Stepping up voltage

[komalbarun]

All i want to know is if this circuit will work? if not, then why will it not work?

 

[BobK]

Define "work"

Bob

 

[komalbarun]

Define "work"

Bob

sorry about that. Will it step up the voltage, assuming it is an ideal step up transformer?

 

[duke37]

It should 'work' but it will depend on the component values as to what you get.

R3 (1k) will limit the power and you may get out less than you put in.

 

[komalbarun]

It should 'work' but it will depend on the component values as to what you get.

R3 (1k) will limit the power and you may get out less than you put in.

thanks

 

[BobK]

A better circuit would use a full H-bridge. DC to AC inverter.

Bob

 

[CDRIVE]

Yes, R3 is an unrealistic value. Way to high.

Chris

 

[komalbarun]

A better circuit would use a full H-bridge. DC to AC inverter.

Bob

what's a full H-bridge?

and yeah R3 wans't supposed to be 1k [ there wasn't supposed to be any value there xD]

 

[duke37]

You will need to specify the power supply and the components.
Fets will be better than npn transistors if they are being driven by logic circuits since they take little power to drive. They are also more efficient.

You can look up H-bridge yourself if you wish. It is a circuit which can change the polarity of the output, often used to reverse DC motors. It will have protection from the case where the two transistors are on for a very short time (shoot through) during switch over. Your circuit does not have this protection which may be a problem.

 

[komalbarun]

I will look up the h bridge. Thnx for the help.

 

[komalbarun]

You will need to specify the power supply and the components.
Fets will be better than npn transistors if they are being driven by logic circuits since they take little power to drive. They are also more efficient.

You can look up H-bridge yourself if you wish. It is a circuit which can change the polarity of the output, often used to reverse DC motors. It will have protection from the case where the two transistors are on for a very short time (shoot through) during switch over. Your circuit does not have this protection which may be a problem.

shoot through = both transistors on?
so this is a possible case?

btw thx to have talked abt h bridge. I knw a lot more now

 

[duke37]

Yes, shoot through is a term used where one transistor turns on before the other is fully turned off. This may not be a problem if the current is limited and it does not occur often.
An invertor running at 100kHz with a 1microsecond overlap will have problems, one running at 10kHz with the same overlap may not have a problem.

 

[komalbarun]

Yes, shoot through is a term used where one transistor turns on before the other is fully turned off. This may not be a problem if the current is limited and it does not occur often.
An invertor running at 100kHz with a 1microsecond overlap will have problems, one running at 10kHz with the same overlap may not have a problem.

lol I understood what you said ! I surprise myself sometimes lol xD

Thanks a lot for the help

 

[komalbarun]

Won't the H-bridge have a shoot-through time too?

 

[CDRIVE]

Yes and so does TTL. With high speed comes shoot-through and subsequent spikes on the supply rails. This is why good supply rail bypassing is important. On an H-Bridge it can be reduced greatly but at the expense of switching speed.

Chris

 

[(*steve*)]

A good H Bridge driver will turn one transistor off before turning the other one on to reduce or eliminate shoot-through. It's especially an issue with mosfets.

Actually CMOS has interesting issues with shoot-through. Sometimes these are called "crowbar currents"

 

[CDRIVE]

I imagine you're referring to a design that comprises more than your basic 2 NPN's, 2 PNP's.

Chris

 

[komalbarun]

I imagine you're referring to a design that comprises more than your basic 2 NPN's, 2 PNP's.

Chris

yes, thanks to all of you for the help