Applying capacitors across AC in various serial/parallel ways using FET's.

jpanhalt · Jan 16, 2018

Why not just build it and report back?

With parts on hand, it shouldn't take more than 30 minutes to do that. This thread has been going on for 7.5 hours already and is nowhere near reaching the simple conclusion a test will give. Total cost of parts for what you show is probably well less than $20.

AnalogKid · Jan 16, 2018

ACharnley said:
The source is not floating, you're forgetting about the FET's internal diodes. Source is AC - 0.7v.

No, it's not.

As you can see *on your schematic*, you have two FETs back to back, so the two internal zener diodes are in inverse serial. Whatever the Vds rating of an individual FET is, figure the zener is at least 10% higher conduction voltage.

ak

ACharnley · Jan 17, 2018

Yes but it's AC, so either way FET source gets power.

The zeners are fine, chips rated to -20Vgs

ACharnley · Jan 17, 2018

Take the top left circuit. Substitute AC2 for 10V, AC1 for ground and the capacitors for a bulb.

D top = 10V
S top & bottom is 9.3v
G top and bottom is 0V due to the 10k resistor.

= Fets turn on, power flows from D -> D and bulb lights.

Apply 3.3V to MCU44. Power from vDirect now saturates the gates, the fet's turn off.

Reverse AC1 and AC2 and the same occurs.

alec_t · Jan 17, 2018

ACharnley said:
Apply 3.3V to MCU44. Power from vDirect now saturates the gates, the fet's turn off.

No. The gate voltage becomes 3.3V -Vbe = 2.7V. The fets remain on.

ACharnley · Jan 17, 2018

How do you figure that, vDirect is 10v and the transistor is passing it. I specifically chose a BJT over an n-fet to avoid this high side issue.

ACharnley · Jan 17, 2018

jpanhalt said:
Why not just build it and report back?

With parts on hand, it shouldn't take more than 30 minutes to do that. This thread has been going on for 7.5 hours already and is nowhere near reaching the simple conclusion a test will give. Total cost of parts for what you show is probably well less than $20.

This is what I normally do, to find out it doesn't work.

kubeek · Jan 17, 2018

ACharnley said:
How do you figure that, vDirect is 10v and the transistor is passing it. I specifically chose a BJT over an n-fet to avoid this high side issue.

That is how a transistor works. For NPN, the emitter is allways 0.7V lower than base. Google emitter follower.
You know, if you stopped saying that the 10VAC or 40VAC or whatever are you using now is irrelevant and showed how it is related to ground and Vdirect, maybe someone could actually draw you a circuit that would work.

ACharnley · Jan 17, 2018

I already did, vDirect is the AC rectified.

OK I didn't know that about BJT's. So I have the same issue as using an N-FET. Sounds like I'll have to stick in a PNP on the upper left/right chip?

NPN -> PNP -> Dual PFET

kubeek · Jan 17, 2018

How is AC related to GND? Is it swinging +/-40V about ground, or something different?

ACharnley · Jan 17, 2018

The AC goes through a bridge rectifier and gives vDirect & GND. It's actually an active rectifier so no significant voltage drop.

ACharnley · Jan 17, 2018

OK so effectively half a FET driver (sorry it's not pretty I have no room on the A4 printout).

I haven't worked with PNP's before, is a resistor needed on the base?

kubeek · Jan 17, 2018

Check this out. VAC is going from 0 to 40V, so I don´t see how you want to feed the gate relative to the source.

ACharnley · Jan 17, 2018

kubeek said:
Check this out. VAC is going from 0 to 40V, so I don´t see how you want to feed the gate relative to the source.

You're using N FET's.

AnalogKid · Jan 17, 2018

kubeek - which color is which net on the schematic? netnames?

tomizett · Jan 17, 2018

Why don't you just use a resistor from gate to source to turn the FETs off, and then a grounded-emitter NPN to pull the gate low to turn them on?

Edit: Thank you for clarifying how you're expecting the circuit to work, but I still think that a clearer diagram would have got us here a lot quicker.

ACharnley · Jan 17, 2018

tomizett said:
Why don't you just use a resistor from gate to source to turn the FETs off, and then a grounded-emitter NPN to pull the gate low to turn them on?

That's what's happening with the bottom two, except I'm using AC->DC as the AC would kill the gates.

It's the top two that are presenting an issue.

alec_t · Jan 17, 2018

Here's what I think you're now proposing. I've added R3 to limit the transistor current.

AnalogKid · Jan 17, 2018

There are patents for circuits that use MOSFETs instead of a TRIAC to control an AC load. A.L.L. of them have the gate drive circuit referenced to the sources somehow.

ak

ACharnley · Jan 17, 2018

Not this one.

Alec_T -> I've been playing with it and it's working save for one issue with the zener, you may have noticed the current is less on one AC cycle. VGS should never drop more than 20V (or 8V in my case). I tried shifting it to the source but that didn't work either. Strange.

Otherwise it works very well.

Edit: I meant to place the resistor between the base. In your case the PNP base can't drop more than 1/2 vDirect, however either way works fine and it does have the benefit of using a little less current through the NPN.

Applying capacitors across AC in various serial/parallel ways using FET's.

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