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When I am using a buck, is the imput current lower than the output curent? Also how would I aljust that on a buck

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    \$\begingroup\$ But some people said because efficiency losses, current cant be increased That is nonsense, don't listen to anyone saying that. What happens is that you get a slighly higher current at the input side due to efficiency losses as in your calculation. \$\endgroup\$ Dec 17, 2016 at 22:07
  • \$\begingroup\$ @FakeMoustache dont you get higher amps scince the equation P=V*A. If u decrease volts u increase amps minus efficiency loss. Accoring to Majenco \$\endgroup\$
    – Ali Ragb
    Dec 17, 2016 at 22:20
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    \$\begingroup\$ You need to read Majenko's answer more carefully, "less a small percentage" is not "you cannot do that". \$\endgroup\$ Dec 17, 2016 at 22:32
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    \$\begingroup\$ Come on. You've got three people that say you can. Why do you need verification from me? Yes. Neither @FakeMoustache nor Majenko nor Spehro Pefhany are idiots. \$\endgroup\$ Dec 17, 2016 at 22:38
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    \$\begingroup\$ Not angry, don't worry. Also, you've only got negative score questions. Please read "how to ask a good question" in the help center. In your case, you like to say things like "I read something" and "some people say". The thing is: if you don't even link to where you found these statements, these are unbacked claims and simply waste your (and our) time. Try to understand the formulas used in these sources and explain the what you have a problem with specifically in your next questions! \$\endgroup\$ Dec 17, 2016 at 22:42

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A 10W output (2A @ 5V) converter with 80% efficiency will draw 1.04A at 12V input.

\$\eta = P_O/P_{IN} = 0.8\$ so \$P_{IN} = P_O/\eta = 10/0.8 = 12.5W\$

However as the battery voltage drops, the current must increase. At 10.8 volts (normally the lowest you'd want to discharge a lead-acid 12V battery), the current will be up to 1.16A. This is a negative resistance effect (increase the voltage and current drops).

Please make sure you are not misinterpreting the battery specification. It is not common to speak of a '1A battery'. Usually batteries are rated in ampere-hours (Ah). The actual Ah will be a bit different for different currents and for the battery condition and temperature.

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This is a typical buck converter:

Buck converter schematic with operation highlighted

picture from: Wikipedia

Now, look at the source side first. When the switch is closed, source is on on-state. So, it is supplying a current of \$ I_{s} \$. When the switch is open, the source current is \$ 0 \$. So, the average source current is:

$$ I_{avgsource} = {T_{on} * I_{s} \over T_{on} + T_{off}} $$

Right?

Now, look at the load side. When the switch is closed, load current is \$ I_{s} \$. (Same as source current). When the switch is open, load current is, lets say \$ I_{L} \$. So, the average load current must be:

$$ I_{avgload} = {(T_{on} * I_{s} + T_{off} * I_{L}) \over T_{on} + T_{off}} $$

As you can see, since \$ I_{L} \$ always greater than zero:

$$ I_{avgload} > I_{avgsource} $$

And that's how the output current is higher than the input current for buck converters! The real hero here is the inductor! It stores energy in magnetic form to release it as current! Where did the energy come from? The source, of course.

The same principles apply for input and output voltages. Input voltage is equal to source voltage at all times. Output voltage is lower than source voltage because it is clipped by the switch.

So for a buck converter, output voltage is lower but current is higher!

Now efficiency is a whole different matter. It depends on many, many parameters but the thing is, the results will hold for a fully efficient buck so you don't need to take efficiency in account when trying to visualize this stuff.

Note: The math used in this explanation may not always coincide with reality.

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Yes. When you reduce the voltage you increase the current, less a small percentage for the efficiency (dissipated as a small amount of heat). Just like a transformer does for AC.

P = VI. Reduce V and I has to increase for P to remain constant.

E.g., 80% efficiency, and 12V to 5V buck drawing 1A from the output:

\$P_{OUT}=V_{OUT} \times I_{OUT} = 5 \times 1 = 5W\$

\$P_{IN} = \frac{P_{OUT}}{0.8} = 6.25W\$

\$I_{IN}=\frac{P_{IN}}{V_{IN}} = \frac{6.25}{12} = 0.52A\$

\$P_{IN} - P_{OUT} = 1.25W\$ dissipated heat.

This is in sharp contrast to a linear voltage regulator which discards the excess voltage in the form of a large amount of heat. Decrease V with I remaining the same, and P decreases (the difference in P is the dissipated heat).

E.g., the same input and output as before with a linear regulator (with, say 20mA quiescent current):

\$P_{OUT}=V_{OUT} \times I_{OUT} = 5 \times 1 = 5W\$

\$I_{IN} = I_{OUT} + 0.02 = 1.02A\$

\$P_{IN} = V_{IN} \times I_{OUT} = 12.24W\$

\$P_{IN} - P_{OUT} = 7.24W\$ dissipated heat.

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