Is it possible to boost current into a load without boosting voltage also?

If a device "D" requires X watts power, and I have a power source Y that is less than the nominal voltage recommended by the device "D", can I obtain the same performance for the device by somehow boosting the current such that Y·(boosted current) = P?

I know this is a very simplified question that doesn't take alot of factors and practical limitations into account, but rather than getting an answer to my specific problem I would like to see the idea generalized. If this can't be generalized, maybe some rules of thumb could be given along with some edge cases where it does not work could be helpful. Note that I won't have any issues with too high current burning out the wire(awg size can be adjusted). My main issue is that I do not have a high enough voltage as the device recommends and wanna know how much I can underpower and still get satisfactory performance.

• You might want to take a look at switch-mode power supplies, especially the boost converter, which exactly provides some wattage that the power source can provide but at a lower voltage than required by the load. Since P=U.I, if you decrease V then you have to increase I by the same ratio to keep P.
– user59864
Commented Dec 24, 2015 at 21:27

Device D needs X watts so providing source Y can provide a 10% more power than device D needs you can use a boost regulator to step up the voltage to suit the voltage required by device D.

The 10% more power accounts for the losses in a boost switching regulator. In other words if Y can supply a bit more power you can convert this (at 90% power efficiency) to give you the right voltage and the right current.

My main issue is that I do not have a high enough voltage as the device recommends and wanna know how much I can underpower and still get satisfactory performance.

Totally unanswerable but who cares - use a boost regulator and get the right voltage.

Along with P = V * I you also need Ohm's Law which states that V = I * R. In the case of a resistive load the voltage and current are proportional. The only way to increase the current is to increase the voltage.

The answer, therefore, is no.

Note that since V and I are proportional that power changes with the square of the voltage or current. So if you drop your voltage to 90% the power will go down to 81%. (0.90 * 0.90 = 0.81.)

Will you get satisfactory performance at 81% power? It depends on the device. A heater might work OK although it will take longer to heat the room. An electronic device may not operate. A motor may not have enough torque to start.

• Mmmm well I don't have my own test bench but if I recall correctly power supplies have option to boost current keeping the voltage the same... And besides I'm pretty sure there are current amplifying circuit's. Commented Dec 24, 2015 at 21:23
• Test bench power supplies not battery, etc. Commented Dec 24, 2015 at 21:24
• @skyfire - "besides I'm pretty sure there are current amplifying circuit's." No. Just no. Commented Dec 24, 2015 at 21:27
• @WhatRoughBeast ? You mean to say that the only signal ever boosted in a circuit is voltage? Its been a minute since school but that doesn't sound accurate. A transistor is a current amplifying device..maybe I miss interpreting your answer. Commented Dec 24, 2015 at 21:34
• @skyfire. Ohm's law is the law (since 1827). You can't increase the current through a resistor without increasing the voltage across it. Commented Dec 24, 2015 at 21:35

can I obtain the same performance for the device by somehow boosting the current

You cannot 'boost' the current. The device will draw whatever current it decides to at the lower voltage.

My main issue is that I do not have a high enough voltage as the device recommends and wanna know how much I can underpower and still get satisfactory performance.

Totally depends on the device. It may still work at a lower voltage, but with lower performance. Or it may draw more current trying to compensate for the lower voltage, perhaps damaging itself. Or it may not work at all.

For example, if the device is a DC electric motor then it will spin slower. You might try to compensate by working it harder, increasing current draw above normal and causing it to overheat.

Yes, you can exchange voltage for current (and vice-versa), but there is always some power loss in the process, so the total input power will be somewhat higher than the power consumed by the end device.

With AC power, this is accomplished by using a transformer. The turns ratio determines the voltage and current ratios from input to output. Some extra "magnetizing" current is required on the input, and there are resistive losses in the coils and the core that generate waste heat.

With DC power, this is accomplished by using a DC-DC converter, which converts the incoming power to AC and then uses a reactive element (coil, capacitor and/or transformer) to make the transformation before converting the power back to DC at the new voltage and current. There are losses associated mainly with the switching elements that generate waste heat.

Simply put, of you can't do what you are asking for. For example, if a device requires a DC input voltage 10 volts and consumes 10 watts, then the input current will be 1 ampere. If you only have a 5 volt supply, you can't force the device to draw 2 amperes so that the total power remains 10 watts. Depending on the device, if you supply it 5 volts it will either draw only 0.5 ampere (if it is a purely resistive load), or more or less current if it is some other type of load. It probably won't work at all or not as well as with the specified input voltage. If you don't have the required voltage, then you need to use some kind of conversion device between your supply and the device to get that voltage. You can't change the characteristics of your device to suit your supply (at least in general).

• rubs brow yeah I suppose I knew it all along... Guess I just questioned the possibility when I could have sworn that in labs at school I could boost the current output holding the voltage output the same. But Transistor is right I can't boost the current to supplement the low voltage, the sum of the voltages in a circuit is zero... Commented Dec 24, 2015 at 22:14
• *sum of the voltage rises and drops Commented Dec 24, 2015 at 22:17