Increase in Current

reduce your resistance (ohms law)

Increase current in what?

connect the source in parallel.

Dude He is talking about the increase in amount of current not the current capacity.

Well, if the voltage source remains the same, the current in a circuit can be increased if we reduce the circuit resistance, by Ohm's law, right?

But can anyone tell me what is a current source or current generator?

On the current source or generator, I guess that while it may be exactly the same as a voltage source or generator, this way of addressing may become relevant when the generation of a high current is of more concern than the exact value of the voltage. Like in the case of current transformers, spot welders, arc furnaces or welding machines?

just put op-amp TDA2030A to your circuit.Gud luck!

Current is a dependant component. It gets drawn by the load in sufficient quantities, depending on the nature of the load and the supply voltage. It is more correct to say that the current is pushed through the load by the voltage applied to the circuit. To change the current, the resistance of the load or the supply voltage has to be changed.

An example is a 12V car battery. While the battery may be structurally and chemically able to deliver 300A on demand, the 24W lamp is satisfied drawing the 2A it needs according to Ohm's law, and no more. There is no way you can increase the current to that lamp without increasing the voltage or dropping the resistance of that lamp.

You can make a current source just as easy as a voltage source. It works like a voltage source in that a voltage source, say a battery is 9V, the voltage will always be 9V and the current is allowed to do what ever it wants to do(it will follow ohms law V=IR). A current source can be set up to deliver or draw say 3 amps. The voltage is then the independent variable and will do what ever it wants (again ohms law). What is your application and what kind of currents to you need?

If you draw too much current from a constant voltage source, the voltage will drop accordingly. A 9-volt battery is not always 9-volts. A 9-volt battery is approximately 9-volts at the beginning of a load, it will immediately begin to decline in accordance with the load placed on the battery. The decline may be inperceptible over some period of time but will occur, and will accelerate. The approximately 9-volts will remain until the source is depleted enough to be detected at approximately 8.9999...-volts, etc..

You would need to increase your current capacity of your source by adding parallel sources, thus allowing you to maintain a relatively constant voltage applied to the load (If your source can supply the increased current demand adequately then this is not necessary). Additionally, you would need to increase voltage or decrease resistance to increase the current in the system. To increase voltage you need to serial sources to step up voltage.

Question is not clear. Conditions ??? Voltage constant?, load???, capacity of power source?? Why you need ie the purpose???

Dear Sir,

What is the load ? Can you use Darlington Pair to boost up the current while

keeping the voltage almost the same.( I am not good in Electronics).

Manroop.Chennai,