Problem with 2N3904 switching 180mA load

[scdvom]

[HELP] Amplifier

Hi. Can anyone help me build a circuit which given input 5V and output 9V, 180m Amp? and when input given is 0V, the output will then drop to 0V as well.

Thanks for help.

Here is my circuit. But it just work when i supply 5V, once I turn off it to 0V, my output still remain the same as 9V, it just varied the current value. Any changes can make my circuit work as what I want? Thanks

 

[keith1200rs]

[HELP] Amplifier

You might get more help with:

a) a better subject heading

b) a better place to show the picture - you can upload it here so everyone can see it!

Anyway, 1.8 ohms on 9V will be 5 amps. The poor old 2N3904 would go up in smoke if you put 5A through it, but you cannot with a 2k ohm base drive to 5V - the transistor doesn't have enough current gain. Even if you changed the load resistor to 50 ohms (9V/50 ohms = 180mA) the transistor would struggle due to the limited current gain.

What are you wanting to drive and does it need to use the 2N3904? Where does your driving signal come from? A microcontroller?

Keith.

 

[scdvom]

[HELP] Amplifier

Hi Keith
haha..thanks for your comments, but I dont know how to show the picture here.

Haha, 2N3904 is just randomly put in, because I'm to electronics, dont know which should I used. Anyway, actually I wan to drive a motor which required 9V and 160-200mA. Thats why I choosed 180mA. And the 5V in my circuit there actually is a signal (5V) comes from my H-bridge which drive by my PIC18F4550. So can you suggest me a trasistor which can work for my condition? And it is good enough if you can show me the whole circuits. Actually the circuits I show you here is workable, but just when my 5V falls to 0, the output still remain as 9V, but I want make it as 0V as well. What can I do now?

Your helping is greatly appreciated.

 

[betwixt]

Re: [HELP] Amplifier

Keith is absolutely right.

There is another issue though, you want the output to go to zero when the input is zero. If you mean voltage, this will never happen in an inverting configuration. think of the transistor as being a switch between collector and emitter. To make current flow you close the switch, to stop it you open the switch.

To close the switch you need current to flow between the base and emitter, that's what V3 and R1 are supposed to do in your schematic. This where two problems arise, applying V3 closes the switch and makes the collector go very close to ground, for practical purposes zero volts. In other words inversion took place. The other problem occurs when you remove V3. The transistor will not switch off completely, at least not in a controlled way. The reason is that an electrical charge is held between the collector, base and emitter junctions and also a small 'leakage' current will exist due to impurities in the semiconductor materials. You should add another resistor between the base and ground to leak away the residual charges. It can be quite high in value as the leakage and charge are very tiny, I would suggest 100KΩ as a starting point.

Brian.

 

[keith1200rs]

[HELP] Amplifier

It is probably easier to use a MOSFET than bipolar transistor. There are literally hundreds to choose from. A couple of suitable ones I picked for 20V are

FDY300NZ
TP0101K

I have attached a suggested circuit. The 50 ohm resistor is your load. As it is a motor drive, you need to include the diode to protect the transistor from blowing up when you switch the motor off.

As mentioned by Brian, this sort of circuit inverts in that 5V on the input will put 0V on the output.

Keith.