# Voltage and current in smps and charge pumps

Status
Not open for further replies.

#### Power_user_EX

##### Newbie level 3
Hi all,

The below doubts are w.r.t DC source/circuits.

#1)I m really confused with the relation of I and V in power supplies and DC DC convertors.

For e.g. given a 5v battery @ 1000mah.

Now if i want to convert 5V to say 50V or 100V then what will be the max current that the any load can take across the output.

#2) I mean is voltage increased at the cost of amperage/current in charge pumps , voltage booster ??? =or= is current and voltage inter-conversion possible ???

#3) also, i want to know more about constant current circuit - plz guide.

#4) Assume that i have a 6V NiMh battery source @ 2500 mah and i have 6V 'DC' supply form an eliminator or SMPS which has 220V AC power source. Whats the difference between the 2 power sources w.r.t current ?????( I believe 6V from eliminator has constant current.)

#5) What is the effect on load like a DC motor if it is subjected to constant current source opposed to a battery source.(consider the dc motor is having full 'load' so that it eats up max amps)

-thx,
Power_user_EX

can any one comment ?

For 1+2, refer to conservation of energy as a fundamental physical law. As a consequence, below equation is valid e.g. for
voltage conversion (Px = Ux*Ix)
Poutput = Pinput - Plosses
Because any real circuit has losses, Poutput is always smaller than Pinput, or Uout*Iout < Uin*Iin

The current unit is mA or A. mAh or Ah is the battery capacity (current multiply time).

DC supplies ("battery eliminator") have mostly constant voltage rathen than constant current.

For e.g. given a 5v battery @ 1000mah.

Now if i want to convert 5V to say 50V or 100V then what will be the max current that the any load can take across the output.

Actually you know, you can get 50V,1A or 50V 200A. It also depends on the type of battery. The unit "ah" dictates as how much current can flow in just one hour. Your above battery can give 5V at 1A for 60 minutes. I can get 5V at 2A for 30 mintes. So this means I can get more current if i reduced the time. So this means I can also get 50V,1000A for a fraction of time? Maths says in this, if you reduce the time you can get more current. You can consume your battery qucikly. But problem is also with the construction of the battery. As soon as you started to draw more current, voltage starts to drop also. It's just because of the physics of the battery.If you get bigger,bulky size battery, suppose 300AH. You can then draw 5V at 10A,5V at 100A or even 5V at 1000A. Your battery voltage won't be reduce, they will remain at the same potential. Now I'm getting more current at the cost of time in bigger batteries, but in smaller battery, you have the cost of time+physics of the battery. So there's no formula in electronic that can say how much current at 50V or 100V. Ofcourse you would be stepping up the voltage from 5 to 50,100V. So you would be needing plentiful supply of current also.Your battery physics must tell you how much current at 50V or 100V.

Status
Not open for further replies.