1
\$\begingroup\$

Could the output current of step-down switching regulator be larger than its input current?

Assuming that the efficiency of your switching regulator is 90% and you are converting 5 V to 1.8 V. If you measure that input current is 100 mA, do you expect that the output current would be 250 mA? (5 V * 100 mA * 0.9 / 1.8 V = 250 mA)

If so, where do the charges come from?

\$\endgroup\$
  • 2
    \$\begingroup\$ Metals are awash in free charges. They just need some motivation to move. That can come from changing magnetic fields. When you are talking about 90% efficiency conversions between voltages, then it is almost always due to the intelligent use of magnetic fields to temporarily store energy efficiently. \$\endgroup\$ – jonk Aug 24 '16 at 4:55
6
\$\begingroup\$

If you're using a 90% efficient DC/DC system (IC, inductor, diode, etc), then output will be the 277mA you calculated *0.9 (250mA).

The "charges" come from the input. Nothing is "gained". You exchange voltage potential for current (and heat, maybe some noise, etc).

1 - Power = Current * Voltage

2 - Power in = P out + P losses

Pin = 5V * 100mA = 0.5W

Plosses = Pin * inneficiency = 0.5W*0.1 = 0.05W

Pout = Pin - Plosses (or Pin * efficiency)= 0.5W - 0.05W (0.5W*0.9)= 0.45W

So, if Vout = 1.8V, Iout = 0.45W/1.8V = 0.25A

p.s.: In step-up systems the reverse is true. The output will be capable of less current throughput. The "original" current does not go anywhere, you just exchange it for more voltage potential (and heat, etc).

\$\endgroup\$
  • \$\begingroup\$ Thank you for the answer. I edited my original questions to fix the error in my calculation. \$\endgroup\$ – SD11 Aug 24 '16 at 6:32
6
\$\begingroup\$

A buck switching regulator has two phases of operation, that it switches between. For example, below are the important components of a 5 V in, 2 V out buck converter.

schematic

simulate this circuit – Schematic created using CircuitLab

In the first phase, the input is connected (usually by a MOSFET) to the inductor, which is connected to the output. The same current flows in both, while the current builds up in the inductor due to the positive 3 V across it.

In the second phase, the input is disconnected, and the input of the inductor is grounded (either by a driven MOSFET, or less efficiently by a diode). Energy storage in the inductor means that output current continues to flow, while the current falls in the inductor due to the negative 2 V across it. Obviously no input current flows during this phase.

If we assume that we switch fast enough that the inductor current 'doesn't change too much', then we can see that output current is always flowing, whereas input current only flows during the first phase. The average input current is the switching duty cycle lower than the output current.

With a buck regulator, it's not so much 'where does the extra output current come from?' It is more 'why is the input current lower?' It's because the inductor current does not flow all the time from the input, but does flow all the time to the output.

If the output current was 250 mA in this case, the input current would be 250 mA flowing for only 40% of the time, or 100 mA on average (lossless case).

Switching converters are always equipped with large enough input and output capacitors, so that the supply and load only see the average current, not the instantaneous switch or inductor current.

How would losses increase the input current? Resistance in the inductor and switches would add voltage drops to the inductor voltage, which would increase the duty cycle required to maintain the output voltage, so it leads to a higher average input current.

\$\endgroup\$
2
\$\begingroup\$

If so, where do the charges come from?

Things that conduct, like wires, are full of charges that can be pushed around. That's what makes them conduct.

You might have an incorrect mental model of how charges flow in a circuit. The charges do have some net movement in the direction of current flow. But largely, they are just vibrating in place pushing against each other.

Think of a giant wooden wheel laying flat with the power source on one side of the wheel turning it and the load on the other side of the wheel applying friction to it. You don't need more wood to make the wheel turn faster or deliver more force to the load. You can make the wood spin as fast and as hard as you want without needing more wood.

When some current flow produces a magnetic field, the collapsing magnetic field cannot deliver more energy than was used to make it. But it can deliver more current if the opposition to current flow is less and the same amount of energy produces more current. This is how buck converters work.

\$\endgroup\$
0
\$\begingroup\$

More generally replying to the actual question

...If so, where do the charges come from?

is indeed quite easy: you don't have to bother about charges. Any circuit, when including all connections coming and going is just a super-node probably better named as closed surface, and hence KCL holds true and tells us charges are just running a closed loop from source.

enter image description here

This can be applied to both to red (Icc=Icc) and blue (IL=IL) surfaces in sketch above.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.