Hi,

I have seen the data sheet of LM7812 which is attached here also.
It can drive inexcess of 1A current.
But, in my circuit, while I am driving a coil in the load (of having 1.1mH inductance and resistance of the coil is 7 ohm) through class AB push-pull amplifier, I am able to get 800mA current, but the supply voltage gets reduced to 6V.

Would you please explain me the fact and how to overcome this?

Thanks & Regards.

 

What is the input voltage to the regulator?
How big a heat sink is it on?

I = E/R = 12/7 = 1.7A, so something is wrong with your math...

 

What is the input voltage to the regulator?
How big a heat sink is it on?

I = E/R = 12/7 = 1.7A, so something is wrong with your math...

Hi,

Thanks for your valuable feedback.
It is the circuit that is giving me this result. Don't know why?

Thanks & Regards.

 

And where is the circuit? post it?

Answers to my questions?

 

I have seen the data sheet of LM7812 which is attached here also.
It can drive inexcess of 1A current.

The datasheet should also indicate the drop out voltage which will give you the minimum input voltage requirement.

But, in my circuit, while I am driving a coil in the load (of having 1.1mH inductance and resistance of the coil is 7 ohm) through class AB push-pull amplifier, I am able to get 800mA current, but the supply voltage gets reduced to 6V.

Are you taking meter resistance into consideration for your current measurement? If 6V was measured without an ammeter in circuit, I=6V/7ohms=857mA.

Would you please explain me the fact and how to overcome this?

1. What is the power source for the regulator (e.g. x volts @ y amps)?
2. What is the regulator input voltage with your load?
3. Is there a heatsink on the regulator?