Understanding Switching MOSFET

[yashrk]

Hey Guys,
                 I was working on a circuit I found on internet and trying to replicate I have understood most of the circuit but what bugs me is this marked (in image) circuit is added to drive the gate of the MOSFET he here has used 3 bjt to drive it, can't we just drive it with one bjt ?  its all about giving it the voltage doubler output and the ground depending on the pwm signal ?

Thanks in advance,
Happy making!! 

[yashrk]

Sorry forgot to enter the img its here now!

[Jony130]

All MOSFET have some input parasitic capacitance. So to speed-up turn-on and turn-off times we sometimes add active pull-up (V3 + 1n1418) and active pull-down V4)
And this additional transistors provides additional current to quickly charge and discharge MOSFET input capacitance.

[yashrk]

Can use this circuit instead of the circuit shown ?

And after what switching frequency I should be worrying about the parasitic capacitance not discharging ?

[yashrk]

Please ignore the resistor values I just quickly made the circuit up !

[PSR B1257]

This circuit won't work at all, because the MOSFET is configured as a so called high side switch.

In your inital circuit there is a base resistor missing for Q2. And the 220k resistors seems rather high - it will most likely not work. Change these to 2k2 (ore there about) instead.

Since you have a micro controller (to invert the output signal*), you can get away with a simpe level shifter consisting of one BJT and one resistor (for f<10kHz it should be fine).

[yashrk]

Sorry this is what I meant and ya the resistor value will be different and I am working on 10KHz PWM freq

[Zero999]

This circuit won't work at all, because the MOSFET is configured as a so called high side switch.

We don't know what voltage Vgate is at. If Vgate is high enough to turn the MOSFET on, then it will work.

It appears to be an attempt at a voltage doubler and the transistors are used as level shifting.

The voltage doubler doesn't look right though.

[PSR B1257]

No, the MOSFET is still high side. And this is even opposed to your initial circuit.

Otherwise is this what I meant.

[yashrk]

The first circuit is not mine but it was a working circuit product which was sold on ebay.

What I want to achieve here is lesser parts count hence trying to control mosfet using one bjt.
If the voltage doubler works and provides sufficient voltage above the source pin than will the single bjt design work ?

if yes than why use 3 bjt in a an commercial circuit ?

[T3sl4co1l]

Hmm, I suspect the negative side of the 1uF is intended to go to the MOSFET source?

Tim

[yashrk]

As I said I didn't design the ckt but yes it's working product of same circuit design

[yashrk]

Sorry  just last time want to confirm that will this circuit will work at 10KHz frequency and the parasitic capacitor will discharge ??

[PSR B1257]

No, it won't (again).
Where is the drain current supposed to flow 

Why do you change to a P-channel MOSFET?

You should use a logic level FET, driven of the 5V rail by a push–pull transistor stage.
Do you really need 10kHz switching frequency?

[T3sl4co1l]

The diode would seem to be superfluous?  Unless you have another good reason for it.

Turn-on time will be fairly quick (in the 100ns range), turn-off can be estimated as t_off = 2 * Rgs * Qg(tot) / Vgs(on).

Switching losses can be estimated as P_sw = F_sw * t_off * Ipk * Vsupply / 2.

Whether this will be sufficient at 10kHz depends on load current, supply voltage, and how well heatsinked the transistor is.

Tim

[nuno]

The voltage after D1 must be within the Vgs limits of the MOSFET. I think that for that MOSFET (didn't check) that is -/+20V maximum (you should stay at least a few V below this limit).

[yashrk]

It is supposed to charge lead acid battery, instead of using a n channel in a push pull config I thought it will be easy to use p channel.

I am thinking of using 10Khz of freq because later I would also try to add buck converter to it.

[Chris C]

In your inital circuit there is a base resistor missing for Q2.

The base resistor isn't missing.  If you limit collector-to-emitter current by putting a resistor on the emitter as shown here, because base current must also flow out of the emitter, you get base current limiting for free.  The ratio between base and collector current tends to be set by the gain of the transistor.  For example, if the gain is 200x, and the resistor permits 10mA to flow from collector-emitter when the transistor is turned on, base current will automagically be limited to about 10/200=0.05mA.

Plus you get something else for free too.  Without a base resistor getting in the way, the MCU can quickly turn off Q2 when it pulls the pin low.  Probably not an issue in this circuit, but in others where speed is required, it eliminates the need for the extra resistors/capacitors/diodes to assist in turn-off.

The diode would seem to be superfluous?  Unless you have another good reason for it.

Looks like it's there to keep power from flowing back through the MOSFET's body diode, should the power supply be turned off when the battery under charge is still connected.

Whether this will be sufficient at 10kHz depends on load current, supply voltage, and how well heatsinked the transistor is.

[yashrk], you should really post these stats.  Otherwise any answer you receive as to whether a one-transistor gate drive is sufficient, will be only a guess!

[PSR B1257]

you get base current limiting for free.

But the output of the PIC will forces this node to 5V. But it can't, hence the base emitter diode will clamp the voltage.

base current will automagically be limited to about 10/200=0.05mA.

Rubbish.
You can put any base current in the transistor with only little or no collector current at all.

Plus you get something else for free too.  Without a base resistor getting in the way, the MCU can quickly turn off Q2 when it pulls the pin low.

That's no plus. It's yust stupid.

[Chris C]

But the output of the PIC will forces this node to 5V. But it can't, hence the base emitter diode will clamp the voltage.
You can put any base current in the transistor with only little or no collector current at all.
That's no plus. It's yust stupid.

My SPICE program disagrees with you.



In which you can see:

1) The base/collector current ratio looking very close to the idealized example I gave.
2) More than "little or no collector current".
3) Limited base current, despite NO base resistor, and an UNLIMITED current source.

As I've built and am currently using many instances of this circuit, I can say the real word disagrees with you as well.

Despite the tone of your response, no hard feelings.  Because I've seen this several times in the past, thought similar things, and only recently realized how it worked.

[PSR B1257]

I do realise how it works, but over the several different circuits I missed the fact, that Q2 works as a emitter follower 
The emitter resistor of course limits the base current and allows the voltage of the emitter electrode itself to rise above GND.
You're right, in this case you need no base resistor, assumed the base voltage stays under (or is at least equal) the collector voltage. Otherwise the base current will be no longer limited.

[Chris C]

I do realise how it works, but over the several different circuits I missed the fact, that Q2 works as a emitter follower 
The emitter resistor of course limits the base current and allows the voltage of the emitter electrode itself to rise above GND.
You're right, in this case you need no base resistor, assumed the base voltage stays under (or is at least equal) the collector voltage. Otherwise the base current will be no longer limited.

Spot on.    Was a simple misunderstanding then.  It is a bit confusing in here at the moment.

[PSR B1257]

Yes, a quite embarrassing misunderstanding...

Quick follow up:
Until the base base voltage reaches V_C+V_BE the current gain drops of like a brick wall. With a couple kOhm no problem, but with a low resistance...better don't try 

[T3sl4co1l]

Ummmmmmm... well yeah, C-B is a diode, so you're shorting that poor voltage source into said brick wall!

This can be a valuable feature, say for a complementary emitter follower / buffer.  The input voltage, as long as it is current limited or supplied through a resistance, will be clamped naturally to the supply rails by this action.

Tim

[yashrk]