Help me design a FET gate driver circuit

Hello,

I am designing 12V to 5V synchronous buck converter. I need to design gate drive for switching MOSFET. Due to cost and design constraints i can not use market available gate drive ICs. I need to design discrete gate drive circuit.

Can anyone please help me to design the same? I tried using push-pull but facing some issues with it.

Thanks in advance

Re: FET gate drive

niks said:

Hello,

I am designing 12V to 5V synchronous buck converter. I need to design gate drive for switching MOSFET. Due to cost and design constraints i can not use market available gate drive ICs. I need to design discrete gate drive circuit.

Can anyone please help me to design the same? I tried using push-pull but facing some issues with it.

Thanks in advance

Click to expand...

So, what are the issues you are facing?

Re: FET gate drive

Please find the circuit and simulated waveform. I have simulatd using TINA Pro from TI.

Re: FET gate drive

Pl. write the problem you are facing with this circuit so that I may help you.

Re: FET gate drive

@Niks,

In your schematics the common point of T2's emitter and T4's collector is connected up to T1's drain, is not a problem? I think, that common point should go the the (negative) ground, right? And the type of D1 diode, 1N4001 is not fortunate, it is a low frequency (not at all fast switcher) type, use here Schottky type or at least a type with max some hundred nanosecond reverse recovery time.

By the way, if you need text on designing syncronous buck converter, you can surely find that by using search engines or text books.

unkarc

Re: FET gate drive

Output waveform from the push-pull is not clean waveform. some places instead of square pulse i get spike. please check the attached waveform with schematics.

As this is FET drive for buck regulator, Push pull needs to be connected between gate and inductor-FET junction. The components used in schematics are only for simulation purpose.

Re: FET gate drive

I have synchronous buck regulator 12 to 5V. How can i drive Main switching FET?

Re: FET gate drive

@Niks,

PLEASE, try to understand the followings:

in your schematics, if we SUPPOSE there is +5V at the output, MOSFET T3 were always open because the smallest possible output voltage from T2-T4 common emmiter output would be also near +5V in that case, ok? Agree with this???

HENCE, there CANNOT BE +5V at the output because T3 were always open so its drain-source low ON resistance would consume all and any output voltage, ok? Agree? (Here I suppose the threshold voltage (Vth) of T3 is between +3 to +4V and I suppose a -3 to -4V threshold voltage in case of T1 too.)

IF you connect T2's emitter and T4's collector to the Ground (to your negative input/output wire), THEN the possible output voltage from T2-T4 common emitters would be able to swing from near zero to near +12V with respect the negative Ground, ok? Agree?

AND if you have a 50% duty cycle square drive to T2's base input, then +6V would appear at the output (with respect to the negative Ground) theorytically for a light load and the two MOSFETs would switch as follows:

--if the DC output is near zero from T2-T4 emitters to the Gates of the MOSFETs (i.e. T4 fully conducts), the P channel T1 opens and charges current to L1. The N channel T3 is closed in this switching phase.

--if DC output is near +12V from T2-T4 emitters to the Gates of the MOSFETs (i.e. T2 fully conducts), T1 switches off and T3 switches ON so takes current from L1.
In fact, the role of T3 is to improve rectification efficiency, without T3 the current from L1 in this switching phase would be taken through the D1 diode anyway, the converter would work without T3!! (The reason it is able to improve efficiency is that when it is switched on, the voltage drop between its drain source resistance (RDSon) is much less than that of any diode.)

This is how this converter is supposed to operate, at least this HOW I THINK it would operate in principle. That is why I suggested modifying your schematics in my first post to you. This now means that you have got a correct drive circuit to switch your main FET.

If you do not agree with any of the above reasonings, please indicate which/what your problems. I invite any member here to comment this converter operation.

unkarc

Added after 28 minutes:

@Nick,

Some more comments:

When designing ONLY the drive circuit in your circuit simulator, first make sure that you include the following components:
the two npn and one pnp bipolar transistors, R2=1K and most important the 200 Ohms resistor between the emitter output and +12V! And check with the scope that at the two emitters output point the voltage changes from about +0.2V to +11.8V with respect to the negative ground, ok? It is possible you have to change somewhat the values of the above two resistors to achieve more easily the nearly full 12V output swing at the common emitters with respect to the negative ground.
Without this output swing, you cannot make the buck converter ever. And of course, connect the first T2 (on the left) transistor's emitter and the T4's collector to the negative pole of the input 12V!!!

(By the way you have two T2 transistors and two R2 resistors, this is not fortunate when speaking of which is which).

And when this is ok, then you can only continue building the rest of the converter components in the simulator and see the results.

unkarc