I need to deliver 2A through a 2 ohm load running at 4V.

The problem is that I have a circuit with a battery and voltage booster that provides 5V@500mah.


I don't want the battery to run low after a few minutes so I thought to boost the current coming from the battery.

Is that possible using a transistor in order to make a constant current source or what else should I use?

 

You can increase the current but the output voltage will be lower, and the available energy from the battery won't be increased.
What exactly, is the battery you are using and its load?

 

You can't boost both the current and voltage from a battery (Or any other power source.) The power after boosting the voltage (At reduced current.) or boosting the current (By reducing the voltage) will allways be a bit less than the power from the source due to losses in the conversion.

Les.

 

You can increase the current but the output voltage will be lower, and the available energy from the battery won't be increased.
What exactly, is the battery you are using and its load?

I have a LiPo battery (3.7V@500mAh), and a 2ohm load.
Will I be able to create with it a 2A current source?

 

You need 4 volts across a 2 ohm resistor to get a current of 2 amps. (Ohms law.)
If you could step the 3.7 volts up to 4 volts with 100% efficiency AND the battery gave its 500 mAH capacity at such a high current for it's size then it would run for about 13 minutes.
Les.

 

You need 4 volts across a 2 ohm resistor to get a current of 2 amps. (Ohms law.)
If you could step the 3.7 volts up to 4 volts with 100% efficiency AND the battery gave its 500 mAH capacity at such a high current for it's size then it would run for about 13 minutes.
Les.

I fully understand that, hence I emphasized that I would not like the battery to drain after a few minutes and asked whether there could be a circuit that will boost the current from the battery to 2A so the battery wouldn't drain as much as fast.

 

Read posts #2 and #3 again.

Les.

 

Read posts #2 and #3 again.

Les.

I'm sorry but the question still remains - based on the information that energy should be preserved, if I>I0 (current boosted) than v<v0 (voltage dropped), if I make this circuit, will it drain the battery slower (since current is multiplied in the transistor)?

will this do the job?

 

4V x 2A = 8W. However you do things this is the minimum power you need from the battery, whatever you do with voltages and currents. Your battery is rated at 1.85W for an hour (not going into the spec sheets for conditionals). The laws of physics dictate this. At absolute best your battery (stated as 500MAH) would only give 2A for 15 mins and at high current output even this would be unlikely. Another point to check is the recommended current drain on the battery. If 2A is well above this then you may damage the battery.

Tracy