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You are here: Home / Electronic Devices and Circuit Theory / Solid State Relay (SSR) Circuit using MOSFETs

Solid State Relay (SSR) Circuit using MOSFETs

Last Updated on January 22, 2021 by Swagatam 56 Comments

SSR or Solid state relays are high power electrical switches that work without involving mechanical contacts, instead they use solid state semiconductors like MOSFETs for switching an electrical load.

SSRs can be used for operating high power loads, through a small input trigger voltage with negligible current.

These devices can be used for operating high power AC loads as well as DC loads.

Solid State Relays are highly efficient compared to the electro-mechanical relays due to a few distinct features.

Main Features and Advantages of SSR

The main features and advantages of solid state relays or SSRs are:

  • SSRs can be built easily using a minimum number ordinary electronic parts
  • They work without any form of clicking sound due to the absence of mechanical contacts.
  • Being solid state also means SSRs can switch at much faster speed than the traditional electro-mechanical types.
  • SSRs do not depend external supply for switching ON, rather extract the supply from the load itself.
  • They work using negligible current and therefore do not drain battery in battery operated systems. This also ensures negligible idle current for the device.

Basic SSR Working Concept using MOSFETs

In one of my earlier posts I explained how a MOSFET based bidirectional switch could be used for operating any desired electrical load, just like a standard mechanical switch , but with exceptional advantages.

The same MOSFET bidirectional switch concept could be applied for making an ideal SSR device.


For a Triac based SSR please refer to this post


Basic SSR Design

basic solid state relay SSR design concept

In the above shown basic SSR design, we can see a couple of appropriately rated MOSFETs T1 and T2 connected back to back with their source and gate terminals joined in common with each other.

D1 and D2 are the internal body diodes of the respective MOSFETs, which may be reinforced with external parallel diodes, if required.

An input DC supply can also be seen attached across the common gate/source terminals of the two MOSFETs. This supply is used for triggering the MOSFETs ON or for enabling permanent switch ON for the MOSFETs while the SSR unit is operational.

The AC supply which could be up to grid mains level and the load are connected in series across the two drains of the MOSFETs.

How it Works

The working of the proposed sold state relay can be understood by referring to the following diagram, and the corresponding details:

positive half cycle SSR working
negative half cycle SSR working

With the above setup, due to the input gate supply connected, T1 and T2 are both in the switched ON position. When the load side AC input is switch ON, the left diagram shows how the positive half cycle conducts through the relevant MOSFET/diode pair (T1, D2) and the right side diagram shows how the negative AC cycle conducts through the other complementing MOSFET/diode pair (T2, D1).

In the left diagram we find one of the AC half cycles goes through T1, and D2 (T2 being reverse biased), and finally completes the cycle via the load.

The right side diagram shows how the other half cycle completes the circuit in the opposite direction by conducting through the load, T2, D1 (T1 being reversed biased in this case).

In this way the two MOSFETs T1, T2 along with their respective body diodes D1, D2, allow both the half cycles of the AC to conduct, powering the AC load perfectly, and accomplishing the SSR role efficiently.

Making a Practical SSR Circuit

So far we have learned the theoretical design of an SSR, now let's move ahead and see how a practical solid state relay module could be built, for switching a desired high power AC load, without any external input DC.

The above SSR circuit is configured exactly in the same way as discussed in the earlier basic design. However, here we find two additional diodes D1, and D2, along with the MOSFET body diodes D3, D4.

The diodes D1, D2 are introduced for a specific purpose such that it forms a bridge rectifier in conjunction with the D3, D4 MOSFET body diodes.

The tiny on OFF switch could be used for turning the SSR ON/OFF. This switch could be a reed switch or any low current switch.

For high speed switching you can replace the switch with a opto-coupler as shown below.

In essence the circuit now fulfills 3 requirements.

  1. It powers the AC load through the MOSFET/Diode SSR configuration.
  2. The bridge rectifier formed by D1---D4 simultaneously converts the load AC input into rectified and filtered DC, and this DC is used for biasing the gates of the MOSFETs. This allows the MOSFETs to get appropriately turned ON through the load AC itself, without depending on any external DC.
  3. The rectified DC is further terminated as an auxiliary DC output which could be used for powering any suitable external load.

Circuit Problem

A closer look at the above design suggests that, this SSR design might have problems implementing the intended function efficiently. This is because, the moment the switching DC arrives at the gate of the MOSFET, it will begin turning ON, causing a bypassing of the current through the drain/source, depleting the gate/source voltage.

Let's consider the MOSFET T1. As soon as the rectified DC begins reaching the gate of T1, it will begin turning ON right from around 4 V onward, causing a bypassing effect of the the supply via its drain/source terminals. During this moment, the DC will struggle to rise across the zener diode and begin dropping toward zero.

This will in turn cause the MOSFET to turn OFF, and the continuous stale-mate kind of struggle or a tug of war will occur between the MOSFET drain/source and the MOSFET gate/source, preventing the SSR from functioning correctly.

The Solution

The solution to the above issue could be accomplished using the following example circuit concept.

The objective here is, to make sure that the MOSFETs do not conduct until an optimum 15 V is developed across the zener diode, or across the gate/source of the MOSFETs

The op amp ensures that its output fires only once the DC line crosses the 15 V zener diode reference threshold, which allows the MOSFET gates to get an optimal 15 V DC for the conduction.

The red line associated with pin3 of IC 741 can be toggled through an opto coupler for the required switching from an external source.

How it Works: As we can see, the inverting input of the op amp is tied with the 15V zener, which forms a reference level for the op amp pin2. Pin3 which is the non-inverting input of the op amp is connected with the positive line. This configuration ensures that the output pin6 of the op amp produces a 15V supply only once its pin3 voltage reaches above 15 V mark The action ensures that the MOSFETs conduct only through a valid 15 V optimal gate voltage, enabling a proper working of the SSR.

Isolated Switching

The main feature of any SSR is to enable the user an isolated switching of the device through an external signal.

The above op amp based design could be facilitated with this feature as demonstrated in the following concept:

How the Diodes Work Like Bridge Rectifier

During the positive half cycles, the current moves through D1, 100k, zener, D3 and back to the AC source.

During the other half cycle, the current moves through D2, 100k, zener, D4 and back to the AC source.

Reference: SSR




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About Swagatam

I am an electronic engineer (dipIETE ), hobbyist, inventor, schematic/PCB designer, manufacturer. I am also the founder of the website: https://www.homemade-circuits.com/, where I love sharing my innovative circuit ideas and tutorials.
If you have any circuit related query, you may interact through comments, I'll be most happy to help!

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  1. Search Related Posts for Commenting

  2. geert says

    The statement that the load current will pass through the body drain diode of one of the mosfets when the circuit is conducting is wrong. If a n-channel mosfet gate is biassed and the channel conducts, the mosfet channel will conduct in both directions. Due to the presence of the diode, it will only block in one direction.

    Reply
    • Swagatam says

      In that case a single FET should satisfy the conditions, as it can conduct in both directions?

      Reply
  3. Seun says

    Good day Swagatam, well done Sir, please I want a circuit using MOSFET to operate as relay to have NO and NC with common function.

    I want to avoid the clicking sound of relay, thanks

    Reply
    • Swagatam says

      Thanks Seun, you can try employing the following design:

      Solid State Relay (SSR) Circuit using MOSFETs

      Reply
  4. wraper says

    This guide contains the same lame error as multiple appnotes from TI. MOSFET body diodes never conduct when MOSFETs are open, and they are always open simultaneously. Here is fixed article on TI forum where audience noticed the mistake and it was fixed. It’s appaling that TI never bothered to fix appnotes for years. https://e2e.ti.com/blogs_/b/industrial_strength/archive/2016/07/26/a-modern-approach-to-solid-state-relay-design

    Reply
    • Swagatam says

      OK thanks for informing! However, the above SSR concept is still intact, and the issue could be easily remedied by adding external diodes across the drain/source leads of the FET.

      Reply
  5. João says

    Hi friend! I really enjoyed the class, it was magnificent; his way of narrating is that of a great Professor.
    I would like, if it were possible a tip on how to create a circuit to keep the soldering iron of 30w, with an ideal temperature, constant. Thank you for the class and Congratulations!

    Reply
    • Swagatam says

      Thank you friend, you can use the second circuit explained in the following post:

      https://www.homemade-circuits.com/fan-speed-controller-for-heatsink/

      Reply
  6. Arvind says

    Hi !

    What is AC load?

    Reply
    • Swagatam says

      Any appliance…

      Reply
  7. steven says

    Hi Swagatam,

    I have few doubts referring to the figure 2 of current flow.

    1) since both the mosfets are turned on at a time the current should flow from both the mosfets rather than the diode, though the current only flows from the drain to source and not from the S to D, it taken a path from the diode please correct me if I’m wrong.

    2) when the gate is powered or triggered with the DC supply voltage and the positive half cycle AC signal is passed from the mosfet, since the negative terminal of the DC supply is connected to the common source then why the positive half AC signal did not taken that DC supply ground path rather than flowing through the diode.

    please let me know the above doubts I’m new to the electronic design thanks.

    Reply
    • Swagatam says

      Hi Steven,
      1) the current will take the easiest path which is through the forward biased diode, and therefore the associated MOSFET will remains completely ignored and turned OFF

      2) Referring to the first diagram, the opposite pole of the AC is across its own supply points not the DC negative, so it will flow towards its own negative/positive points not the external DC negative. The points indicated as “AC Input” are referenced with each other, not to an external DC

      Reply
  8. Trishpota says

    Dear Swatagam
    I love your site and its amaze me everytime thank you for opening site like this
    Can you give a specific value of diodes?
    Dc out? How much voltage needed for circuit first stage(capacitors,resistor,part)?
    You say for fast switching add opto coupler so triggering opto coupler need another voltage input?
    Thank you for your patience and answering <3

    Reply
    • Swagatam says

      Thank you Trishpota, The diode rating will be twice of the load current rating. If load is 2 amp, then diodes must be 4 amps. Only 12V to 17 V DC will be enough for the MOSFET switching. Yes the opto will need to be switched externally from an external trigger source, because in an SSR the input and output is supposed to be completely isolated.

      Reply
      • Trishpota says

        What about zener diode?

        Reply
        • Swagatam says

          zener diode is 15 V, please make the last circuit which will ensure correct working of the SSR. Once confirmed you can break the red line and introduce an opto based switching through that line.

          Reply
  9. Tim says

    Hey Swagatam,
    i am wondering how your rectifier can supply the Zener after the SSR has been switched on.
    Imagine we are in the positive Half of the Sinewave.
    If one of the Mosfets is swicthed on (lets say T1) then D3 of T2 is also in conducting mode. So lets say over T1 is 0.3V and over D3 is 0.7V then you have 0.3V+0.7V=1V for your Zener diode (after rectifier Diodes…), which tries to hold the OnState of the SSR. So, please correct me, but this will not work.
    Best regards Tim

    Reply
    • Swagatam says

      Hi Tim, During the half cycle before T1 can conduct the capacitor will be charged and this will hold the supply across the zener diode and also for the MOSFET gates.

      Reply
      • Swagatam says

        Here’s the original link from where I referred the concepts:

        https://www.ti.com/lit/ug/tiduc87a/tiduc87a.pdf

        Reply
        • Tim says

          Thanks for your answer. Please have a look at page 12; “2.4 Undervoltage Lockout Design Theory”.
          The cap will discharge as long as the SSR is in switched on and after some time it must be switched off to reload the cap again. So this is a good solution for something, that is switched on only for a short time, like a thermostat or something. The Texas guys seem to switch it at a certain point off and on again, so the cap reloads. Well its not a problem, but one have to know it before.

          Reply
  10. jooch says

    Nice article, how would you drive this from an Arduino if the load is 400 watts?
    Then a totem pole would be needed to drive T1 & T2?

    Reply
    • Swagatam says

      Glad you liked it, I think the last concept would be suitable for interfacing with an arduino

      Reply
  11. Jay says

    Hello, great tutorial, but I had a question. Since half of the time, the one mosfet has current going through its body diode which often has a voltage drop of 1-2V in the models I’ve seen, wouldn’t the mosfets get very hot for higher loads? Or do they not because they are only in this state half of the time? Thanks.

    Reply
    • Swagatam says

      Hi, to ensure an efficient working you will have to complement the body diodes with external parallel diodes rated to handle the specified amount of current and reduce MOSFET dissipation.

      Reply
  12. Paul Ely says

    This is exactly what I want. However I need to switch AC on/off by means of 12v dc. If I connect +12v to the combined gates I don’t think it will work, will it? So how do I do this please?

    Reply
    • Swagatam says

      It will work if you connect the the 12V DC across the gates and the sources of the MOSFETs as indicated in the first diagram.

      Reply
      • Paul Ely says

        You mean -12 connected to 240ac? Also if the gates are left floating will it switch of?
        Tks for your help

        Reply
        • Swagatam says

          yes that’s right….

          Reply
        • Swagatam says

          gates must not be kept open…

          Reply
          • Paul Ely says

            I disconnected +12v from gates and have a 10k resistor between gates & sources but it’s still on,
            so how do I switch it off?

            Reply
            • Swagatam says

              Can be difficult to judge without seeing your circuit practically??

  13. Joel says

    Hello Sir,
    I have a question about this project. I like this project and may use it. However I would like to use something like this that would take the place of a standard SPDT relay for DC. Basically to replace the relays in one of your other circuits used to charge two 2 batteries flip-flop style. Thank you much for all you do.

    Reply
    • Swagatam says

      Hello Joel, for a SPDT DC version you can probably try this concept:

      https://www.homemade-circuits.com/12v-dc-solid-state-relay-ssr-100-amps/

      Reply
    • Joel says

      Thank you I found it and it looks like it will fit just the way I wanted. I enjoy your site and play with your circuits to draw pc boards with. Just for fun.

      Reply
      • Swagatam says

        Glad you are enjoying my site, please keep up the good work!

        Reply
        • Joel says

          I have a question about the circuit that you directed me to. I have used your split 12v charger and the above mosfet circuit you directed me to, for a project I want to build. My question is by looking at what you showed me it seems that the mosfets switch the ground and not the positive side? If that is true the common is nothing more than the positive side of the battery? Y or N. Have not done much work with mosfets which is why I ask.
          Thanks much for your help.

          Reply
          • Swagatam says

            yes, that’s correct, even in a relay you have one of the load terminals connected to + or the – line.

            Reply
            • Joel says

              Ok I thought so just wanted to make sure I was seeing things right. Thanks again

  14. Jeff says

    Swagatam: Thanks mate. In my 50+ years of work in the IT industry I have been taught many a lesson by the old “240”. Many faulty power switches, many incorrectly wired extension cables, many mistakes by unknown “dodgy” sparkies. I am still here, but I wonder if the fibulation I suffer from can be attributed to those “shocking” events…
    Jeff

    Reply
    • Swagatam says

      That’s awesome mate, I appreciate your knowledge and help! Please keep up the good work!

      Reply
  15. Evans mworeh says

    Sir can you give a system for a full bridge topology using 4 N channel mosfets?

    Reply
  16. Evans mworeh says

    Is this system suitable for powering inductive loads? The reason why I am asking this is because,power is rectified to DC before reaching the load.

    Reply
    • Swagatam says

      All types of AC load can be powered, please see the arrow diagram.

      Reply
      • Evans mworeh says

        I can see the arrows show that power passes through the forward biased diodes in either cycles,this will in turn be rectified to DC. Am I right or wrong?

        Reply
        • Swagatam says

          The DC is rectified and used for the gates, not for the load. The load will be switched with the input AC only.

          Reply
      • Evans mworeh says

        If I connect the gates directly without a switch,is there any issues?

        Reply
        • Swagatam says

          The MOSFETs and the load will be permanently switched ON then.

          Reply
  17. Jeff says

    Thanks you for publishing this, however I must make one very important point:
    When dealing with the AC Grid mains, you must provide some sort of isolation between the low voltage trigger input and the mains. Maybe an opto-isolator or a reed switch?
    Also, what is the purpose of the second circuit (with the 12 volt DC power supply?) How do the mosfets get triggered to control the AC load?
    Note, that circuit requires isolation for the low voltage DC as well.

    Thanks

    Reply
    • Swagatam says

      Thank you, the SSR explained above is designed to derive the gate voltage from the load itself, not from any external source, therefore isolation is not necessary. Moreover the entire configuration is supposed to be enclosed inside a sealed module. Also the Opto coupler would require more current than a 100k resistor so that may not be appropriate for an SSR design.

      The first two designs are for explaining the concept and the conduction path of the module. The battery sign at the mosfet gate shows a biasing DC input for the gate.

      Reply
    • Swagatam says

      I am sorry…Yes an opto coupler should be added for an external low current switching, in my earlier diagram I completely missed that if the gates were connected with the 100K/zener output that would keep the SSR permanently ON. I have changed the design now for your reference.

      Reply
  18. Evans mworeh says

    Sir swagatam,any diagram with p channel mosfets?

    Reply
    • Swagatam says

      pchannel is not recommended due to its higher resistance compared to N channel

      Reply
  19. Evans mworeh says

    Hello sir swagatam,help me understand;
    1. 100k and zener act as voltage divider/step down?
    2. I understand that during the positive half cycle,power goes through D1,100k zener and D3 back to load, so my question is, Can 12v zener diode and 100k hold a huge load say 5000w load ac?
    3. Can I this idea be used to switch more than 200A DC? How?
    4. Can this circuit be used as a 12v step down for inverter change over? Also how?

    Reply
    • Swagatam says

      Evans, the 100k and the zener are for biasing the MOSFET gates, the load is switched through the MOSFETs and their body diodes and therefore can be dimensioned for switching any desired load, please read the article carefully.

      Reply
      • Evans mworeh says

        Understood sir swagatam,I think was missing a point but got it. Now please correct me if I’m wrong, is it true that the load will be powered by dc since power goes through the mosfets internal body diode? If so , isn’t there any way this can be made to pass ac to the load?

        Reply
        • Swagatam says

          The devices will conduct alternately for switching the AC input, please see the arrow diagram

          Reply


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