gate to get the same effect. – Majenko Dec 19 '11 at 12:01

It's an H-bridge. I'll explain for the left half, the right half is identical.
If the control line is high Q1 and Q5 are conducting, and by pulling Q2's base low Q5 shuts it off. As a result this relay pin is switched to ground by Q1.
If the control line is low Q1 and Q5 are not conducting, and Q2 is conducting; R1 feeds the base. Via Q2 the relay pin is switched to V+.
You switch the relay on or off by placing opposite signals to the control pins. Since it's a latching relay you can remove the voltage across the coil after it's switched by making both relay pins the same voltage.

As said this is a H-Bridge.

Q5 and Q6 work as inverters.
Q1 and Q2 are never on at the same time. The same is true with Q3 and Q4.

So if current is going through the relais it is either that way (Q2 and Q4 on)

   V
|
+--R--+
|
V


or this way (Q1 and Q3 on)

         V
|
+--R--+
|
V


All transistors are just used a switches.
Still I think you really need 2x3 transistors (unless you want to use more control lines or other devices (e.g. inverters from IC))