Low side N-channel mosfet gate driver desing question from 24V-with a twist - Page 1
Forum section: electronics / beginners

author: Atom
author_rank: Regular Contributor
timestamp: November 07, 2022, 08:16:00 am
content: hi, so lets get straight to the point, i have to drive a n channel low side mos, from and only from a 24v psu.i dont have any other voltages available to drive the mosfet gate so this is the problem, no 5V no 3.3 no 12, only thing i have is 24vdc.and i also dont want to derive a lower voltage from the 24v supply because i dont want to pull unnecessary current from the main voltage rail since this thing will be always powered, so i wanted to keep the standby consumption lowthe main reason to not wanting to derive a lower voltage from the 24V is that every solution would be bulky (linear reg or sw reg) would take too much space in the pcb, that has to be very small , like 10*20mm(i have to fit the mos and a soic8 ic in this space already).so the question is, how do i drive a low side Nchannel mos only having a 24vdc source?by searching this forum i found a similar question , see second image, the guy proposes a totem pole driver with a voltage divider on the output to bring the voltage down to 12V, my question is:didnt we use a totem pole driver to increase the current that we could sink and source from the gate ? by putting that voltage divider we are limiting by a lot the current that can be sourced or sinked, so we'll have slow turn on and turnoffs (maybe turnoff its ok because we have a diode).Thanks in advance to everyone that will answer or contribute, maybe my question might seem stupid but i'm here to learn, have a nice day!

author: magic
author_rank: Super Contributor
timestamp: November 07, 2022, 08:52:22 am
content: What generates the turn-on signal, where is it taking power from, what's the level of this signal? Could it drive the MOSFET directly?BTW, there are MOSFETs rated for up to 30V gate voltage.How fast and how often does it need to switch?

author: jpanhalt
author_rank: Super Contributor
timestamp: November 07, 2022, 10:42:57 am
content: What value will R3 be? For "low standby consumption," it presumably is fairly high as the mosfet will be off only when the control transistor is on. You also don't have a pull-down/turn-off resistor on the gate. Adding such a resistor will give you a voltage divider and reduce the gate voltage to a tolerable level.EDIT: You can also add silicon diodes in series to drop the voltage. 3x1.7V(each) = 5.1VEDIT2: Clearly the 1.7V is wrong. I have no idea how I came up with that. The usual value is 0.6V or so.

author: Atom
author_rank: Regular Contributor
timestamp: November 07, 2022, 12:44:09 pm
content: Quote from: magic on November 07, 2022, 08:52:22 am1 the control signal is generated by an IC that has an open collector desing as its output, so i can put whatever i want there, it can sink up to 18ma. the level of the signal can be whatever you want since the output is a open collector.if i put R3 = 10k i have an squarewave that goes from 0.7V to +24V, i can halve that by putting 2 10k resistor in series and get 0.7V to 12.7V, so its flexible.i dont think the output from the ic could drive the mosfet by itself but maybe im wrong2 the IC is powered by an internal LDO (i cant tap into that), the ldo is fed by a resistor attached to 24V (it consumes only 2mA)3 switching should be 400hz,(not really sure since chinese ic manufacturer doesn't say anything) as a rule of thumb surely its under 10khz.

author: Atom
author_rank: Regular Contributor
timestamp: November 07, 2022, 12:50:02 pm
content: Quote from: jpanhalt on November 07, 2022, 10:42:57 amR3 its 10k so 2.4mA are flowing when the transistor is on.i can add a resistor and obtain a voltage divider but my question is, does the voltage divider have enough capability to charge the gate of the mos in a reasonable time ?i see drivers ic that claim 1A and more sink and source from the mos gate, im not really sure if some mA can turn on a mosfet in a decent time.silicon diode dont have a drop of 0.7V? i would need to add a lot of them to the pcb and i dont know if i have the space.

author: Psi
author_rank: Super Contributor
timestamp: November 07, 2022, 01:10:10 pm
content: what about just putting a 9V zener in series with the gate to drop 9V?You'd need a pulldown resistor on the gate to turn it off though.Which I guess is then wasting power so maybe that wont work.You'd waste power when NPN is on through its own pullup and when its off through NFet pulldown.

author: jpanhalt
author_rank: Super Contributor
timestamp: November 07, 2022, 02:02:33 pm
content: Re: voltage dividerIf the turn-off/voltage divider resistor is 47K, when the transistor is on, the 47K is effectively shorted. When the transistor turns off, the gate capacitance effectively shorts that resistor too initially (i.e., its apparent resistance and the 47K are parallel). Of course, your turn-on time will increase some. I haven't bothered to calculate it. The maximum gate voltage will be limited to approximately 19.8V.You could also use a much larger turn-off resistor and diodes as suggested earlier too.NB: This analysis ignores the gate resistor of unknown value you show. It also needs to be considered. If you want to speed up turn off, you can add a "speed up" diode across the gate resistor with its anode closest to the mosfet gate. You already have so much resistance in the gate circuit, a gate resistor may not be needed.

author: magic
author_rank: Super Contributor
timestamp: November 07, 2022, 05:14:51 pm
content: Quote from: Atom on November 07, 2022, 12:50:02 pmDepends on the gate charge of the MOSFET, but typical values under typical operating conditions are some nanocoulombs. With 1A current, the gate charges in nanoseconds. With 1mA current, in microseconds. This may be good enough for a few kHz. It could be a problem if it's a large FET conducting lots of current and even a few µs in linear region heats it up significantly.Since the output is open collector, you have no choice but to pull it up with a resistor. This may be a low value resistor to switch the FET directly or a high value to save power, then use the NPN/PNP circuit to multiply drive current by β of the transistors.For less voltage you could use a divider (or a resistor and zener, but certainly don't bother with a string of diodes) and connect the middle point to the open collector output. But this consumes power even when the output is off (and the FET is on). If this is a problem, use higher value resistors.The simplest way out of the gate voltage problem is to look if there are 30V tolerant FETs that fit your application.editPutting a resistive divider between the emitter followers and the FET makes no sense and brings the exact problems you described.The right way is to put the divider before the emitter followers.