Totem Pole question

Thread Starter

MatheusLPS

Joined Jul 16, 2011
34
Good morning all.

I am learning how to use a toten pole circuit to drive a MOSFET. At this time, I can make it work like the image below:

http://img.photobucket.com/albums/v222/ahhh/toten_teste.jpg

The input signal is a PWM signal. On the image above, i am not sure but I guess it has 10% in duty cicle.

So, the problem is that the circuit is working in reverse.

With 10% of the PWM I have 90% of the current flowing on the FET.

I want to have 10% on the PWM and 10% of the current, 90% on PWM, 90% in current. That need to be direct proporcional.

I tried to use a PNP transistor but did not worked.

bye
 

Thread Starter

MatheusLPS

Joined Jul 16, 2011
34
I tried that but when i removed that resistor and transistor, my gate signal is 5V and not +12V.

I want to keep the +12V to avoid the ohmic region.

bye
 

crutschow

Joined Mar 14, 2008
23,361
Just a nit, the circuit shown is a push-pull stage, not a totem-pole. A totem-pole uses two transistors of the same polarity. ;)
 

Ron H

Joined Apr 14, 2005
7,012
Just a nit, the circuit shown is a push-pull stage, not a totem-pole. A totem-pole uses two transistors of the same polarity. ;)
Carl, you're an old guy like me. I had the same thought. A little Googling convinced me that a push-pull emitter follower has come to be known as a totem pole stage.
International Rectifier even does it.
I know, it ain't right, but the same thing has happened to a lot of other terms over the years.
 

Thread Starter

MatheusLPS

Joined Jul 16, 2011
34
You need to invert your input signal.
Where does your PWM signal originate?
What is the frequency?
If i invert the signal, the output is inverted too.

The PWM is from a microcontroler PIC @ ~ 25KHz

Just a nit, the circuit shown is a push-pull stage, not a totem-pole. A totem-pole uses two transistors of the same polarity. ;)
Yeahhh. You are right. push-pull. I googled it...

Sure I can make it work the way it is. I just need to remember that when I apply 10%, i will have 90% on the load. OK...

But if I can make it right......
 

JoeJester

Joined Apr 26, 2005
4,077
Apparently IR is fast and loose with their definition. Wiki agrees with us old farts ...

Totem-pole push-pull output stages
Two matched transistors of the same polarity (or, less often, Vacuum tubes) can be arranged to supply opposite halves of each cycle without the need for an output transformer, although in doing so the driver circuit often is asymmetric and one transistor will be used in a Common-emitter configuration while the other is used as an Emitter follower. This arrangement is less used today than during the 1970s; it can be implemented with few transistors (not so important today) but is relatively difficult to balance and so keep to a low distortion (the highly non-linear TTL circuits such as the 7400 use this arrangement).
 

tubeguy

Joined Nov 3, 2012
1,157
Sure I can make it work the way it is. I just need to remember that when I apply 10%, i will have 90% on the load. OK...

But if I can make it right......
I see your dilemma.. No matter what, the output current is inverted.
Because, the Mosfet always inverts it.
I think you need a P-channel Mosfet with the load 'below'. ??
 

tubeguy

Joined Nov 3, 2012
1,157
Type too fast, think too slow. (Happened again Ron) :)

What I was thinking of was a follower circuit. So, the N-channel would work with the load 'below'. But then I think you need an extra 4-5 volts positive to drive the Mosfet fully.

I started in electronics with tubes (of the vacuum type).
:D I still like them. :D
 
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