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I want to switch 50V AC voltage. Maximum drained current will be 5A. Frequency is 50Hz. Switching speed is not important, can be real slow, it is not a problem in my application.

I wanted to use solid state relay at first for this purpose. But as soon as I started to search an SSR, I saw that their prices are too high. For a cheaper alternative solution, I want to use MOSFET transistors (can be a different transistor type as well) instead of solid state relay.

Can you suggest me a MOSFET equivalent circuit of the solid state relay with the specifications I gave above?

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Three ways of making an SSR follow :

The 1st two use FETs and can be switched off and on throughout an AC cycle as required. Switching speed need to be understood. The floating gate versions have an RC time constant that controls turnoff unless extra care is taken to avoid it.

The TRIAC circuit turns on when fired and off at the next zero crossing. It can be fired as soon as the zero crossing has passed but again, can then not be turned off until the next zero crossing. So you can get whole-half-cycles or part half cycles extending from a firing point to the end of that half cycle. Inductive loads complicate this slightly but are outside the basic discussion.

(1) Place a MOSFET inside a 4 diode bridge as the "load". Ac to bridge AC input is "shorted " = on for AC when FET is on Gate is floating so you need to get voltage to gate. Not hard but needs thought. Rough diagram - better later maybe. Transistor shown here is bipolar but MOSFET does same job. MOSFET always sees DC. Load sees AC switching. Drive gate with opto. Derive power by eg resistor feed from drain to a reservoir cap to drive gate via opto.

enter image description here


(2) Two eg N channel MOSFETs in series - connect source to source and gate to gate. Inputs are 2 x drains. Drive gate +ve to source to turn on. Gates to source to turn off. Again, gates and sources float so you need to get drive to them but not hard - just needs thought.

enter image description here


The circuit diagram below shows an example of a practical implementation of this principle.
Note that the FETS are both N-Channel and that the Sources of both FETs are connected and the Gates of both FETs are connected. This circuit works because MOSFETS are two quadrant devices - that is, an N channel FET can be turned on by a positive gate realtive to source regardless of whether Drain to Source voltage is +ve or -ve. That means that the FET can conduct "backwards" if driven in the normal manner. Two FETS are required connected in "anti series" (opposite relative polarity) because of the "body diode" inside each FET which conducts when the FET is biased oppositely to usual. If only one FET was used it would conduct when the FET was turned off when Drain was negative relative to source.

enter image description here

Note that "isolation" and level shifting of the on/off signal to the floating gates is achieved by the 2 x 100 pF capacitors. Consider the circuitry at right as potentially at mains potential. The right hand 74C14 forms an oscillator at about 100 kHz and the two inverters between them provide opposite polarity drive via the 2 capacitors to the 4 diodes which form a bridge rectifier. The rectifier provides DC drive to the floating FET gates. The gate capacitance is probably ~ a few nF and this is discharged by R1 when the drive signal is removed. I'd guesstimate drive removal would occur in tenths of a mililsecond but do the calculations yourself.

The circuit is from here and notes

  • The circuit uses an inexpensive C-MOS inverter package and a few small capacitors to drive two power MOS transistors from a 12v to 15v supply. Since the coupling capacitor values used to drive the FETs are small, the leakage current from the power line into the control circuit is a tiny 4uA. Only about 1.5mA of DC is needed to turn on and off 400 watts of AC or DC power to a load

(3) TRIAC CIRCUIT

You specifically mentioned MOSFETs.
A TRIAC is also commonly used in AC SSRs.
Below is a typical TRIAC circuit.
L1 may not be used.
C1 & R6 form a "snubber" and values depend on load characteristics.

enter image description here

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  • \$\begingroup\$ 5A x 2 diode drops = 1 hot bridge. Neat trick, just not sure I'd want to push all that much current through it. Anyway +1 on you. \$\endgroup\$ – JustJeff Aug 31 '11 at 23:56
  • \$\begingroup\$ I think this should be split into multiple answers \$\endgroup\$ – endolith Sep 1 '11 at 5:28
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    \$\begingroup\$ @endolith - serious question - why split it? This is all on his topic AND forms a single point resource for others. \$\endgroup\$ – Russell McMahon Sep 1 '11 at 6:10
  • \$\begingroup\$ @Russell McMahon: Thank you very much for sparing your effort of writing this long answer. My circuit will switch the voltage on the secondary winding of a transformer. This voltage will directly be switched to a filter (capacitor), then to a regulator stage. In your circuit (1), can I use a PNP BJT or P-channel MOSFET, and connect the filtering capacitor (and the regulator stage parallel to it) on the low side? In your circuit (2), what did you mean by "gates are floating", gate driving looked OK to me. What improvements should I do on that circuit? Thank again, and thanks in advance. \$\endgroup\$ – hkBattousai Sep 1 '11 at 7:14
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    \$\begingroup\$ @RussellMcMahon: Because there are several different solutions here. They should be in different answers so we can vote each one up or down and comment on each independently. \$\endgroup\$ – endolith Sep 1 '11 at 13:33
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Solid state relays are opto-coupled back to back SCRs in their simplest form. You can duplicate that yourself, but it gets a bit messy. Since solid state relays are opto-isolated, the output side can float with respect to the input side, just like a real relay.

If you really need isolation, then it gets complicated to do this yourself. You say switching speed is low, so why not a regular mechanical relay?

If you don't need isolation, then there are various possibilities. One is to use a triac and control it directly from your circuit. For details, we need to know more about how this 50V AC is referenced (or not) to whatever power supply you have available.

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  • \$\begingroup\$ 5A, 50V, and not particularly fast? I agree, at least consider going mechanical. \$\endgroup\$ – JustJeff Sep 1 '11 at 0:03
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You're absolutely right, SSRs are expensive. The most simple alternative is to roll your own using an opto-triac + power triac:

SSR schematic

This costs 80% less than the equivalent SSR.

The MOC3041 switches at the voltage's zero-crossing, so that may be an advantage. If you don't need that, the MOC3051 is a random switching opto-triac. A disadvantage of using a triac may be that there's a voltage drop of a few volts, and when the voltage to switch is just 50V the loss is more in comparison than for for instance 230V.
A MOSFET as switching element may sound like a better idea, but if you use it in the bridge like in Russell's solution you'll have about the same voltage drop anyway, but this time across the diodes.

The best solution regarding voltage drop is the good old electromechanical relay. Depending on the type of load you'll have to derate the relay, so that for switching 5A you may need a 16A version. Price for the 16A relay is comparable to the DIY SSR.

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  • \$\begingroup\$ Why do we use such a complicated circuit instead of using just a triac? For electrical isolation between the control signal and the circuit to be controlled? Is there any other reason? \$\endgroup\$ – hkBattousai Sep 1 '11 at 7:06
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    \$\begingroup\$ @hkBattousai - hey, come on, it's not that complicated! :-) The MOC3041 takes care of the electrical isolation, but these opto-triacs can't handle the high currents you want to switch, so that's why it's used to switch a second triac which does the actual hard work. (The NAND gate at the left is not necessary if your \$\mu\$ can drive the opto-triac's LED directly.) \$\endgroup\$ – stevenvh Sep 1 '11 at 7:12
  • \$\begingroup\$ Note that the minimum drive requirements for such optos must be met or they may not switch at all. Designs should always meet worst case specs but sometimes things work when you don't. Im this case the difference can be 100% dead below required drive level. \$\endgroup\$ – Russell McMahon Sep 2 '11 at 0:20

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