How to amplify current using NPN transistor

[Reynard]

Hi all

I have fixed up the circuit as shown in my attached picture and i measure the amplified current (that i wanted) between the LED and the Collector using an ammeter.

I used a power supply set to 120mA and when i input 5V, i measured 0.433mA as the base current and got 8.199mA as the collector current which gives me a current amplification of aboput 19times. Is this the correct way to measure the amplified current?

However this is only for testing. If my actual input is about 40mA @5V and i want to achieve 100mA @ 5V what should i modify my circuit to?

Cheers

 

[bloki]

First, I am not sure you understand settings on your power supply. Current setting on PS is nothing but setting load current limit for PS. If you want to draw more current than preset value it will not be possible because output voltage will drop.
Second, DC current gain of transistor is measured correctly. It is simply Ic/Ib.
Third, you can decrease base resistor to increase collector current. But the maximum collector current will be determined by supply voltage, 470ohm resistor, LED voltage drop and Vcesat. That is:

Ic-max=(5-Vled-Vcesat)/470

Your max collector current can be about what you measured, abt 8mA.

If you want 100mA collector current you should first set PS current limit above 100mA and reduce 470ohm resistor. But when you do this your LED will probably die. Maximum LED current is about 20mA for ordinary LEDs. For LED you are using find datasheet and read maximal ratings.

 

[RadioRon]

Your method is correct, but you are measuring the current amplification for the circuit as it is, but not the MAXIMUM current amplification that the transistor is capable of. And often it is the maximum that we are interested in. In this particular circuit you are pushing so much current into the base that the transistor is what we call "saturated". It is operating in that part of its transfer characteristic where there is so much current flowing through the LED and the collector that all the voltage available from the power supply is being dropped across that 470 ohm resistor and the LED. When it is operating like this, it doesn't matter how much more current you push into the base of the transistor, no more current can flow through the collector because there is no more voltage available.

In order to measure the maximum current amplification that this transistor can deliver, you would have to do one of two things, either reduce the current that you are pushing into the base, or change the collector load resistance (make it lower) so that there is some voltage left for the transistor. Since it is risky to raise the collector current too much, the safe thing to do is lower the base current. You can do this by simply raising the value of the base resistor. Raise it until you measure a collector voltage of approximately 2.5 volts. Since we often expect a typical NPN transistor to give a DC current amplification of about 100, I would expect that you need to raise the base resistance to 270Kohms or thereabouts. I hope your meter is able to measure values as low as 15 uA.

Now, having said all that, let's consider how to modify your circuit so that you can put 40 mA in and get 100 mA out. Of course, I'm curious why you want to do that because most people don't want to put that much current into the base (40 mA that is). Well, anyways, if you want to put 40 mA into the base, you would simply reduce the base resistor value to 107 ohms. To get 100 mA to flow through the LED in this case, you need only reduce the 470 ohm resistor down to about 42 ohms. That is, assuming the transistor doesn't burn out first, or the LED doesn't burn out. Or, for that matter, you might burn out that 42 ohm resistor too.

 

[audioguru]

You should have more than only 8.2mA for a collector current, if your battery voltage is actually 9V (it doesn't drop to 6V when the LED is on?). Maybe the forward voltage drop of the LED is about 5V which is pretty high even for a white or blue LED.

Calculation:
9V battery.
2V red LED.
0.2V transistor saturation voltage.
Therefore the 470 ohm resistor has 6.8V across it and the collector current is 14.5mA.

 

[silviaS15]

Electronics Help!!

hi guys

i am building a rc boat an i have a 9V battery, is there any way i can connect the 9V to a transistor ore something simple so it puts out 12V, i just started electronics at school so i am new to it, if anyone knows anything helpful or similar please help

Thanks

 

[JBrock]

Transistor gain is far from constant. What is the part number of the transistor? Are you sure it is a bipolar NPN silicon?
Why are you operating it in common-emitter mode? A series resistor alone is a better power-dissipating device with a lower temperature co-efficient.

 

[silviaS15]

well idk wat im talking about reakky i just wanna make sumthing simple to get 12v or higher out of a 9v

 

[silviaS15]

i just want to have the light weight of a 9V but get the power of a 12V or higher, can i use a transformer, amplifier, or something, if you know just tell me the simple alterative

Thanks

 

[Hero999]

You don't want an amplifier, you need a boost DC-DC converter.

Amplifiers can't exceed the supply voltage - the only way to do that is to use a switching regulator.

How much current do you need?

How big is the 9V battery?

Wimpy 9V batteries can't supply much power for very long.

 

[Roff]

hi guys

i am building a rc boat an i have a 9V battery, is there any way i can connect the 9V to a transistor ore something simple so it puts out 12V, i just started electronics at school so i am new to it, if anyone knows anything helpful or similar please help

Thanks

Please don't hijack someone else's thread. If you have a question, start a new thread by going to the forum's home page and clicking on the New Thread button.

 

[Hero999]

Whoops, I didn't notice that.

I shouldn't have responded.

If you have a question start a new thread, it's rude to interrupt.

 

[picbits]

If you have a question start a new thread, it's rude to interrupt.

And bump a 2 year old thread with a totally unrelated question.