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I've been trying, unsuccessfully, to dim a 60W incandescent light bulb with two back-to-back MOSFETs, similar to this web page:

http://easy-electronics4u.blogspot.co.uk/2012/02/switch-ac-loads-using-mosfets-as-relay.html

My circuit looks like this: circuit diagram (there is also a 2A fast blow fuse and a 7D471K MOV on the mains input)

Below is the output from my scope: oscilloscope image

  • Channel 1: Live_230VAC
  • Channel 2: DIM_GPIO
  • Channel 3: Drain of Q5
  • Channel 4: Source of Q5 and Q6

Q6 gets burning hot, Q5 stays cool. The lamp (obviously) flickers as we are only getting every other half wave. Thinking about the circuit, I feel that it will be impossible to get a Vgs voltage that is higher than the voltage at the source pin, since the source pin will see the peak of the 230VAC?

Is there a way (preferably simple & cheap :)) of solving this kind of double MOSFET dimmer?

Edit 1: Added revised schematic as per suggestions from @RoyC for consideration (and so that RoyC doesn't have to click my pasteboard.co link) Rev3

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  • \$\begingroup\$ How is the emitter of the optocoupler related to GNDA and how is that in turn related to producing a Vgs for each MOSFET? \$\endgroup\$ – winny Jan 1 '17 at 10:43
  • \$\begingroup\$ The emitter is connected to GNDA via R18 (20K) but I'm guessing you already know that since I don't understand the second part of your question :) \$\endgroup\$ – englund Jan 1 '17 at 11:43
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On the right of the diagram swap live and neutral. This is a safety thing although you should never assume neutral is anywhere near earth it is likely to be lower than live. Now your switch is floating on the neutral line not the live.

The drive voltage for your gates has to float referenced on the source connection between Q5 and Q6. Short out R19.

Remove the bridge rectifier it is not needed instead connect D7 to the point that you have labelled in your current diagram as Neutral_230VAc. The ground end uses the body diode of Q6.

Here is a rough diagram component values same as yours

enter image description here

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  • \$\begingroup\$ I'm not quite following you on the safety of switching live and neutral coming in to the lamp and Q6. With the setup in my original schematic the lamp will only be connected to neutral when the (dual mosfet) switch is open (off). If I swap them the lamp (and its socket) will always have live connected to it which sounds bad. Have I understood your corrections with the following schematic? !http://pasteboard.co/auIAhrD3.png \$\endgroup\$ – englund Jan 1 '17 at 18:59
  • \$\begingroup\$ also added my revised schematic to the original post so you don't have to leave stackexchange to view it.. \$\endgroup\$ – englund Jan 1 '17 at 19:17
  • \$\begingroup\$ Nearly there D7 should be connected to Live not neutral. It was neutral on your original diagram \$\endgroup\$ – RoyC Jan 1 '17 at 20:36
  • \$\begingroup\$ Put a rough diagram in the answer \$\endgroup\$ – RoyC Jan 1 '17 at 21:23
  • \$\begingroup\$ @englund re the safety thing I was trying to put most of the circuit bits that you are working on and making measurements on at or near the neutral line this is a little safer than having them on the live line. I have personally learned this lesson the hard way. \$\endgroup\$ – RoyC Jan 2 '17 at 13:17
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It appears that your MOSFETs are not driven from a solid (i.e. low impedance) source.

Your driving voltage of 15 V, regulated by D8 and buffered by C5, is passed to Q5 and Q6 via the optotransistor, a 1 kΩ resistor and back to GNDA via a 220 kΩ (!!!) resistor. This means the resistors alone introduce 221 kΩ into your gate driving circuit. I wouldn't trust this configuration to properly drive a MOSFET's gate-to-source capacitance. It is good design practice to keep the impedance of a MOSFET driving circuit well below 50 Ω to avoid driving the MOSFET in the linear region, i.e. neither fully off nor fully on. I don't see a reason for putting a 220 kΩ resistor between the MOSFETs' sources and GNDA.

In order to fix this circuit, I suggest the following:

  • Make sure you don't kill yourself with the mains voltage, I will not cover this very important part of the job here, because it can't be done Q&A-style - so please ask a professionally trained person to assist. At the very least, use a safety transformer to prevent yourself from getting a shock or blowing your scope when testing.

  • Think about the driving circuit: Do you get a stable supply of 15 V? Do you get proper gate voltages at Q5 and Q6? What's the job you're trying to achieve with R19? How are the MOSFETs sources bouncing with regard to GNDA?

I know I'm using some questions to answer, but I am sure that these questions are of better help than just answering with a snotty "gate driving circuit is flawed".

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  • \$\begingroup\$ Thanks for the quick reply! The gate is charged to ~15V (confirmed stable at that) through the 1K and then discharge through R18 (20K) when the optocoupler turns off. This is to create a soft turn off to minimize EMI as described in this appnote from ST: goo.gl/7QmYrl Connecting the sources to GNDA via R19 is something I thought would allow me to use the rectified 15V as a gate voltage but that apparently didn't work :) Indeed the gate driving circuit is flawed, that kindof was my initial question although I might have phrased it vaguely.. \$\endgroup\$ – englund Jan 1 '17 at 11:55
  • \$\begingroup\$ There is nothing wrong with driving the gate of a switching MOSFET from any impedance source. 50R is very low and not a target. What matters is that the gate capacitance can be charged/discharged fast enough for the MOSFET to not dissipate excessive power when changing between saturation and cut-off. 'Excessive' is a combination of wasteful and destructive defined by the application. 'Fast enough' can be derived from max' tolerable switching power but often easy to switch it much too fast for problems anyway. Take Cin(max) of both FETs to total 2.8nF from datasheet. Sorry, no time to do maths. \$\endgroup\$ – TonyM Jan 1 '17 at 12:20

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