Push-pull driving a fet at atleast 1A!

Thread Starter

ElectronicNewbie0

Joined Oct 18, 2025
54
Hello, i have been working on a push-pull stage that has tò drive a fet at 1A.

Now what im doing Is sizing R1 and R2 , from what i have understood, there Is got the two diodes, in order to reduce crossover distorsion, where R1 and R2 sets only the current that will make the diodes always be conduction, i dont know exactly the voltage split we need to have, but i have seen that usually they choose R1>R2( so most of voltage will drop on R1 ), i have chosen to have an ID of 10mA ( typical forward current )

Where R1= VR1/ID and R2=VR2/ID

By doing calculations with a vcc of 12V, i found that
R1=10/0.01=1k
R2=0.6/0.01=60 Ohm

With this part done, now i got another question, in a push-pull amplifier the current will depends on the load right? Like what matters for us Is that out power supply ( vcc ) Is able to give us atleast 1A in output and secondly that the chosen transistors can handle 1.5 as IC current no?
for the npn i chose a BD139 and as pnp i chose the BD140
( For now thats what i have done , as next i will be choosing the MOSFET that the push-pull output will drive )

U got any feedback or suggestion to give me on the design? Am i thinking in the right way?

Thanks.
 

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BobTPH

Joined Jun 5, 2013
11,483
Is this a logic circuit or linear amplifier?

Biasing and crossover distortion is something to worry about in linear circuits.

Is this supposed to be a MOSFET gate driver? That is a digital circuit.
 

Thread Starter

ElectronicNewbie0

Joined Oct 18, 2025
54
Is this a logic circuit or linear amplifier?

Biasing and crossover distortion is something to worry about in linear circuits.

Is this supposed to be a MOSFET gate driver? That is a digital circuit.
Ur right here It doesnt really matter, then what u suggest me then? having the input directly applied to the base of the transistors? with having low resistor value attached tò VCC?

I could replaced diodes with a BJT npn transistor also, or am i wrong? or even in that case its kinda useless?
 

BobTPH

Joined Jun 5, 2013
11,483
Crossover distortion occurs because there is a delay between on side turning off and the other side turning on. But this is exactly what you want in a gate driver! It is called dead time, and good gate drivers do it purposely.
 
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