How to design a circuit of INPUT 7V (2ma) to OUTPUT 5V (360ma)

good day to all!!

im a newbie here and i would like to seek for your help

were having a project about the voltage of kamias fruit

we found out that the voltage of kamias in our experiment is 7V, 2ma

we would like to design a dc to dc converter that will make that 7v,2ma input to 5v, 360 ma output...probably for a cellphone charger.

1. Is this project design possible?
2. Is there a design available without using ic (problem is the availability of ic here in philippines)

THANKYOU SO MUCH!!!!!!

Basic physics applies everywhere: conservation of energy ..
Input: 7V * 2mA => 14mW
And that (minus efficiency of dc-dc conversion) you can have at the output ..

If the dc-dc converter has to output 5V @ 360mA, that’s roughly 1.8W, you have to “inject” at least 1.8W + efficiency of conversion from an external source (eg. Mains power) ..

IanP
;-)

TO: IanP

so much thankyou...i get your point there...

sorry for being so newbie in electronics..but how can i design it?

i mean what components best fit this design...

once again thankyou!!!!!!

Short answer - you can't !

What IanP is telling you is that you can convert between low voltage at high current and a high voltage at low current although you have to accept some loss in the conversion.

The power is calculated by multiplying the voltage by the current, you can't make more power appear from nowhere. If you start off with 7V and 2mA, the power available is 0.014 W ( = 14mW) you can change the V and I but when multiplied together the result can't be more than 0.014.

So even if you ignore the conversion losses, to get 5V out the maximum current you could get is 0.0028A ( = 2.8mA) because 5 x 0.0028 = 0.014.

In other words, you can to some degree exchange voltage for current but you can't make the power any greater. Even at 100% efficiency input power and output power must be the same. In reality, the conversion efficiency will be much lower so you would get even less power out.

Brian.

i see....our project is not possible then?

what if a current multiplier is added on the circuit to make that 2ma amplified?

hankyou brian.

hoping to learn more from you guys!

If you use a current multiplier the problem of the power remains the same, being voltage * currente.
If you increase the current, increases the power on the same amount.

Sorry but is umpossible to create power, anyway, if you can find your generator with 7 vot and higher current, I suggest a low drop voltage, because the efficency of the DC-DC converter is similar the one of the low drop.
This is true if the input-output voltage difference is low, if it increases, the advantage of the DC-DC converter is evident.

mandi