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 internal body diodes across the drain/source of the respective MOSFETs.
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 drain/source 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.
Here's an excerpt from the datasheet of the article.
Video showing the testing of the above SSR circuit
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 internal 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.
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
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 the internal body diodes and therefore can be dimensioned for switching any desired load.
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
Sir swagatam,any diagram with p channel mosfets?
pchannel is not recommended due to its higher resistance compared to N channel
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.
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.
Sir can you give a system for a full bridge topology using 4 N channel mosfets?
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!
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
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, 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 mosfets with more number if parallel mosfets
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
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.
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.
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) 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
Hi Swagatam
I have been looking for ideas to allow the use of FETs on AC lines, and these ideas have sparked my interest.
But isn’t Stephen correct that both FETs will be switched on simultaneously, and the (FET) diodes are redundant as FETs conduct current on both directions? (I am aware that TI indicates current flow though a FET diode, but with both FETs on the diode would be bypassed.)
Diagram Labelled ‘The Solution’
I am a little confused about the voltages on the 741 OpAmp pins. The capacitor will have the peak AC voltage minus 1 diode voltage drop, and a little ripple depending upon current draw. The OpAmp power supply voltage will be around 15V, limited by the zener. The non-inverting input will have the same voltage as the positive power supply. The inverting pin will have the rectified and smoothed AC line minus a couple of diode voltage drops. For a 230Vac supply, that peak would be 325V.
Forgive my stupidity, but I do not understand the operation of the OpAmp, the voltages on the pins of the OpAmp and timing of the FET switching.
Take care,
Mike
Hi Michael,
I have tried to explain what I have studied and understood in the ti.com article, so these are not my own ideas. You are correct, after going through some online artciles it seems the mosfets act like low value resistors, and the current pass through the mosfets rather than their body diodes.
But then I have a few questions which are as follows:
1) If the diodes are not used then why are they shown inside most mosfets? What s their actual use?
2) During an avalanche reverse current flow through mosfets why these diodes are referred to as the protective elemts, which protect the mosfets from avalanche current.
3) If mosfets can conduct either ways without the body diodes then what is the need of using two mosfets for the bidirectional SSR, why not do it using a single mosfet?
To answer your second question….the peak across the zener diode will never reach 310V even if 230V is used as the AC input. It will be always restricted by the zener diode voltage, because the moment the zener voltage reaches 15V, 741 pin3 becomes higher than its pin2, and the 741 output becomes positive, this switches ON the mosfets which conduct and bypass the current to the op amp and the zener, cutting off the voltage to the op amps until the next cycle arrives.
Hi Swagatam,
The diodes are intrinsic to the MOSFET, i.e. they are created as part of the manufacturing process. So they are present whether we like it or not. That is why two MOSFETs are placed back-to-back, so that one intrinsic diode is reverse biased on the positive half cycle and the other is reverse biased on the negative half cycle ensuring the FET switch combination does not conduct until switched on via their gates.
If 1 MOSFET was used then its intrinsic diode would be forward biased on one of the half-cycles, no matter which way around it was connected. If there were no intrinsic diodes then only 1 MOSFET would be required.
The intrinsic diodes are often of use when switching inductive loads, such as in an H-bridge. An external protective (free-wheeling) diode can be omitted but sometimes the intrinsic diode does not behave as well as an external diode so an external diode can be used in addition to the intrinsic MOSFET diode.
As both sources are connected together and both gates are connected together, then the gate-source voltage is always the same on both transistors. Therefore they are both either switched on at the same time, or off at the same time.
Thank you for the explanation of the OpAmp voltages. Would it be possible to add some diagram of the voltage waveforms, so idiots like me can see this better?
Take care,
Mike
Thank you very much Mike,
That’s very well explained, and has cleared all my doubts. Nevertheless I wouldn’t change the diagrams in the article, as it somehow helps to understand the conduction process of the mosfets with the help of the intrinsic diodes, and moreover modifying the article would be a lot of hard work for me.
Regarding the op amp question, I think you are correct, due to the presence of the 10uF capacitor, the voltage across the DC line would eventually reach 310V peak, and this would also reach the pin3 of the IC 741.
To rectify the issue we could probably do a few changes in the diagram.
We can allow pin7 to get the 15V from the zener, clamp the pin2 with a separate resistor and a 12V zener diode, so that pin2 now is fixed with a 12V reference, and next add an additional 15V zener across pin3 and ground.
Also the 10uF filter capacitor will now need to be rated at 400 V.
The waveform now can be a 310V DC across the DC out lines, and a AC 230 V across the AC load.
Do you think it makes sense now?
Hi Swagatam,
Thanks for the work you have done on this.
But I am horrible when it comes to ‘visualising’ signals.
I will ask one question.
What is Ground?
That means different things, in different circumstances.
I would argue here that ‘Ground’ is whatever voltage is on the FET sources. And this voltage is the mains input, which is altering between +325Vpeak and -325Vpeak (for 230V RMS).
And I don’t think we can say that ’15V is applied’ to OpAmp pin 7. All we can say is that pin 7 will have a voltage 15V greater than the FET sources.
As the voltage on the FET sources is AC – varying from -230V (RMS) to +230V (RMS), then the voltage on pin 7 of the OpAmp will (or might) also vary from -230V +15V (325peak+15, or 340V) to -230V +15V (-325+15, or -310V).
But…
But the voltage cannot be greater than the input peak voltage, so the voltage on pin 7 of the OpAmp can never reach +340V. (Which also means that the gate voltage cannot be 15V greater than the source to turn on the FET at this point.)
And when the input (AC) voltage swings negative, the voltage on the sources will go to around -325V peak. With +325V still on the capacitor…
Well, I’m not sure pin 3 of the OpAmp would like that very much.
And when the FETs are off, what voltage is on the sources?
But I might be missing something. I often do.
But there must be a way to create a gate switch on voltage that would track the varying voltage on the sources. Hopefully a simple way. Hopefully a way that does not need to be ‘topped up’ at regular intervals, because this would interfere will a configuration where the FETs need to be switched on for a long time.
Take care,
Mike
Thanks again Mike for your detailed explanation and expert hints on the subject.
Yes the ground here refers to the negative DC line which joins the sources of the mosfets.
I designed the op amp based ssr by referring to the ti.com suggestions, in which they have said that the 4 diodes work like a bridge rectifier, regardless of whether the mosfets are turned ON or OFF. And I assumed that the DC from this bridge rectifier could be used to power the mosfet gates.
I am actually having difficulty understanding and simulating your explanation regarding the voltage swings across the various points in the circuit.
However after reading you explanation I now feel that there are certainly some confusions in the op amp ssr design and the design may have some flaws.
As you said there should be a simpler way of creating the SSR design, after some thinking I could come up with the following simple design. I hope this should do the job as expected from an SSR. Let me know your thoughts on this.
Hi Swagatam,
I took a quick look at the link you provided (https://www.homemade-circuits.com/wp-content/uploads/2022/03/basic-ssr-theory.jpg).
Might an idiot like me make a few comments?
1) The text seems to make two claims that contradict each other
(a) The control voltage is provided by the FETs, in conjunction with two separate diodes
(b) The control voltage is used to switch the FETs
If there is no control voltage (at the beginning) to switch the FETs, how can the FETs create the control voltage?
2) In your mind, remove the 2 FETs. The remaining 2 diodes form a full-wave rectifier. The FETs are not needed for this, despite what the text suggests.
3) Both FETs switch together, as both gates and both sources are connected. Therefore the FETs would short out the AC supply when switched on.
But perhaps I am just being particularly stupid today.
Thoughts on your new design…
1) There could be a large inrush current if the supply was connected (switched on) at the peak of the AC waveform. A small value series resistor – maybe 100 Ohm 1/2W – might be useful
2) it might be good idea to put a bleed resistor across the 400V capacitor, to bleed off any stored voltage should the supply be disconnected (switched off) at or near the AC waveform peak.
3) The ‘Neutral’ of the AC supply is at or near the ‘Earth’ voltage. (These are connected together at the point where the mains enters the building.) Connecting the FET sources to another ‘Ground’ and connecting this to the bridge output ‘ground’…
Well, I am not good at visualising waveforms as I said earlier, but this doesn’t look quite right.
Hi Michael, You are right the FETs would switch together and short out the AC supply, so that makes the working even more confusing with respect to the 4 diodes. Honestly I am finding it difficult to simulate the whole process in my mind.
Regarding the new design, I was also worried about the inrush current that’s why I used a small value 0.1uF capacitor which might create an inrush only for microseconds and the 1 watt zener diode might be quite effective in neutralizing it, but as you said adding a 100 ohm resistor would be safer option. A bleed resistor across the capacitor also recommended in order to discharge the capacitor and safeguard the user from a possible momentary shock, when the unit is unplugged from the mains.
I am quite confident that this circuit will work. According to me the bridge rectifier does the trick here and it effectively creates a separate DC for the MOSFET gates. I wish somebody builds it and tests the results.
Hi Swagatam,
Instead of trying to visualise signals and waveforms, let’s create an imaginary circuit.
Let the upper AC input be the AC Live (Hot) and the lower AC input be the Neutral.
Let there be a load that takes 1A.
Let the ‘on’ resistance of the FETs be 0.1 Ohm.
Therefore the drain-source voltage drop will be 0.1V (when ON).
As the AC supply voltage swings between -325Vpeak and +325V peak, this 0.1V drain-source voltage drop is effectively nothing.
That means the voltage on the FET sources is the same voltage as the mains Live, assuming Live (or ‘Hot’) is connected this side.
The bridge ‘ground’ is connected to the FET sources. So the bridge ground is as the same potential as the AC Live. Further, the bridge +15V output must be 15V greater than the bridge ground, or +15V greater than the supply, to be able to turn on the FETs.
But this cannot be.
Now let’s make the upper AC input the Neutral, and the lower the Live. (Swap the 400V capacitor onto the ‘new’ Live also.)
The FET sources are now almost at Neutral (0V) potential.
The bridge output ‘ground’ is now 1 diode voltage drop higher than Neutral. If connected together, large currents can flow between the bridge ground and the FET sources. (Maybe this could be ameliorated with the use of a diode-resistor between these.)
So, this might work if use as a ‘low-side’ AC switch, i.e. if placed on the Neutral.
But, please, please, please, don’t take my reasoning as the law of the land.
These are just my ramblings.
I, myself, have been considering how to achieve this another way.
What we really need is an independent, ‘floating’, power source, where one side can be
connected to the sources and the other switched onto the gates as needed.
Would it be possible to use a transformer? A very small 230V-9V transformer? The rectified, smoothed, output of such a transformer would be around 12V, which would be perfect. The current required would be very small. Maybe such a transformer could be the size of an 8-pin DIP (such as a dual comparator or OpAmp), or maybe even smaller.
Research has, so far, shown that small magnetic cores seem to be all ferrite, which might not be the most suitable material for a 50Hz transformer. So I got to thinking…
How about stacking steel washers and using this as a core? Basic checks show that washers with a 3.2mm ID, 7mm OD and thickness of about 0.5mm are available. 50 of those stacked would be about 25mm (1″) tall, and about 7mm in diameter. The inner diameter would allow about 200 turns of 0.1mm wire without overlapping.
So…
Any thoughts Swagatam?
Take care.
Mike
Thanks Mike, I read your explanation, it makes sense and seems to be correct. I understand the working to a great extent now.
Surely a transformer based DC can be used and designing it is quite easy. But whether it is a ready-made transformer or a custom made transformer, these can be quite cumbersome and bulky. It is not required according to me.
An SSR is supposed to be very sleek, light weight and compact, so a transformer may not be a recommended option here.
A capacitive based power supply looks a better option to me.
Hi Swagatam,
Yea, transformer *can* be bulky. But the power requirement of the transformer would probably be less than 100mVA, and maybe a tiny fraction of that. It would be interesting to see how small this could be fabricated. If it could be made with the same PCB footprint as an 8-pin DIP or a bridge rectifier, or even smaller, then the solution would be no larger than those that use an OpAmp or a bridge rectifier.
I need a pair of switches to isolate the output of a grid-tie inverter from the grid. Unfortunately the bridge-zener solution would not work for this.
Take care
Mike
Hi Michael,
I agree a transformer would be a very clean method of implementing the ssr switching, and also it can be designed to be very small in size, but still we would have to calculate the number of turns, which could be a lot complicated, unless we do it do through some trial error.
If you know how to accomplish the number of turns then the problem can be quickly solved.
Hi Swagatam,
Sorry for not responding sooner, but I have been brutally tired of late.
I had hoped, based on your earlier responses, that you might be best placed to look at such a transformer.
Such a transformer – if possible – would also allow the isolated monitoring of mains level voltages by lower voltage control circuitry, which might be useful.
By the way, the ‘Reply’ button to your latest comments has disappeared, and I am using the ‘Reply’ button from earlier comments to respond.
No problem Michael,
Unfortunately I do not have the information regarding how to build a small low current transformer. I have a related article but that involves E/I core and the whole procedure looks quite complex. I think only a professional transformer maker would be able to provide a proper solution to this problem.
How to Make Step Down Transformers
The comment section is set to hold only 6 threaded comments, so may be after 6 replies the reply button disappears.
Hi Swagatam,
Perhaps it could be thrown out to the community for suggestions?
Spec:?
A toroid based small, low power transformer 230V – 9V 50Hz, maybe around 50mVA
Toroids can be stacked. The height of the resulting transformer can be as high as the tallest other device and it will take up no more enclosure space. (In my case, I have FETs mounted onto 50mm high heat sinks, with an overall height of about 60mm. So ‘small’ stacked toroids could be up to 60mm high and take up no more board space than a single toroid.)
Identifying the most suitable toroid material might not be easy, and is certainly outside my areas of expertise.
All the best.
Take care
Mike
Thanks Michael,
Yes I hope somebody will know how to build small customized transformers, and guide us through.
A alternative simple way is perhaps to stack up many small iron washers and wind around 300 turns for the primary and 20 turns for the secondary, using a thin copper wire, and then adjust the turns by checking the results.
Iron core will be most suitable since the AC mains frequency we use is 50 or 60 Hz.
Hi Swagatam,
My winter vanished over the space of a couple of days, and it is now time for me to put aside my winter projects and take back up with those outside. I have much to do there.
However, over these warmer months, I hope to be able to spend a little time ordering the items I would need for testing the concept, including the item needed for an inductor current saturation tester.
Perhaps we’ll meet again next winter.
Best wishes.
Mike
Thanks for the update Michael, hope you enjoy the summer months. Please keep up the good work.
Hi !
What is AC load?
Any appliance…
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!
Thank you friend, you can use the second circuit explained in the following post:
https://www.homemade-circuits.com/fan-speed-controller-for-heatsink/
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
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
Thanks Seun, you can try employing the following design:
Solid State Relay (SSR) Circuit using MOSFETs
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.
In that case a single FET should satisfy the SSR conditions, as it can conduct in both directions?
Commenter is correct … MOSFET Drain-to-Source is directional (and if you think of it in terms of the Field Effect on the device Channel, it makes sense that the channel cannot sense the applied polarity) … BUT … the intrinsic diode is what prevents a single device from functioning as an bi-directional SSR. When the SSR is ON, both MOSFETS conduct bi-directionally. When the SSR is OFF, it relies on whichever intrinsic diode is reverse-biased to block, since the other will be no help (forward-biased). Which FET’s diode does the blocking depends on the polarity being applied to the SSR.
Hi Swagatam,
Do we need snubber circuit for this model (the last schematic)? I dont have sufficient knowledge to explain myself when the load is inductive and does spike kill the Mosfet?
Ngoc
Hi Dang, I don’t think a snubber would be required as long as the reverse diodes are rated adequately high. Still for better safety you additionally add a 1uF/40V capacitor right across the inductive load terminals.
Thanks Swagatam,
I am thinking of making a home switch with back2back Mosfet to turn on up to 700W AC loads (any type). Since I was noticed that MOSFET in saturation mode will have very low resistance. Will it be possible to make mosfet relay without heatsink?
It seems that MOSFET relay should be more expensive than triac Relay?
THank you very much for your attention!
Hi Dang, MOSFETs will heat up when the load wattage is high enough….so if your load wattage exceeds the MOSFETs optimal threshold, then the device will heat up.
Yes MOSFET SSR is expensive than triac SSR.
Hi
you have great ideas.
need you help to design bi directional switch (MOSFET) . i need to pass thru in both directions signal (analog) that might have spikes up to 300V ,RDS On for both MOSFETS together is less than 10 Ohms and SW time is less than 80uS.
will appreciate your help
Thank you! the required circuit is already explained in the above article….you can customize it according to your own needs!
thanks, my problem is that i dont know which components to use in the design.
like part numbers of components and which design to select (i think that the NMOS because its low RDS On)
will appreciate any help here 🙂
many thanks
eli
You can try the first circuit initially using a battery for the gate/source switching, which will confirm the basic working of the bidirectional mosfets. You can use IRf540 for the mosfets.
If everything works fine, then you can assemble the last circuit and check the response!
many thanks !
Hi Swagatam,
Looking at the back-to-back MOSFET relay for AC load you have provided in this example, How do you think about its advantage over Triac Relay on Inductive/Capacitive load? Can MOSFET relay survive without a snubber circuit if an inductive device is wired up to the load size?
Thank you very much for your attention.
Dang Dinh Ngoc
Hi Ngoc, the MOSFETs already have an in built diode so the effect of reverse EMF is minimal for these devices, moreover today’s mosfets are much more advanced with their protection features than the triacs.
Hi,
What voltage the capacitor should be rated?
I simulated your circuit, and V on capacitor terminals raises almost up to the AC source’s peak.
Is this right or have I something wrong in my simulation?
I have some more questions if you what kindly answer:
1. Why the capacitor is only 10uF in the circuit with de op amp?
2. Why you lowered the resistor over the zener from 100K to 47K?
3. What does the new 100K resistor going to pin3 of the op amp do, is just for discharge the capacitor?
Thank you.
Hi, yes that’s right, the output voltage may be almost equal to the input peak voltage, however this may not be true also, because the moment 15V is reached across the filter capacitor, the MOSFETs would conduct causing the voltage across the capacitor to drop to almost zero. So the effective voltage across the output could be 15V DC
1) I reduced the value to 10uF since the charge value will be sufficient to hold the pin3 voltage to 15V for many milliseconds. Also this 10uF voltage can be increased upto 300V if required.
2) Because the current through 100k may not be sufficient for the zener to conduct.
3) It keeps the pin3 of the op amp referenced to the relevant supply line which is a standard practice for all opamps and CMOS ICs.
Hello mr. Swagatam, no comment but a question.
Considering usage of your basic bidirectional SSR circuit, I take it that there N- Mosfets are being used.
Checking my ham shack spare parts I only find P-Mosfets.
My question: Is it OKay tot use these P-Mosfetd instead with the only difference to reverse the polarity of the Gates trigger signal ?
If not, what more should be done ?
73 PE0DVD
Hello Ad, yes it is possible to use P channel mosfets also….you can connect them in the following manner, and see how it responds:
Good morning mr. Swagatam,
Thank you very much sir
71 Ad
You are welcome Ad.
Dear Sir!
Can the switching of a mechanical morse be replaced by FETs?
To prevent the 2 connected parts from opening each other during ON-OFF? Thanks for the reply!
Hi Jozsef, yes, FET or BJT can be tried to replace the morse mechanical contacts.
Swagatam, is there way to insure that switching always occurs at a 0V crossing (or near zero crossing)?
Hi Wes, which circuit are you referring to?
Can we use this circuit as a Lithium-Ion Battery Cutoff switch? If we can do, then which changes would be required in this circuit.
I am not sure how the above circuit can be used as a Li-Ion battery cut off switch.
hi dear sir i was looking a circuit that i can control through a 0 to 10 volt input,such as a PLC output .that i can creat variable voltage based on triac device with 220 voltage
i look forward to help you
Hi Sedigh, I do not have proper knowledge about PLCs, so it might be difficult for me to understand the concept and solve your query.
hello dear i do not mean circuit relation to PLCs
it was only to convey the meaning of the my request.Lamp light control circuit with 220 v and through triac but intead of using a potentiometer,use an input voltage 0 to 10 volts to control the light of the Lamp
Hello Sedigh, understood! You can try the following circuit, it should work: