Transistor amplifier voltage gain

[king.oslo]

Hello there,

I need to make an ultrasonic power amplifier. I have looked at class A and B and AB.

I have 150V 7A adjustable PSU single supply, I need to amplify a sine from function generator to 70Vrms 5A or higher amplitude out. That's easy.

My problem is that I want V_OUT to swing really close to V+ and GND, so that V+ can be as low as possible. That way i can dissipate as little heat as possible/high efficiency. How do I design the gain stage to achieve a voltage swing close within the limits of the supply?

Thanks.M

[Mechatrommer]

I have 150V 7A adjustable PSU single supply, I need to amplify a sine from function generator to 70Vrms 5A or higher amplitude out. That's easy.

can you share the schematics to us?

[king.oslo]

So far I don't have any schematics. I will make the schematics when I know how I am going to make the gain stage.

Thanks.M

[SeanB]

You will need an auxiliary rail then to power the output stage, generally only about 10V or so. Look at bootstrapping audio amplifiers, as they make the auxiliary supply by using the output stage to derive this extra supply using a capacitor and a diode.

[king.oslo]

You will need an auxiliary rail then to power the output stage, generally only about 10V or so. Look at bootstrapping audio amplifiers, as they make the auxiliary supply by using the output stage to derive this extra supply using a capacitor and a diode.

Dear SeanB,

I am certain that you are correct.

My understanding of bootstrapping is taking the output of an amplifier and feeding it into the input to get more gain. Is that correct?

I drew up an amplifier, where exactly would you wire in the capacitor to get the extra gain?

Thank you for your time.

Kind regards,
Marius

[SeanB]

http://en.wikipedia.org/wiki/Bootstrapping_%28electronics%29

http://www.ti.com/lit/an/slyt077/slyt077.pdf

Will help

[olsenn]

My understanding of bootstrapping is taking the output of an amplifier and feeding it into the input to get more gain. Is that correct?

You're thinking of a darlington pair

[king.oslo]

No, I was thinking of taking the output of an amp, and feeding it back into its own input. Positive feedback.M

[SeanB]

In this application it is charging a capacitor to provide a rail higher than the supply so that a power device can be driven closer to the power supply, with the actual drive being able to briefly be higher than the supply rail so that it is still saturated.