Urgent High Side FET help! - Page 1
Forum section: electronics / beginners

author: unit998x
author_rank: Contributor
timestamp: October 24, 2015, 05:43:19 pm
content: Hi all, quick problem that I'm sure is just down to my own stupidity.I am trying to drive a high side MOSFET with +12V system voltage, so to produce the Vgs voltage needed to turn on the FET I used a bootstrap circuit. The driver uses discrete parts, and the STP14NF10, for both power and logic because I don't have any others lying around. But I only get 3V Vgs, not suitable.Attached is the driver circuitry and Vgs waveforms for the circuit, at 2V/div, I have tried both 20kHz and 40kHz both at 50% duty.Due to deadlines for the project I do not have time to order a MOSFET driver, I know this is the easiest solution, but it's just not possible.Many thanks in advance!

author: bktemp
author_rank: Super Contributor
timestamp: October 24, 2015, 05:48:29 pm
content: Connect R3 to the bootstraped power supply (C1) not to 12V.

author: Andy Watson
author_rank: Super Contributor
timestamp: October 24, 2015, 05:54:31 pm
content: The voltage on the drain of Q4 is limited to +12V because it is pulled up via R3 - so any extra boost voltage from the 1uF is not available for the gate (even if we ignore losses). Try pulling R3 up to the boosted voltage on capacitor.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 24, 2015, 05:56:14 pm
content: Also, replace Q4 with a 2N7000 at the worst... the poor thing will take forever to switch at that current.Tim

author: unit998x
author_rank: Contributor
timestamp: October 24, 2015, 06:19:21 pm
content: Quote from: bktemp on October 24, 2015, 05:48:29 pmThanks, I thought there was some stupidity in this. I've fixed but the bootstrap supply voltage is 4.3V, and the (Q1) gate-source voltage is now one transistor drop below the bootstrap voltage (3.2V). Could there be another reason the voltage isn't being increased enough? Are the bias resistors too high?Quote from: T3sl4co1l on October 24, 2015, 05:56:14 pmYeah, unfortunately it's the only FET I have available at the moment, switching losses don't concern me for now, I can replace Q4 with a smaller FET later.

author: unit998x
author_rank: Contributor
timestamp: October 24, 2015, 07:35:38 pm
content: Attached is the new schematic, is there any kind of reference design for high side driving without an IC?Also this circuit exists on a breadboard, could the stray capacitance make any difference?

author: bktemp
author_rank: Super Contributor
timestamp: October 24, 2015, 07:39:49 pm
content: I have simulated the circuit, it works fine, so the problem is not in the schematic.Measure the voltage across C1. It should be about 11.5V. If it is much lower the load does not swing to 0V during off state and therefore the capacitor does not fully recharge.

author: unit998x
author_rank: Contributor
timestamp: October 24, 2015, 08:09:21 pm
content: Hi, I've used a 330 resistor in place of the load. The voltage across C1 is 5.0V. Vgs(Q1) is 3.9V, at 20kHz 50%. Vgs(Q4) is 12.0V, 20kHz 50%. Waveform across Bootstrap diode and GND is shown in attached picture (10V/div, 50us/div, DC).Many thanks!

author: bktemp
author_rank: Super Contributor
timestamp: October 24, 2015, 08:25:07 pm
content: If this is the voltage at the positive side of C1 then your diode seems to be not working correctly. Maybe a 1N4148 can not handle the high peak currents.

author: Andy Watson
author_rank: Super Contributor
timestamp: October 24, 2015, 08:27:55 pm
content: How are you driving Q4. Both sets of waveforms suggest that Q4 is turning-off relatively slowly and hence the boost circuit is supplying power to Q4 rather than the gate of the output transistor.

author: unit998x
author_rank: Contributor
timestamp: October 24, 2015, 08:41:55 pm
content: I know the pictures are poor, but I've tried using a 3A 1N5819 instead of the 1N4148, but still only 5V across C1. (D1 and Q2 are not touching leads although it looks like it on the picture)Q4 is driven from a 4011, 12V output. Attached is waveform for Vgs(Q4), 5V/div, 5us/div. So turn off time for Q4 looks to be about 1us. Is this reasonable?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 24, 2015, 09:50:06 pm
content: Then use another BC107, with the appropriate base current limiting resistors.Tim

author: IconicPCB
author_rank: Super Contributor
timestamp: October 25, 2015, 04:34:06 am
content: Don,t waste your time reinventing the wheel.Look to automotive high side switches.Temperature and over current protected with a logic level input.Siemens have them so do SGS and a number of other manufacturers.