# Controlling an AC load with a MOSFET

#### Holz1

Joined Jan 21, 2020
58
I came across this post:
Controlling an AC load with a MOSFET

Can someone explain how this works in simple words or with a circuit diagram.
I don't understand how this works.
I want to switch a modelrail Märklin M turnout, 2 solenoids 16 Volts AC current, with a TWIN COIL SWITCH machine driver board which is built for DC turnouts or modelrail semaphores twin solenoid wingsignals, 12 Volts DC.
The twin coil driver board is TTL 5 Volts DC containing a SN74LS123 One Shot MVV.
greetings
Henry

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#### LowQCab

Joined Nov 6, 2012
4,312
What type of Control-Voltage will be used to turn the "AC-Switch" On and Off ?
( and, will an AC-Voltage ever be used to activate the AC-Switch ?)

What is the Maximum-Current, and Maximum-Voltage, that will be switched via the "AC-Switch" ?
.
.
.

#### Holz1

Joined Jan 21, 2020
58
What type of Control-Voltage will be used to turn the "AC-Switch" On and Off ?
( and, will an AC-Voltage ever be used to activate the AC-Switch ?)

What is the Maximum-Current, and Maximum-Voltage, that will be switched via the "AC-Switch" ?
.
.
.
The resistance of the turnout solenoids is approx 60 Ohm (between the solenoid INPUT leads), the max voltage 16 V AC :
To switch the turnout approx 16V~ 500mA

#### BobTPH

Joined Jun 5, 2013
9,300
That is an interesting technique, very clever.

To understand how it works, draw on your knowledge of how a diode bridge works to convert AC to DC. In one half cycle, the current flows through two of the diodes. On tje other half, it flows through the other two diodes, which reverse the direction, so the + output is always positive with resoect to the - output. In that application, the load is in the DC part of the circuit.

Now, take the load and put it in the AC part of the circuit, and simply short the DC side of the circuit with a MOSFET. The current flows the same way, as AC through the load, but as DC through the MOSFET.

The only drawback is I see is that the AC will be distorted, but this will not matter for many loads.

#### LowQCab

Joined Nov 6, 2012
4,312
What type of Control-Voltage will be used to turn the "AC-Switch" On and Off ?
( and, will an AC-Voltage ever be used to activate the AC-Switch ?)
.
.
.

#### Holz1

Joined Jan 21, 2020
58
That is an interesting technique, very clever.

To understand how it works, draw on your knowledge of how a diode bridge works to convert AC to DC. In one half cycle, the current flows through two of the diodes. On tje other half, it flows through the other two diodes, which reverse the direction, so the + output is always positive with resoect to the - output. In that application, the load is in the DC part of the circuit.

Now, take the load and put it in the AC part of the circuit, and simply short the DC side of the circuit with a MOSFET. The current flows the same way, as AC through the load, but as DC through the MOSFET.

The only drawback is I see is that the AC will be distorted, but this will not matter for many loads.
Is this diagram correct?

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#### Holz1

Joined Jan 21, 2020
58
What type of Control-Voltage will be used to turn the "AC-Switch" On and Off ?
( and, will an AC-Voltage ever be used to activate the AC-Switch ?)
.
.
.
The turnouts are switched with 16V AC: the Yellow port of the trafo is power supply for the turnout (16V AC); 2 blue cables connected to the solenoids and the BROWN trafo output AC is for switching.
When this is not the answer you expect, be more clear please

#### BobTPH

Joined Jun 5, 2013
9,300
No, the bridge is wired wrong.

The brown wire needs to go to the corner opposite the yellow wire.

#### BobTPH

Joined Jun 5, 2013
9,300
What circuit is controlling the MOSFET? It must have a common ground with the MOSFET source, and cannot be derived from the AC power. If it’s power is derived from the AC power, you need an opto-isolator and you still need a separate power supply for that.

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#### Holz1

Joined Jan 21, 2020
58
What circuit is controlling the MOSFET? It must have a common ground with the MOSFET source, and cannot be derived from the AC power. If it’s power is derived from the AC power, you need an opto-isolator and you still need a separate power supply for that.
See above. In order to be complete, diagrams are always helpfull, here the complete drawing. The only things I want to know is:
1. Can this be done with this circuit diagram
2. And when yes, how to attach the leads.
3. When no, what should be done
4. How to wire the bridge rectifier in this diagram
5. other suggestions are welcome
6. I posted links above of the TWIN coil driver: specs can be found here

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#### djsfantasi

Joined Apr 11, 2010
9,188

#### AnalogKid

Joined Aug 1, 2013
11,209
That is an interesting technique, very clever.
I'm not a big fan of it. Back when, it was less expensive than two power MOSFETs, especially if switching mains at a couple of amps. But I never liked the power dissipation in the bridge. It wipes the power savings / efficiency / reduced cooling requirement of using a MOSFET for the switch instead of a TRIAC. A power MOSFET might have less than 0.1 V across it, but there still is a 2 V drop in the bridge. I've been playing around with this and other circuits for controlling banks of fluorescent bulbs. The single-FET control circuitry is less complex, but that extra power dissipation bugs me, especially in small, enclosed spaces.

In this application, a round-number guess is that the power dissipation in the bridge is 100x the power in the FET. Not a problem because both numbers are so low, but still . . .

ak

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#### Holz1

Joined Jan 21, 2020
58
There are small 1 A and 2 A solid state relays in DIP packages using MOSFETS that should work in this application. OptoMOS

https://www.vishay.com/en/solid-state-relays/

Here is the schematic for the driver board:

https://github.com/SethNeumann/MRCS-Twin-Coil-Driver/blob/master/MRCS-Twin-Coil-Driver-Schematic.png

ak
Thanks Ak and all responders!
Oké, this post is clearing the "fog". Because AK pointed out to me to have a look at OPTOMOS i saw on the side HOME_made_circuits the attached diagram and tried to visualize this in my TWIN driver. PNG enclosed
As additional proof of this possibility :
https://www.renesas.com/us/en/produ...een-photocouplers-and-optical-coupled-mosfets.

I made some circuitdiagrams and gathered some information:
1. the diagram with the HOME_Ciruit layout connected to the TWIN driver board (named sheet 3)
2. the diagram of the PCB bareboard of the TWIN driver
3. the diagram with the diode placings and connections of the TWIN driver
4. PDF Home made circuits with the explanation of the circuit

I think the info is sufficient enough to have everybody satisfied and that this can lead to a simple solution.
If not, I would be grateful to receive the specs with the details. THANKS in advance

Note: the pulsed DC is achieved by the ONE SHOT MVV.
Brief explanation: I make use of Infrared gates along the railroad track. When the IR gatebarrier is passed, the signal is a "RISING EDGE" and triggers the 1st OUTPUT of the TWIN driver. ( the semaphore wingarm is switched). As long as the train is between the IR barrier (see last photo of the IR Led and Transistor) the logic level is HIGH. When the train leaves the IR gate, the "FALLING EDGE" triggers the 2nd OUTPUT and goes to logic level LOW and the wingarm is switched back. I have tested the circuit with the IR gate as shown in the photo. Because the SN74LS123 is an ONE shot, the high and low level is not a continuoussignal but a brief pulse of less then 1 seconds which is necessary for the solenoids not to burn

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#### AnalogKid

Joined Aug 1, 2013
11,209

1. The images on pages 3 and 4 are incorrect. It never is the case that one FET is on while the other is off. They are either both off, in which case the back-to-back diodes block all current; or both on, in which case both diodes are "shorted out" by their respective FETs. This is where the real efficiency gain comes from. The voltage drop across the FET channel could be 10%, or even 1% of the voltage drop across the diode.

2. It skips over the only difficult part of the approach - that little battery symbol between the sources and the gates. This is where there has to be an electrical relationship and physical connection between the DC-powered control board and the AC-powered load. At low voltages this is not a big deal, but at mains voltages it can expose a low-power circuit to lethal voltages.

Don't let my personal bias skew things here - As long as the AC lines to the solenoid are not doing anything else, such as powering the control board, then having the FETs on the control board switch the loads through bridges will work just fine and is simple, reliable, and cheap. You've already paid for the FETs, so let them do the work.

ak

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#### Holz1

Joined Jan 21, 2020
58

1. The images on pages 3 and 4 are incorrect. It never is the case that one FET is on while the other is off. They are either both off, in which case the back-to-back diodes block all current; or both on, in which case both diodes are "shorted out" by their respective FETs. This is where the real efficiency gain comes from. The voltage drop across the FET channel could be 10%, or even 1% of the voltage drop across the diode.

2. It skips over the only difficult part of the approach - that little battery symbol between the sources and the gates. This is where there has to be an electrical relationship and physical connection between the DC-powered control board and the AC-powered load. At low voltages this is not a big deal, but at mains voltages it can expose a low-power circuit to lethal voltages.

Don't let my personal bias skew things here - As long as the AC lines to the solenoid are not doing anything else, such as powering the control board, then having the FETs on the control board switch the loads through bridges will work just fine and is simple, reliable, and cheap. You've already paid for the FETs, so let them do the work.

ak
Thanks AK
re: 2 Yes, I understand that the AC lead must NOT be connected to the DC board.
I use 2 transformers:
1 Märklin 16V AC trafo for powering the turnouts (16V AC Yellow= for the lightning, BROWn= 0 , this contact for "mass" being the rails, Blue for switching over the MASS/rails.
2. 12V DC trafo for the TWIN board which will regulated back to 5V DC TTL.
The only connection is between the 2 powersupllies is the GND from the DIODES with the SOURCE of the MOSFET.
Is this a PROBLEM?
Must the SOURCE of the MOSFET disconnected from the Twin driver circuit?
I read your remark Battery symbol: Can this be meant as the "space" between the gate and the coupling of the Pmos and Nmos in the drawing? I read the article of Robert Keim (AAC) of 15th Mai 2024 When you lookat that drawing, that is also a "battery" symbol btween Vin and Pmos and Nmos

#### AnalogKid

Joined Aug 1, 2013
11,209
I read your remark Battery symbol: Can this be meant as the "space" between the gate and the coupling of the Pmos and Nmos in the drawing?
No.

That is meant to indicate a DC potential of somewhere between 4 V and 20 V (for most normal power MOSFETs). This is listed as Vgs on a datasheet. 20 V is a very common maximum value. The low voltage varies from part to part, and can change depending on how much current you are pushing through the FET compared to its maximum current rating. 10 V is often the minimum voltage guaranteed to fully enhance ("turn on") the FET. This is the condition where the channel resistance is at its lowest possible value, Rdson.

There is a class of FETs called "logic-level". For these parts, Rds reaches Rdson with only 5 V or less Vgs. Lotsa little acronyms there, but it's worth the time to learn how to speak FET.

ak

#### MisterBill2

Joined Jan 23, 2018
19,473
Is this diagram correct?
Not quite correct. One AC feed connection is in the wrong corner, where an output is already connected.

#### MrAl

Joined Jun 17, 2014
11,734
That is an interesting technique, very clever.

To understand how it works, draw on your knowledge of how a diode bridge works to convert AC to DC. In one half cycle, the current flows through two of the diodes. On tje other half, it flows through the other two diodes, which reverse the direction, so the + output is always positive with resoect to the - output. In that application, the load is in the DC part of the circuit.

Now, take the load and put it in the AC part of the circuit, and simply short the DC side of the circuit with a MOSFET. The current flows the same way, as AC through the load, but as DC through the MOSFET.

The only drawback is I see is that the AC will be distorted, but this will not matter for many loads.
That circuit has been around since the 1980's. I used that to drive a transformer one time because I did not like the way a triac worked for that. A triac can only be turned off at the zero crossing, while this circuit can be turned off any time. That does bring up its own problems however.

In pseudo theory, you can force the AC wave into a pulsed AC wave and therefore control the power getting to the load and at the same time keep the distortion down. The only problem is there has to be a snubber with it because the inductive kick back can be high and destroy some of the components.

In a regular PWM circuit that drives inductive loads, there is usually a return path built into it so that the kick back power returns to the filter caps and therefore stays lower. With this kind of circuit there is no filter cap so there is no easy way to recover the kickback energy.

The whole idea is that if you short out the DC side of a full wave bridge rectifier you short out the AC side too. That can be used as a switch with a unipolarity switching device like a transistor.

#### MisterBill2

Joined Jan 23, 2018
19,473
That circuit has been around since the 1980's. I used that to drive a transformer one time because I did not like the way a triac worked for that. A triac can only be turned off at the zero crossing, while this circuit can be turned off any time. That does bring up its own problems however.

In pseudo theory, you can force the AC wave into a pulsed AC wave and therefore control the power getting to the load and at the same time keep the distortion down. The only problem is there has to be a snubber with it because the inductive kick back can be high and destroy some of the components.

In a regular PWM circuit that drives inductive loads, there is usually a return path built into it so that the kick back power returns to the filter caps and therefore stays lower. With this kind of circuit there is no filter cap so there is no easy way to recover the kickback energy.

The whole idea is that if you short out the DC side of a full wave bridge rectifier you short out the AC side too. That can be used as a switch with a unipolarity switching device like a transistor.
Really, that circuit has been around since the early seventies, but as common then because the bridge modules were much less common way back then. That circuit avoided needing to drive back-to-back SCRs, so it was a real saving at the time.