5v ---> 13v DC amplification.

[sonaiko]

I have a 5v voltage source pure DC with no ripple or AC components.
I want to amplify this source into 13v DC also without any AC components.

I know mostly im gonna use a common emitter transistor, But i actually dont know if i must use BJT or FET and i dont know the exact model # i should use. Also help me in choosing the values for the resistors i need and the biasing voltage i must use.

thanx.

 

[Nigel Goodwin]

I have a 5v voltage source pure DC with no ripple or AC components.
I want to amplify this source into 13v DC also without any AC components.

I know mostly im gonna use a common emitter transistor, But i actually dont know if i must use BJT or FET and i dont know the exact model # i should use. Also help me in choosing the values for the resistors i need and the biasing voltage i must use.

You give no where near enough information, I don't have the faintest clue what you want to do!.

For a start, is it a signal or a power supply you are talking about! - this basically determines if it's easily possible!.

 

[audioguru]

Hi Nigel,
You know what Sanaiko is talking about. He doesn't know how much output current he needs, nor how much his source can supply, but that can be determined later.
He heard that a common-emitter BJT or common-source FET has voltage gain. So with the appropriate amount of negative feedback, then 13VDC will come out of its collector or drain like majic! :lol:

Wouldn't that make our lives much easier?
Imagine powering just about anything from an AA cell and a single transistor as its voltage amplifier! :lol: :lol:

 

[sonaiko]

no guys thats not the story. Im actually gonna use that voltage to control my House's electronic door using my PC. Parallel port supplies 5v DC voltage and the door needs 13v DC voltage to open. The point is that i dont know how much current the PC supplies nor how much the current the door needs.

What can i do for u to help me?

 

[Nigel Goodwin]

no guys thats not the story. Im actually gonna use that voltage to control my House's electronic door using my PC. Parallel port supplies 5v DC voltage and the door needs 13v DC voltage to open. The point is that i dont know how much current the PC supplies nor how much the current the door needs.

What can i do for u to help me?

That's makes rather more sense now!.

Firstly you don't 'amplify' anything really, you use the logic output from the PC to work a 'switch', that switches the 13V supply to the door.

If you look at my PIC tutorials, under 'Hardware Extras', I give a circuit for feeding a relay from a PIC output pin - this is essentially the same as a parallel port output pin, and you can use exactly the same circuit.

It doesn't matter about what the PC supplies (it's just a TTL logic output), but it DOES matter what the door requires - you need to pick a relay that has contacts that can handle the door current - so you really need to measure it, or find it's specifications.

 

[Styx]

assume 100mA, not useful at all
use that 5V signal to drive something else connected to a 13V supply

 

[mstechca]

Firstly you don't 'amplify' anything really, you use the logic output from the PC to work a 'switch', that switches the 13V supply to the door.

Thats a easy way out, BUT, wouldn't you need to attach 13V power source to the collector, the data to base, and the emiter receives the output?

What he wants is 13V from 5V.

I think the best answer is to use what is known as a "charge-pump" circuit. If you hook it up to a ladder of capacitors and diodes (Johnson or cockroft ladder. I not sure the name of it), the voltage multiplies quickly, but the current might be less.

If you find that insufficient current is being applied to the door, then look for any resistors in the door circuit and lower their values, but don't lower them too much or else something in the door circuit will malfunction.

 

[jrz126]

The parallel port wont be able to provide enough current. Assuming it can handle 100mA at 5V .1mA * 5V =.5watts, .5W/13V = 38.5 mA. And that 100mA is a pretty big assumption.

I think the best thing to do is find out how much current the door draws. If the current is pretty low (<5amps), then just use the +12V already available in the PC's Power supply. Then just use a transistor or something like everyone else suggested. I dont really see the 1V difference affecting it that much.

 

[Nigel Goodwin]

Firstly you don't 'amplify' anything really, you use the logic output from the PC to work a 'switch', that switches the 13V supply to the door.

Thats a easy way out, BUT, wouldn't you need to attach 13V power source to the collector, the data to base, and the emiter receives the output?

Presumably he already has that?.

What he wants is 13V from 5V.

Not from his second (more sensible) post, he wants to control a door from a PC.

I think the best answer is to use what is known as a "charge-pump" circuit. If you hook it up to a ladder of capacitors and diodes (Johnson or cockroft ladder. I not sure the name of it), the voltage multiplies quickly, but the current might be less.

It's called a Walton Cockcroft multiplier - of no use whatsoever for this application! - and (due to the laws of physics) the current halves as the voltage doubles (in fact more than halves, due to losses), and it also only works off AC.

The power available from a PC parallel port will light an LED, and that's about it! - NO WAY CAN IT FEED A MOTOR.

If you find that insufficient current is being applied to the door, then look for any resistors in the door circuit and lower their values, but don't lower them too much or else something in the door circuit will malfunction.

That's the biggest load of rubbish I've seen yet!, all you are going to do is kill the parallel port - YOU MUST NOT CONNECT IT DIRECTLY TO THE PARALLEL PORT!.

 

[audioguru]

MStechca,
Why use an emitter-follower with its resulting max output voltage of only 4.3V?
Just use a transistor with its emitter grounded and a current-limiting resistor feeding its base from the parallel port. Its collector will swing nearly the full 13V and its saturation voltage will be less than 0.1V, if the load current is only 100mA.

 

[mstechca]

Since there seems to be no solution, another one may be available.

Use some sort of wireless technology. Radio technology may be OK, if you can prevent the motor from causing interference. Maybe use a microcontroller in the door lock and when a certain bit pattern is received, the motor turns on.

The only way I can see the wireless method fail to work is if there is nearby interference, or if you are sending data and having it processed asynchronously.