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A High Side MOSFET Switch - Redux




 
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Author Topic: A High Side MOSFET Switch - Redux  (Read 1181 times)
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KK4YY
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« on: June 13, 2019, 02:17:19 AM »

I tried, and failed, in a previous thread...
http://amfone.net/Amforum/index.php?topic=44879.0
...to create a, relatively simple, high side MOSFET switch for high voltage. Now, I try again.

Circuit description:
ZD1 provides a voltage drop to the Drain so that a higher voltage can be used to bias the Gate ON. R6 holds the voltage up on the Emitter of the opto-isolator to prevent VCEO breakdown. ZD2 protects the opto-isolator from any transient voltages encountered during switching.

So, how'd I do this time? Think it'll work???


Don



* High Side MOSFET Switch 6_13_19.png (54.11 KB, 917x412 - viewed 221 times.)
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« Reply #1 on: June 13, 2019, 03:36:23 PM »

I am afraid  not. The zener will conduct in the off mode and supply 750 V to the gate. So you will get 750V out in the off mode
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KK4YY
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« Reply #2 on: June 13, 2019, 03:58:51 PM »

Nico,

Okay, I added a couple of blocking diodes, maybe that helps? My understanding is that the gate must be higher than the source by VGS(th) for turn-on. If the Source is floating, how can the gate be higher? The whole 'fet is pretty much floating, no?


Don


* High Side MOSFET Switch 6_13_19 a.png (54.77 KB, 917x412 - viewed 96 times.)
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steve_qix
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« Reply #3 on: June 13, 2019, 11:17:13 PM »

There is still a current path from the HV, through the zener across the opto, and back to the gate.

Sorry for the hand drawn schematic, but something like this may do the job.  Requires 2 MOSFETs, but no opto.  I moved the bleeder resistor (120k at 24w) to use as the pull-up.

There is a charge pump type of thing that will charge up a 1000uF capacitor when not in transmit mode.  When going to transmit, the charged up cap will raise the gate approximately 12V higher than the source, giving a "full on" switch.

Not tested, but conceptually should be OK.

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KK4YY
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« Reply #4 on: June 14, 2019, 12:23:27 AM »

Steve,

Thanks for your reply. Yes, I could use a discrete component charge pump circuit, as you've drawn, or perhaps one of the integrated charge pump/mosfet drivers that are available. As an exercise, apparently in futility, I've been looking for a simpler way to make an N-channel mosfet high-side switch. I think my luck is running out and I may have to concede that it can't be done.

It takes a little of the fun out of building the thing if I'm just doing it the way everyone else does. My goal is to find, and put forth, something new. Something that others may find useful to make a contribution. This, not for any personal accolades I might receive in doing so, but as pay-back for all that others in our community have done for me. I owe so much... and I don't like being in debt. It looks like I may also have to concede that I'll never get to pay it all back. But I haven't given up trying I'm not done yet.


Don
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PA0NVD
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« Reply #5 on: June 14, 2019, 10:27:23 AM »

Simply a source follower with 1MOhm from the gate to the drain. Than set the opto from the source to the gate shorting the gate to the source when ON. It reverses the ON/OFF signal (opto on is no output, opto off is output) That should work. You still can put a zener across the opto (from gate to source of the FET) for surge protection. KISS. Don't complicate too much. The FET will dissipate approx 6 x I output, which seem quite acceptable to me.
The only disadvantage is that the output is still approx 0.8 mA when ON, that may be a problem but can be solved by a small load, eventually a switched load.
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MikeKE0ZUinkcmo
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« Reply #6 on: June 15, 2019, 07:22:37 AM »

This basic circuit is what I've used on several BA receivers that switch B+ to mute.   I just mount them to the chassis apron with a mica insulator and some grease.  If no such chassis is available, you need a small heat sink to dissipate the heat of the pass device.




The series pass is a 11N 90, they are cheap and can handle up to 900 Volt supplies.   The gate driver is also an 11N90 for the same reasons.

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KK4YY
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« Reply #7 on: June 15, 2019, 10:21:37 AM »

Mike,

Excellent! I think that's just what I've been looking for. Smiley

1 diode
2 fets
2 zeners
3 resistors

The elegance of simplicity.

Thanks,
Don
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KK4YY
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« Reply #8 on: June 15, 2019, 04:05:48 PM »

Mike,

When the bottom 'fet turns ON, its Drain goes close to ground, with the 390K resistor as a load. Then, the Switched B+ Out, from the Source of the top 'fet, has a 10V zener going from B+ to ground. That doesn't look right. What am I missing here?


Don
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« Reply #9 on: June 15, 2019, 05:27:21 PM »

Seems very nice to me, simple and straight forward, nice solution Mike. The only thing that may improve reliability in case of transients etc is adding a resistor of e.g. 1K in the source of the lower FET. That prevents big current spikes that can blow the zener and upper FET
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KK4YY
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« Reply #10 on: June 16, 2019, 02:08:16 PM »

Okay. I think I got this now. Let's see...

It looks like it's inverting. When the control 'fet is ON the pass 'fet is OFF. When the control 'fet is OFF the pass 'fet is ON, or mostly ON. To me, it looks like the pass 'fet is not operating in full saturation. Source voltage would be lower than Drain by Vgs(th), perhaps 5V. With small currents drawn by high voltages, this may not be a problem. 5V at 200mA = 1W of dissipated power. I could live with that. If I were switching 16V at 10A (160W) a 5V drop across the 'fet would dissipate 50W and I'd only get 11V output. But with 800V at 200mA (160W), I get 795V output. I can live with that as long as the output voltage doesn't vary much with current demand. It may need a cap at the output to "slow it down" a bit.

I'll have to prototype this thing and see what it does.


Thanks again,
Don
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« Reply #11 on: June 16, 2019, 02:25:06 PM »

A cap at the output will not affect the switching speed much, the impedance of the FET when ON is low. When you slow down the switching at the input or the gate, the dissipation during the transient may be high, so better keep it like it was. A simple switch or relay also switches fast.
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KK4YY
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« Reply #12 on: June 16, 2019, 03:36:41 PM »

And yet another proposed circuit...

This variation inverts the inversion, and adds 5KV isolation as a safety factor by using an opto-isolator. A 12V PTT signal at the opto-isolator input produces a HV output from the pass 'fet. An open or grounded input turns the HV off. That's the theory, anyway.

This should be good for high voltages up to 950V (the rating of the 'fet).


Don


* High Side MOSFET Switch 800V with Isolation 6_16_19.png (888.58 KB, 751x403 - viewed 102 times.)
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« Reply #13 on: June 16, 2019, 11:01:05 PM »

A few things:

1) ALWAYS provide a safety margin.  In other words, don't run the FET at its maximum voltage.  If you want to switch 950V, use a 1200V FET, etc.

2) To invert the "sense" of the control signal, just use a common (grounded) emitter 2n3904 transistor ahead of the gate of the "control" (not the switch) MOSFET, with the drain hooked to the 1K resistor that goes to the MOSFET gate.  Put a 10k resistor in series with the base of the 2n3904.  Apply the 12V control voltage, which will cause current to flow in the base-emitter junction, which will pull the collector low, and turn off the control MOSFET.

Simpler than using the opto.   Just a thought...

As pointed out, this circuit does not drive the "switch" MOSFET into saturation.  The other circuit (the hand drawn one) does.  But, as a practical matter, if you're dropping 5V at 300mA, it's only 1.5 watts.
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KK4YY
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« Reply #14 on: June 17, 2019, 09:26:16 AM »

Steve,

Isolating the internal HV from my external 12V PTT circuit loomed large in my thoughts, so I opted for the opto.

The interesting thing that I learned, though this whole torturous exercise, is that a separate isolated power supply is not needed to turn the 'fet "ON" when high voltage/low current is used. The loss incurred by running the 'fet close to saturation is negligible. This was not mentioned in any of my readings. In fact, it's said to be a great difficulty in using an N-channel 'fet as a high-side switch. In low voltage/high current applications, this is certainly true, but not-so-true for my application. Smiley

~~~~~

So, I finally got from here, to there. My thanks to all who held my hand as I wandered through the mosfet desert.

Don
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