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AC Voltage control using MOSFET

I am using a similar circuit to speed control a 60 W AC fan using phase control. Unlike a TRIAC, supply to fan is given at the start of the cycle. I thought it would minimize the switching noise usually heard in TRIAC controls.

PWM is from 0 to 10 milliseconds. In low PWM, the MOSFET heats a lot with an inductive load, but not with a resistive load. A snubber using a 0.1 µF capacitor and 100 or 39 ohm resistors is connected across the MOSFET's source and ground pins.

What should I do?

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    \$\begingroup\$ It might be that on an inductive load the phase relationship between voltage and current is 90 degree phase shifted i.e. when you trigger the MOSFET on a zero cross it is likely that the current is at maximum and switching losses are therefore greater. \$\endgroup\$ – Andy aka Jan 23 '14 at 13:21
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    \$\begingroup\$ Is this an inductive load or actually an induction motor? Quite different characteristics especially when the rotor is stalled (or at very low speed) : closer to a short circuit than an inductor! \$\endgroup\$ – Brian Drummond Jan 23 '14 at 14:08
  • \$\begingroup\$ @ Brian Drummond : It's a AC Ceiling fan 230V 50Hz \$\endgroup\$ – Mitz Jan 23 '14 at 15:02
  • \$\begingroup\$ check out this link : freescale.com/files/microcontrollers/doc/ref_manual/DRM039.pdf regards Dennis \$\endgroup\$ – user60179 Dec 10 '14 at 11:57
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See this AC PWM Dimmer for Arduino article on instructables, which says:

problems start, because he is feeding the gate from the MOSFET, with a voltage that is shorted by that same MOSFET. In other words, if the MOSFET is fully opened the DC voltage coming from the rectifier is completely shorted. Therefore there will be no voltage anymore to put on the gate and the MOSFET will block again. This effect might not be so outspoken by a low dutycycle (= lamp on a low intensity), because of the presence of C1, that will retain its charge for a while and will be receiving new charge thanks to the low dutycycle, but at 25-80% dutycycle the voltage on C1 just cannot be sustained anymore and the lamp may start to flicker. What's worse is that at moments that the voltage on the gate drops, for a while the MOSFET will be still conducting, but not be fully saturized: it will slowly go from its nominal 0.04 Ohm resistance to infinite resistance and the slower this goes, the higher the power that needs to be dissipated in the MOSFET. That means a lot of heat. MOSFETS are good switches but bad resistors. They need to be switched ON and OFF fast. Currently the circuit heavily relies on D1 to keep the voltage on the gate of T1 at acceptable limits while the voltage is swinging between 0 Volt and Full peak At peak the rectified voltage is 230x1.4=330V The average rectified voltage is 230x0.9=207V

If we forget about the smoothing effect of the capacitor for a while and presume the optocoupler to be fully open the average voltage on the capacitor would be 22/88 * 207 =52 Volts and in peak 22/88 * 330= 83 Volts. That it is not is because of D1 and the fact that the MOSFET will short the Voltage.

If the optocoupler is not in saturation and its impedance therefore infinite, the capacitor C1 would charge up to full rectified voltage if not for D1. On average 3mA will flow through R3,R4 and R5 (207-10)/66k which equals a power consumption of 0.6 Watt in the resistors R3,R4, R5

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First of all, this circuit cannot be used to control inductive loads. T1 is switched asynchronously with the mains frequency and this can cause DC current to flow. The reason that you can see this effect in low PWM is that the voltage across D1 remains the same (10 V) to about 90% of the duty cycle span. So T1 conducts a little longer than you would expect from PWM. At a higher duty cycle the voltage drops and T1 starts to conduct sufficiently.

In addition, the snubber dissipates power as heat. The snubber will have a different effectiveness at different frequencies. You need to choose the values for R and C to suit the frequencies you want to work with.

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  • \$\begingroup\$ @ GR Tech : Have any idea to modify this circuit for inductive load? \$\endgroup\$ – Mitz Jan 24 '14 at 14:14
  • \$\begingroup\$ OK. I can understand. This is a circuit with minimum components, cheap & good BUT for light switching since you don't know exactly when to switch load ON/OFF. An idea is to eliminate D1C1 and to add an inductor in series to the gate of T1, to turn on at 90 degrees where the voltage is at its peak..... OR replace T1 with a coolMOS like 20N60S5.... OR use a zero crossing opto i.e MOC3031M. Sorry but I don't have time to solder all above. Just idea! \$\endgroup\$ – GR Tech Jan 24 '14 at 20:28
  • \$\begingroup\$ oops...edit time passed... but you can look here techome.de/manuals/85829_DI200AB_KM_UM.pdf \$\endgroup\$ – GR Tech Jan 24 '14 at 20:47
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    \$\begingroup\$ If the PWM frequency is much higher than mains, I don't think DC current would pose a problem; a bigger issue I see is that inductive loads need to be switched between "current flows through supply to load and back to supply" and "current flows through load, bypassing supply" modes. A flyback diode can provide that latter function when driving a DC load, but for obvious reasons won't work with AC. \$\endgroup\$ – supercat Jul 18 '14 at 20:13
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For inductors,

$$V = L \frac{di}{dt}$$

PWM is an on-off style switch and cutting the supply current from the inductor instantaneously will generate a tremendous reverse voltage that will most likely break your MOSFET.

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    \$\begingroup\$ You understand that this is an AC application, not DC, right? Where would you put such a diode? \$\endgroup\$ – Dave Tweed Jan 23 '14 at 13:30
  • \$\begingroup\$ @ Andy aka: As I mentioned, for resistive load MOSFET is cool. Even at high pwm MOSFET is cool for inductive load. @ Pyxzure: As this inductive load is using AC, I am not sure of using a flyback diode. If possible would you pls show a schematic. \$\endgroup\$ – Mitz Jan 23 '14 at 13:32
  • \$\begingroup\$ Sorry, deleted to prevent misunderstanding \$\endgroup\$ – Pyxzure Jan 23 '14 at 13:42
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schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Schematic stripped down to essential current path with MOSFET represented by a switch and re-organised for clarity.

Figure 1 may help to understand the problem.

  • T1 is represented by SW1.
  • When L is positive and T1 on current flows through D3 and D4 to the lamp. (Figure 1b.)
  • When L is negative and T1 on current flows through D2 and D5 from the lamp. (Figure 1c.)

In a DC circuit Figure 1b would have a snubbing diode wired in parallel with LAMP2 and pointing upwards (anode to N). Figure 1c would have it pointing downwards (cathode to N). It should be clear that we can't have the diode pointing both ways and so we can't use a snubbing diode for an inductive load.

Your options would be to use an RC snubber but we don't have enough information to help you with that.

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If you use this to drive an inductive load, you're very likely to fry T1.

When the PWM signal goes low, T1 will attempt to interrupt the current while the load tries to maintain it. Result: high voltage will be induced till something breaks.

You could use a big ass zener (avalanche diode actually) across the transistor as a snubber. This will limit the back-EMF voltage from the load to safe levels.

Having some capacitance in parallel with the inductive load would be nice too.

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This dimmer outputs a rectified AC voltage which is basically unfiltered DC. An inductive load current with a DC source is only limited by the Resistance of the coil. This creates a high current through the components which causes an overheating and ultimately the destruction of the motor and the Mosfet.

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  • \$\begingroup\$ Improve your short answer to include issues with phase shifting in some motors for rotation, back EMF, and needs a clean sine wave to work correctly. Light dimmers and many PWM schemes do not work well with single-pole AC motors. \$\endgroup\$ – Sparky256 Jun 29 '16 at 22:12
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    \$\begingroup\$ "This dimmer outputs a rectified AC voltage ...". No it doesn't. The MOSFET is inside the rectifier but the load is outside it and receives AC supply. Edit required. Trace the way the current flows on each half-cycle. Welcome to EE.SE. \$\endgroup\$ – Transistor Jun 29 '16 at 22:28

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