i have dc power surce of 45V DC and 40 Milliamps current. my output current is in mili's and i need atleast 3.5 Amp at output..how I can arrive to the 3.5 Amp ? any Electronic circuit can do that?

 

You need to obtain a power source that can deliver at least 3.5A. There's no way to increase the power output of a power supply with extensive modifications.

 

i have dc power surce of 45V DC and 40 Milliamps current. my output current is in mili's and i need atleast 3.5 Amp at output..how I can arrive to the 3.5 Amp ? any Electronic circuit can do that?

If you could do with 0.5V then 3.5A will be theoretically possible. Anyway, you will never get more power out than the initial 1.8W.

 

And always get a power supply that can provide more than what you need. Its never a good idea to run them at 100% capacity.

 

You could get 3.5A out of it if you were willing to settle for only ~437mV out.
40mA @ 45v is 1.8W, so 1.8W/3.5A = 514.3mV; figure ~85% efficiency in the power conversion and you get ~437mV.

Another thing you could do is to charge a capacitor for a period of time, until it got to nearly 45v, then apply the load. You can wind up with a good bit more than 3.5A that way, but it's only for a short period of time; the instant that the connection closes.

 

45V DC, but only a few tens of mA: does that ring any bells with anyone?

This sounds rather like someone trying to nick some power from a phone line. I would be delighted to hear some other explanation.

 

Yep, caught me half asleep this time.

 

Thank you SgtWookie

I can leave it down to 12V as my intantion is to be able to charge a car battery and I need to stay with 12v to 13v with 2 or more amps I like your Idea with the capacitor is there a drawing or more information on how to do it? and how after I charge the capacitore I can amplifier the current? please help.
Thank you

 

With only 1.8 Watts available, you will only get a maximum of 132mA at 13.6v; the float voltage for a 12v lead-acid battery. That is if a converter were 100% efficient; and you will not get 100% efficiency out of a converter.

You cannot use the telephone line as a power source, or the telephone company will turn off your telephone service and charge you extra for the power that you have tried to steal from them.

 

With only 1.8 Watts available, you will only get a maximum of 132mA at 13.6v; the float voltage for a 12v lead-acid battery. That is if a converter were 100% efficient; and you will not get 100% efficiency out of a converter.

You cannot use the telephone line as a power source, or the telephone company will turn off your telephone service and charge you extra for the power that you have tried to steal from them.

I am not using the phone line.....lol I have a solar panel that I am getting indor 45 volt and no Current at all only 40 milliamps I am trying to increase the current in any possible way. any one please help?

 

OK, maybe I jumped to conclusions about the phone line, but some people do come here asking about that idea.

You really need to understand though that there is a basic principle called conservation of energy, as a result of which the mean power output from your system cannot ever be more than the input power, but generally less due to inefficiencies.

Capacitors or other energy storage systems can allow short bursts of higher power, but then the system needs to recharge so that the long term average power is not increased. Amplifiers function by allowing a weaker input signal to control energy delivered by some other power source. They do not create power from nothing.