# Increasing amps in a circuit

I was wondering if it was possible to increase the amount of amps in a circuit. Say if I wanted to get 12 volts at 1 amp to something like 6 volts at 2 amps. Is this even possible and if so, is there a feasible way of doing it?

I am confused by your question but the answer is probably "Yes, but...".

You see, you seem to be assuming 100% efficiency. What you want to do could possibly be accomplished with a DC to DC converter but the efficiency might be closer to 80%, so perhaps you would get 6 volts with 1.6 amps available.

Is this theoretical or do you have a real need?

This is theoretical, and I was just wondering if it could be done. I will see what I can learn about DC to DC converters

podsol:
Say if I wanted to get 12 volts at 1 amp to something like 6 volts at 2 amps. Is this even possible and if so, is there a feasible way of doing it?

LM2596 ultra-cheap module from eBay can do that, you can buy ten of them for peanuts.

Every computer motherboard has a powerful switchmode converter dropping 5V down to the processor
voltage (which is usually less than 1V these days). So 5V 15A becomes 0.9V 80A or something like that,
using MOSFETs capacitors and inductors.

There are many ways to do DC-DC conversion, usually you'll see switchmode buck converters, but
there's also switched-capacitor schemes, motor-generator sets, or inefficient linear regulators which
can't increase the current, and throw away the extra power as heat (but are electrically much quieter).

I used to have a shoebox size motor-generator. Put in 12 volts at about 1/2 amp, get out hundreds of volts at a small fraction of an amp. Of course, I never got out quite what I put in (there was noise, there was heat, the motor had to keep spinning despite friction) but I didn't care because the result was amazing.

Look at the Pololu DC-DC regulators here, High voltage to Low voltage (step-down), and Low voltage to High voltage (step up). And some that will do Step-down until the source voltage gets too low, and then switch to Step-up to hold the same output voltage.

even asked such a question. Obviously, it is because you were unaware of the following:

A. Power(in Watts, expressed as the unit “W”) = Current (in Amps, expressed as the unit “I”) x V (volts)

So for your case , we :
Let V = 12 V
I = 1A

CASE A: Inefficient Monolithic Linear Voltage Regulator

Primary Power source V1 = 12V

Power source V2 (LM7806 (see page 24)) = 6V

Power source V3 (LM7806 (see page 24)) = 6V

The 7805 voltage regulator is a great device if you want a simple way of bringing a voltage down to 5V. It’s a three-pin, one-component solution that puts out five volts and a lot of heat. … Depending on the input voltage you might see 50% efficiency. Going to a switch mode supply, that efficiency shoot up to about 90%.May 13, 2015

PEfficiency = POUT/PIN

If V2 & V3 are 50% efficient, then

PV2 = IV
= 1
6 = 6W (Total)
3W LOST AS HEAT

PV3 = IV
= 1
6 = 6W (Total)
3W LOST AS HEAT

V2 & V3 = 6V @ 1A (50% of total power lost as heat)

For 80% efficient switching regulator/DC to DC Converter,

0.82 = 1.6A
1.6A/2 = 0.8A
P12V = 1
12 = 12W

12W/2 = 6 W

0.8*6W = 4.8W

P = I * V

I = P/V

= 4.8W/6V

I = 0.8A (800mA) (Expected current delivered to 6V load @4.8W)

If you need more current, use a 12V supply that has a higher rated current (ie 2A)

vaj4088:
I used to have a shoebox size motor-generator. Put in 12 volts at about 1/2 amp, get out hundreds of volts at a small fraction of an amp. Of course, I never got out quite what I put in (there was noise, there was heat, the motor had to keep spinning despite friction) but I didn't care because the result was amazing.

Note that the Chev was 6 V and had plenty of space under the dashboard. 