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Well.I am new to electronics. I have just started creating small circuits. Coming to the point,I have a circuit with 5v input and current in the circuit is 2ma but I have a requirement to increase it to 200 ma in a circuit. When I googled for the solution, I found threads like increase voltage or decrease resistance (using V=IR) to increase the current But those values are to be fixed in my circuit. Basically I Studied that CC transistor has high current gain so I thought to use it in my circuit but people on internet says it amplifies signals current(probably an ac signal i guess so) and I am unable to understand what does that mean.

What does voltage gain and current gain mean in a Transistor? Does it amplify only ac signals? If I apply 5v dc input to a CB configuration that has high voltage gain, Wont it make it as 20v at the output? Similarly if I input 10 ma current of a DC circuit to CC amplifier, Does not the CC make it as 1 Amp at the output?

After seeing the comments I added the circuit picture. Now I want to increase the current in the circuit where R is 2.5Kohms and E1 is 5volts and the current is 2 ma but I want the current to be increased to 200 ma but keeping the R1 and E1 fixed. But Im unable to figure it how.Please kindly clear me this. and let me know what should i prefer if i want to increase current in a circuit keeping the voltage and resistance values constant in a circuit.Thanks

  • \$\begingroup\$ How would the current have to increase 100-fold if the resistance is fixed? \$\endgroup\$
    – stevenvh
    Sep 30 '12 at 10:41
  • \$\begingroup\$ @stevenvh yes how the current to be increased if the resistance and voltage source is fixed. \$\endgroup\$
    – niko
    Sep 30 '12 at 10:42
  • \$\begingroup\$ I think you'll have to explain where the value of 200 mA comes from. What do you have connected to the 5 V source now, and how do you need to change it? \$\endgroup\$
    – stevenvh
    Sep 30 '12 at 10:45
  • \$\begingroup\$ Show us your circuit. \$\endgroup\$
    – Dave Tweed
    Sep 30 '12 at 11:13
  • 1
    \$\begingroup\$ I have a feeling this is a homework problem of some sort, in which case they are probably looking for him to know to put a resistor in parallel. I can not image this problem in a real work place unless you were stuck with a load that needed X current and you needed to make it work with a certain supply. \$\endgroup\$
    – Kortuk
    Oct 1 '12 at 6:54

You can't. Actually it's all Ohm's fault:

\$ I = \dfrac{V}{R} \$

The current is defined by voltage and resistor. If you have 2 mA at 5 V then your resistor will be 2500 Ω. End of story.

If you want to increase the current you'll have to 1)increase your voltage, 2)decrease your resistance, or 3) a combination of both. For instance:

\$ I = \dfrac{V}{R} = \dfrac{50 V}{250 \Omega} = 200 mA \$

or also

\$ I = \dfrac{V}{R} = \dfrac{5 V}{25 \Omega} = 200 mA \$

A transistor can be used to increase current. You'll have a low current path, from base to emitter in an NPN, and a higher current path from collector to emitter. The collector current will be a multiple of the base current if the circuit allows it. That means that the voltage source at the collector side must be high enough, and the load resistance low enough to get the 200 mA. Again Ohm's Law: if you have a 1 kΩ collector resistor you will never get more than 5 mA with a 5 V supply, no matter how hard the transistor will try to draw more.

About your comment on controlling a LED. First 200 mA is a lot for a common LED, too much actually. A value like 20 mA will be more than enough for an indicator LED. Your LED will have about 2 V across it, that's a voltage you subtract from the 5 V. The remaining 3 V will be across the series resistor. So we apply Ohm, again:

\$ R = \dfrac{V}{I} = \dfrac{3 V}{0.02 A} = 150 \Omega; \$

If you use a 150 Ω resistor in series with the LED you'll get 20 mA. You don't have to amplify anything for that; it's the resistance which dictates the current. If you would simply use the 150 Ω resistor without the LED then the full 5 V will be across the resistor and the current will be 33 mA = 5 V / 150 Ω. Again no need for amplification, just changing the resistance or voltage will do.

  • \$\begingroup\$ 20mA is a great steady state value, but he could get more brightness out of the LED if he PWMs it, but that's jumping a little ahead here... I think I've done over 100mA before, but can't remember what the important LED spec is to look for. "Trial and error" keeps popping into my head, but I'm pretty sure that's not on the datasheet. :) \$\endgroup\$
    – Dave
    Sep 30 '12 at 15:59
  • \$\begingroup\$ @Dave - Ooooo, that's bad. Granted I have done this also, but it's not good. That 20 mA LED may have an Absolute Maximum Rating of 30 mA, and then 100 mA is a Bad Idea™. You will get the same brightness at 100 mA for a 20 % duty cycle, but you'll reduce the LED's life. \$\endgroup\$
    – stevenvh
    Sep 30 '12 at 16:14
  • \$\begingroup\$ :) Got it. I learned this in school many many years ago, and that's right, we were told the LED lifespan will suffer. But for the one time I did it for a school project, it worked out okay. I'll definitely keep that in mind and not do that in the future! \$\endgroup\$
    – Dave
    Sep 30 '12 at 16:39

Now I want to increase the current in the circuit where R is 2.5Kohms and E1 is 5volts and the current is 2 ma but I want the current to be increased to 200 ma but keeping the R1 and E1 fixed.

Solution 1: (Parallel Resistor)

Put another resistor in parallel to R1 with a rating of 15Ω. Your E1 & R1 will remain fixed and the current increases. ;)

Let the parallel Resistor value be R:-

\$ I = \dfrac{V}{Req} ..........(equation 1)\$

\${Req}= (R1 * R)/(R+R1)\$

putting \$ I =200mA , V=5V , R1=2500 \Omega \$

\$ R = 15 \Omega\$

Solution 2 (Another Voltage Source):

Put in series another voltage source of 495 V(sounds more Theoretical).

In \$ equation 1 \$ set Req = \$ 2.5 k\Omega \$ and \$ V= 500 V\$
\$ I \$ becomes \$ 200 mA \$

  • 1
    \$\begingroup\$ Please don't actually put in series another voltage source of 495 V! \$\endgroup\$
    – kvanbere
    Feb 11 '15 at 9:44

You can get 200 mA to flow, only not through that resistor.

You can keep R1 and E2 in that circuit where they are and provide a parallel path in which 200 mA is able to flow. Thereby you satisfy the requirement that R1 stays at the same value and in the same configuration relative to E1.

R1 could be involved in a transistor circuit such that approximately the same 2 mA flows through R1, and this turns on the transistor such that 200 mA flows.

Take a look at base bias under http://en.wikipedia.org/wiki/Bipolar_transistor_biasing under "Fixed bias".

Under this arrangement, the collector resistor RC might be chosen so that it limits the current to 200 mA when the transistor is fully turned on (saturated). Your original R1 resistor plays the role of RB, the base resistor.


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