Non-inverting NFet gate driver

Thread Starter

cmartinez

Joined Jan 17, 2007
7,246
Here's a dumb question. I've been playing around with this NFet low-side gate drivers (no, I do not want to use a chip for this, but rather discrete components), but I can't find a way of making them non-inverting. That is, I'd like the NFet to be turned on when the driver signal is high, and not low as it's doing in this circuit. Driver signal is 5V, btw.

upload_2018-4-20_11-23-37.png

Any suggestions?
 

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OBW0549

Joined Mar 2, 2015
3,566
I can't see what Q1 and D1 are there for; what are they supposed to be doing? The way you've got Q1 connected, it will never turn on (same applies to Q3 and D2). Remove these two components.

To make the circuit non-inverting, connect the left end of R2 to +5V, disconnect the emitter of Q2 from ground, and apply your input signal to the emitter of Q2. (In other words, operate Q2 in common-base mode rather than common-emitter.)
 

Jony130

Joined Feb 17, 2009
5,213
I can't see what Q1 and D1 are there for; what are they supposed to be doing?
The emitter follower is here to quickly charge the MOSFET input capacitance. And D1 and Q2 will quickly discharge the capaciance.
But TS should add a small resistor in series with this diode.

connect the left end of R2 to +5V
And do not forget to place a capacitor in parallel with this resistor (R2).
 

crutschow

Joined Mar 14, 2008
26,407
I can't see what Q1 and D1 are there for; what are they supposed to be doing? The way you've got Q1 connected, it will never turn on (same applies to Q3 and D2).
If you add the gate capacitance to the schematic you will see that Q1 turns on to charge that capacitance when Q2 turns off.
The diode isolates Q1's base signal from the emitter when it is charging the capacitance.
But TS should add a small resistor in series with this diode.
Why?
That will just slow down the fall-time at the MOSFET gate.
 

OBW0549

Joined Mar 2, 2015
3,566
The emitter follower is here to quickly charge the MOSFET input capacitance. And D1 and Q2 will quickly discharge the capaciance.
But TS should add a small resistor in series with this diode.
Oops! You're right. My mistake...

And do not forget to place a capacitor in parallel with this resistor (R2).
Good point; a 100 pF (or thereabouts) capacitor will speed things up.
 

crutschow

Joined Mar 14, 2008
26,407
If you use a common-base configuration as OBW0549 suggested, not that the discharge current for the gate capacitance must be sunk by the input signal emitter drive signal.

Likely a better way is to just add a inverting stage in front of the MOSFET driver.
 

Thread Starter

cmartinez

Joined Jan 17, 2007
7,246
I can't see what Q1 and D1 are there for; what are they supposed to be doing?
:oops: ... I dunno ... I just thought they'd look pretty in the schematic ... :D

Seriously, when I simmed it I saw that those components were indeed participating in the circuit. And I didn't understand why until crutschow was kind enough to explain it. I didn't design those circuits, I copied them from an old post by someone else at another site after doing some searching.

upload_2018-4-20_12-55-24.png

I've introduced the changes you suggested, and it's now doing things the way I want :) ... funny, but in my head I always see a transistor as being controlled through its collector... that's why I couldn't think of your tweak before ... I need to train myself some more to think outside the box.

Thanks for your help too, Jony130
 

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Thread Starter

cmartinez

Joined Jan 17, 2007
7,246
If you use a common-base configuration as OBW0549 suggested, not that the discharge current for the gate capacitance must be sunk by the input signal emitter drive signal.

Likely a better way is to just add a inverting stage in front of the MOSFET driver.
you're right about that ... the signal is going to be generated by an MCU (an AT89LP4051) output pin, and driving the NFet's gate capacitance might be a bit too much to ask from it
 

Thread Starter

cmartinez

Joined Jan 17, 2007
7,246
Also, please ignore the zener (D3) in my last circuit. I put it there because I'm a little obsessed about protecting the fet's gate. But I doubt it's doing any useful work.
 

OBW0549

Joined Mar 2, 2015
3,566
:oops: ... I dunno ... I just thought they'd look pretty in the schematic ... :D

Seriously, when I simmed it I saw that those components were indeed participating in the circuit. And I didn't understand why until crutschow was kind enough to explain it. I didn't design those circuits, I copied them from an old post by someone else at another site after doing some searching.
Scratch my original comment about Q1 never turning on; it was a brain-phart (probably old-age-related). Jony130 and crutschow set things straight.
 

Thread Starter

cmartinez

Joined Jan 17, 2007
7,246
Scratch my original comment about Q1 never turning on; it was a brain-phart (probably old-age-related). Jony130 and crutschow set things straight.
thanks ... your help is always thoroughly appreciated ... anyway, I'm gonna start mulling how to produce the non-inverting switching I want without having the MCU's output pin sinking the NFet's gate capacitance. Crutschow is right about needing an inverting stage. I wish I didn't have to add more components to this thing, though. I wanted to keep things simple.
 

kubeek

Joined Sep 20, 2005
5,744
Thanks, I already tried that. But I couldn't get the gate to rise above 4.5V using that circuit. Here's proof
You need one more transistor stage to translate the 0-5V input to 0-12V output. Still, the steady state current through the first sttage will have to be about 1/10 to 1/100 of the final gate drive current, which is not very efficient. You could have another totempole before that, but that loses another 0.7V from the output range limits.

So why not use an actual dedicated CMOS gate driver?
Brewing your own will generally lead to more board space used and a lot more time spent fiddling with it. Unless you are re-designing some chinese toy to even less production cost because you make tens of thousands of them, jsut use something that gets the job done right and focus on the other more important aspects.
 

Thread Starter

cmartinez

Joined Jan 17, 2007
7,246
I think I did it :) . I used an extra transistor (Q3) to drive Q2, this acts as an inverter and makes things work the way I want them to. Too bad I couldn't do it with just two transistors.

upload_2018-4-20_18-15-10.png
 

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Thread Starter

cmartinez

Joined Jan 17, 2007
7,246
Brewing your own will generally lead to more board space used and a lot more time spent fiddling with it ... jsut use something that gets the job done right and focus on the other more important aspects.
You're right about that ... but I like a challenge, and I'm trying to learn too :) ... On the other hand, the cost of the three transistors, plus the diode and the three resistors is still significantly less than a, say, IR2101. I also consider it a sad waste to use only the low-side driver of that chip. As for board space, you're right about that. But wouldn't this circuit be more reliable and last longer than the IR2101? I could be wrong, of course, but I'm under the impression that the 2n3904 and/or 2n3906 are less susceptible to ESD or spurious surges in voltage than a driver ic.
 

kubeek

Joined Sep 20, 2005
5,744
Yes, that is sort of what I had in mind (please delete teh top 5V source and connect R4 directly to the 12V rail). But anyway, if M1 is to be off, then you have 25mA flowing through R1, which is roughly 300mW power loss on that resistor. Call me conservative, but to me that is quite a lot of wasted power.
You could add a few more drive transistors and make the top final transistor a pnp and the bottom one npn, make sure they don´t conduct at the same time and have a nice efficient driver out of it, but I doubt it can compare in effort to any standard gate driver you can buy. Price, maybe.
 
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