1
\$\begingroup\$

I know the diodes facing each other are for ESD protection, but what do the diodes facing opposite do ?

The circuit is basically a 24V (usually used in industry) digital output.

enter image description here

\$\endgroup\$
5
  • \$\begingroup\$ It looks like the dotted line encircles an optocoupler/isolator? It may be a good idea to explain what you are showing us. \$\endgroup\$
    – Bort
    Commented Jul 11, 2018 at 13:31
  • \$\begingroup\$ @Bort, it's the digital output circuitry on an industrial controller. \$\endgroup\$
    – Salman
    Commented Jul 11, 2018 at 13:37
  • 2
    \$\begingroup\$ They probably represent the body diodes of what seems to be an optical mosfet. \$\endgroup\$
    – dim
    Commented Jul 11, 2018 at 13:38
  • 3
    \$\begingroup\$ Also, this is not a push-pull configuration. Push-pull is when you have a digital output that either source power from the positive supply rail, or sink power to the ground. Here, you have no positive supply rail, or even ground. The output here acts more like a relay. There are two FETs because it is bidirectional, but not push-pull. \$\endgroup\$
    – dim
    Commented Jul 11, 2018 at 13:41
  • \$\begingroup\$ There are many variations of this Opto-coupled SSR or BOSFET design ranging from 250V to x kV and 20mA to x Amps and $1 to $xxx mouser.com/datasheet/2/427/vor1121a6-1076514.pdf \$\endgroup\$ Commented Jul 11, 2018 at 16:21

2 Answers 2

8
\$\begingroup\$

The two diodes within the dotted line represent the bulk parasitic diode in parallel with each MOSFET as per this diagram of a solid state relay (SSR): -

enter image description here

what do the diodes facing opposite do

When both MOSFETs are activated the MOSFETs will offer low on-resistance and shunt their respective parasitic diodes. You need two back-to-back MOSFETs in a SSR so that you avoid half-wave conduction on AC - that's the problem of having a bulk diode in parallel with a MOSFET.

MOSFETs will operate as switches with a negative supply voltage but this is restricted to about -0.6 volts because of the bulk diode: -

enter image description here

Picture source.

Notice the light blue curve for Vgs at 0 volts (device off) - conduction in reverse (due to the diode) starts at a shade over -0.5 volts but closer to 0 volts the MOSFET can act as a negative voltage switch and has a very similar characteristic when the voltage is positive or negative. So, the back-to-back MOSFET configuration acts as a true AC switch with an on-resistance double that of a single MOSFET. So basically, the diodes are not called into conducting much current at all; when the MOSFETs are off the diodes are blocking and when the MOSFETs are both on, the diodes are shunted by their respective drain source on-resistances.

\$\endgroup\$
5
\$\begingroup\$

This is not a "push-pull" configuration. In a push-pull the common node between the 2 transistors would be the output. Then one transistor can "pull" current while the other can "push" current.

You should show the source of this drawing for completeness.

My interpretation of the two MOSFETs in series is that the component in the dotted line supports AC current, a current can flow in either direction. In theory it is possible to use only one MOSFET for that but in practice there will be an issue with properly biasing of the substrate of the MOSFET to make the MOSFET go off when it needs to.

The practical solution is to allow only one direction of the current through each MOSFET. That then requires the use of two MOSFETs in series. The substrate (bulk or body) of a MOSFET is then connected to its source. If you look up how MOSFETs are constructed and then you'll learn that both drain and source have a diode to the substrate. The diodes in your schematic are the drain-substrate diodes of each MOSFET. They cannot be avoided, they're an integral part of the MOSFETs.

Here these drain-substrate diodes actually help as they conduct the current preventing the MOSFET from having to work with a reversed current.

So when the current flows from top to bottom the top MOSFET and bottom diode conduct. When the current flows from bottom to top the bottom MOSFET and the top diode conduct.

\$\endgroup\$
2
  • \$\begingroup\$ thats what I thought earlier, in positive cycle, upper mosfet and lower diode conducts and in negative cycle lower mosfet and upper diode conducts, but according to the wiki page "solid state Relay" and Andy's answer, the issue with one mosfet is that when it's off, the diode will conduct in -ve cycle, so they put two mosfets in series back to back, so their diodes face opposite, now if the mosfets are off, no current can flow through diodes, when on, both mosfets will conduct and when off both mosfets will be off, it's not possible for one mosfet to be on and other to be off in these devices \$\endgroup\$
    – Salman
    Commented Jul 11, 2018 at 18:06
  • \$\begingroup\$ yes, you are right it's not push pull, the configuration looks similar to push-pull at a first glance, I'll change the title to remove confusion, thanks \$\endgroup\$
    – Salman
    Commented Jul 11, 2018 at 18:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.