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
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:
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.
- It powers the AC load through the MOSFET/Diode SSR configuration.
- 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.
- The rectified DC is further terminated as an auxiliary DC output which could be used for powering any suitable external load.
If you find the above concept having difficulty in switching, you can test the following improved SSR driver concept using an op amp
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.
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
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.
Hi Steven,
1) your first assumption is correct.
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
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
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.
What about zener diode?
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.
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
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.
Here’s the original link from where I referred the concepts:
https://www.ti.com/lit/ug/tiduc87a/tiduc87a.pdf
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.
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?
Glad you liked it, I think the last concept would be suitable for interfacing with an arduino
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.
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.
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?
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.
You mean -12 connected to 240ac? Also if the gates are left floating will it switch of?
Tks for your help
yes that’s right….
gates must not be kept open…
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?
Can be difficult to judge without seeing your circuit practically??
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.
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/
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.
Glad you are enjoying my site, please keep up the good work!
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.
yes, that’s correct, even in a relay you have one of the load terminals connected to + or the – line.
Ok I thought so just wanted to make sure I was seeing things right. Thanks again
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
That’s awesome mate, I appreciate your knowledge and help! Please keep up the good work!
Sir can you give a system for a full bridge topology using 4 N channel mosfets?
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.
All types of AC load can be powered, please see the arrow diagram.
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?
The DC is rectified and used for the gates, not for the load. The load will be switched with the input AC only.
If I connect the gates directly without a switch,is there any issues?
The MOSFETs and the load will be permanently switched ON then.
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
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.
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.
Sir swagatam,any diagram with p channel mosfets?
pchannel is not recommended due to its higher resistance compared to N channel
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?
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.
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?
The devices will conduct alternately for switching the AC input, please see the arrow diagram