bjt totem pole

[wakoko79]

I'm using a bjt totem pole to drive a MOSFET. It seemed to work but I'm not really very familiar with the way it works. This has been bugging me for some time and I remembered to post it here this time.
View attachment 63297

The totem pole can source or sink current right? that is why I think it is a good driver for MOSFET for switching speed reasons.
But here, the input to the totem pole is from practically 0V to around 11V.

For an example: the initial state of the MOSFET is ON, so Vgs => Vth(on) (say 10V)
Then I turned the input signal to 0V. At some time, the lower part of the totem pole will turn on untill Cgs is drained. After Cgs has been drained (or Vgamma of pnp is below 0.7V), both npn and pnp of the totem pole will be in cutoff, am I right? As the things stand, isn't the gate of the MOSFET left floating? Or will it be held as the same voltage as the input (0V)?

 

[ronsimpson]

Your two circuits are not the same but.....
What you call a totem pole "MOSFET" driver has a voltage gain of 1 and a high current gain. If we forget the 0.6 volt Vbe loss the input voltage and the output voltage are the same.

Using your example; (collector of bottom transistor on 0 volts, collector of top transistor on 12 volts) the bases have 10 volts, the emitters will have a voltage near 10 volts. If the output is pulled above 10.6 volts the bottom transistor turns on and pulls the output down. If the output drops below 9.4 volts the top transistor will turn on and lift the output. It is true that when the output is very close to 10 volts the base-emitter voltage on the transistors is less than 0.5 volts and both transistors are off. You could say the MOSFET gate is left floating. The gate has a large capacitance and very high resistance so it will tend to stay at 10 volts. (10 +/- a little)

If the bases are now driven to 0 volts the bottom transistor will pull the gate down with high current. When the gate capacitors are discharged to about 0.6 volts the bottom transistor will no longer have current. The gate is left 'floating' at maybe 0.5 volts. (+0.5 to -0.5v)

Read about 'emitter follower'.

 

[crutschow]

The input must be high to turn off the MOSFET. For that condition the MOSFET gate will be held at about one diode drop above ground. It's not really floating.

 

[wakoko79]

I'm sorry that I confused you, what I was talking about in the post is the picture above, not the one below. I first posted the high side driver but then I realized Its easier to make an example out of the low side driver that is referenced to the ground. Sorry about the mix up.

 

[ronsimpson]

You are talking about the top picture!
What is the voltage of the collector of Q10?
Q1 collector is at about 0.2 volts when low and 13.5 volts when high.
Q5, Q10 emitters will also want to be at about 0.5 volts and 13 volts.

Q8 D5 I can not see and do not understand the reset of the circuit.
If you think this is a high side driver you will be unhappy with it.

 

[wakoko79]

You are talking about the top picture!
What is the voltage of the collector of Q10?
Q1 collector is at about 0.2 volts when low and 13.5 volts when high.
Q5, Q10 emitters will also want to be at about 0.5 volts and 13 volts.

Q8 D5 I can not see and do not understand the reset of the circuit.
If you think this is a high side driver you will be unhappy with it.

Q10 collector voltage is 0V, it's grounded.

Q8 is the switching element. D5 is there because cathode of D5 can be connected to a microcontroller, it is also connected to an LM393 which may be destroyed if too much negative voltage is applied to the input terminals.

NO, this is the LOW side driver. LOL =)

 

[Ubergeek63]

you can NOT put 23V on a 20V MAX gate!

 

[audioguru]

you can NOT put 23V on a 20V MAX gate!

The first schematic used R1 and R8 as a voltage divider so the max gate voltage was +10.8V.
R8 disappeared from the second schematic.