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We are designing a buck using discretes. I have a problem with OFF time of Control MOSFET(U1) and driver(T1). In OFF time, Gate of U1 is rising very slowly to 12V which makes U1 always ON. Which in turn makes output to 12V always.

Buck using disretes -> p-MOSFET drivign problem I have seen few BJT Totem pole drivers for p-MOSFET. But, i want to use this same circuit. any modification suggested so, that we can make U1 gate to rise faster.

VG1 is 1.8V,100K, 50% duty cycle square wave.

Vgate of U1 and other node waveforms

I tried to change the R1 2K to 500 and VG1 1.8V,90K,50% and found.

Improvement when R1=500, switching freq reduced..

How to still make the Turn OFF of T1 fast, so, that Vgate of U1 rises to 12V very quickly.

Edit:

I have changed the 2k to 0.5k and i am getting FET swtiched OFF and getting 5V output as intended, but Switching frequenxy is very less close to 80K and it is not following Buck formula Duty cycle ~ Vout/Vin. As OFF time has notches and slow. Any tweaks we can do on the same circuit to make it more faster. ....

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    \$\begingroup\$ Now you're finding out why most people use totem-pole drivers! \$\endgroup\$ – Dave Tweed Apr 30 '14 at 12:04
  • \$\begingroup\$ . I have changed the 2k to 0.5k and i am getting FET swtiched OFF and getting 5V output as intended, but Switching frequenxy is very less close to 80K and it is not following Buck formula Duty cycle ~ Vout/Vin. As OFF time has notches and slow. Any tweaks we can do on the same circuit to make it more faster. .... \$\endgroup\$ – user19579 May 2 '14 at 7:48
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The reason the FET turns off so slowly is because you only have 2 kΩ pulling it high. Take a look at the datasheet for that FET. It should show you the actual and effective gate capacitance when switching. The relatively weak 2 kΩ pullup is working against that capacitance.

Here is a trick I sometimes use in this situation:

The double emitter follower is basically a impedance buffer. The current to charge and discharge the gate capacitance is handled by Q1 and Q2, so the signal at GATE can be much higher impedance and still switch the FET quickly. Note that this double emitter follower will loose 700 mV or so on each end. That is fine for a FET that switches over a 10 V gate range. It will still be "off" with 700 mV on the gate just as well as with 0. At the other end, its usually easy enough to drive GATE the extra 700 mV further than what you want to make sure shows up on the gate of Q3.

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  • \$\begingroup\$ Author using transistor with very low cutoff voltage (0.5 ~ 1.5 V). Emitter follower may fail to deliver cutoff voltage to it (due to Vbe). Though, resistor will help \$\endgroup\$ – Vovanium Apr 30 '14 at 13:59
  • \$\begingroup\$ @Vova: As I pointed out in the answer, this trick is for FETs intended for a large gate drive, like 10 V. Those are usually still "off" at 700 mV gate drive. If the OP's FET has a low gate threshold, then either don't use this trick or get a different FET. \$\endgroup\$ – Olin Lathrop Apr 30 '14 at 14:08
  • \$\begingroup\$ I have mentioned i want to use the same circuit. I don't want to modify the circuit. I have changed the 2k to 0.5k and i am getting FET swtiched OFF and getting 5V output as intended, but Switching frequenxy is very less close to 80K and it is not following Buck formula Duty cycle ~ Vout/Vin. As OFF time has notches and slow. Any tweaks we can do on the same circuit to make it more faster. ....(If i changed the R1 to 10k then p-MOSFET will not be OFF at all and output is Input 100% Duty cycle.) \$\endgroup\$ – user19579 May 1 '14 at 2:46
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Power MOSFETs have large gate capacitance, so when switching it consume large currents. That's why you should not drive with resistive pullup. You need push-pull (totem-pole as @Dave Tweed said) amplifier after generator.

schematic

simulate this circuit – Schematic created using CircuitLab

Here's sample schematic which include inverting amplifier with two complimentary MOSFETs (M1M2). These should have small gate capacitance while deliver ehough charge/discharge current for power MOSFET.

Gate capacitance of your transistor is 1070 pF typical and threshold voltage is 0.85 V. You need not more than 17 ohm high drive to obtain reasonable switch time of 50 nS. Peak curret would be about 0.7 A. That's why you need active high drive. (17 ohm = 50 ns / 1070 pF / ln(12V/0.85V) based on discharge time of RC circuit)

Using power transistor with higher threshold voltage (e.g. ~ 2-4 V as in IRF9530) may also help -- transistor with higher threshold will get cutoff state earlier.

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