# How do you drive a high current load from a moderate current source?

If I have a power supply that can supply 5A @ 12V and I need to be able to supply an inductive load with 20+A @ 12V, how would I go about this?

I need a burst of current for about 2-5 seconds.

Will a capacitor bank and high current relay do it?

• Car battery. Possibly RC model battery pack to save weight. Overkill but maybe cheaper than supercapacitors... – Brian Drummond Dec 4 '13 at 23:21
• The Car Battery will be very simple and portable! I would be tempted to rig up a PC power supply. There are also specific modules that do this. We have a ~50 amp 12 volt supply somewhere @ work... The PC Power supply is the least expensive. – HL-SDK Dec 6 '13 at 1:18

Will a capacitor bank and high current relay do it?

Ignoring the inductive load for the moment since that raises an entirely different question, simply assuming that you need to supply 20A @ 12V for 5 seconds, the energy required is:

$$12V \cdot 20A \cdot 5s = 1200J$$

Let's do a quick calculation to get a feel for the size of capacitor bank to deliver this energy over 5 seconds.

Say that you have a capacitor bank charged to 12V and you wish to supply 1200J of energy. Understand that, as the capacitor bank discharges, the voltage across must decrease. So, for example, assume the capacitor voltage will decrease from 12V to 11V during the discharge. The equation for the required capacitance is:

$$\dfrac{C}{2}(12^2 - 11^2)V^2 = 1200J \rightarrow C = 104F$$

That's an enormous capacitance but there are ultracapacitors available that fit the bill.

For example, two of these in parallel yields 116F @ 16V but it will set you back about \$300. I'm quite certain that you could find a power supply that can deliver the 20A @ 12V for considerably less than that. • Ouch! I will look for a better suited PSU. Thanks! – Hair_of_the_Dog Dec 5 '13 at 0:58

I think Alfred Centauri has the right conclusion: simply obtaining a better suited power supply is the most economical option. However, I don't know that the calculations tell the whole story. Indeed, the simplest thing to do is throw some capacitors in parallel with the power supply: simulate this circuit – Schematic created using CircuitLab

And it's true, your transient load of 20A at 12V for 5s requires 1200J of energy. And, if C1=104F, then you can get that 1200J by discharging them from 12V to 11V. However, this overlooks a few things:

• over the course of this 5s load, your 12V, 5A power supply can supply $12V \cdot 5A \cdot 5s = 300J$ of energy. This is energy you don't need to store in the capacitors.
• Even after being discharged to 11V, the proposed 104F capacitor still contains $\frac{1}{2}CV^2 = \frac{1}{2}104F(11V)^2 = 6292J$ of stored energy. This is quite wasteful, given that you only needed 1200J.

The trouble is that to get at the remaining stored energy in the capacitors, you need to discharge them below 11V. Your load, which requires 12V, won't be happy.

Fortunately, there's a solution: a boost converter. This allows you to step up any voltage to any other voltage, at the expense of drawing more current from the source (otherwise, energy would not be conserved). In our particular case, it would allow us to get at all the stored energy in the capacitor, discharging it all the way, while supplying a constant 12V.

So, if we assume now (to simplify) that we have ideal components available, we just need to find capacitors that can store 1200J of energy, less the 300J that the power supply can deliver. We could do that with a small capacitance charged to a high voltage, or a large capacitance charged to a low voltage, but to keep our charging circuit simple, let's say we are going to charge the capacitors to 12V. How much capacitance do we need?

\require{cancel} \begin{align} 1200J - 300J &= \frac{1}{2}C(12V)^2 \\ 900J &= \frac{1}{2}C \cdot 144V^2 \\ \frac{900J \cdot 2}{144V^2} &= C \\ \frac{1800\cancel{V}As}{144\cancel{V}V} &= C \\ 12.5F &= C \end{align}

This is still a really big capacitor. And, since we assumed ideal components, we will need actually more energy storage to account for losses. And, a bigger PSU is still cheaper than a 13F, 12V capacitor and a boost converter. And, a bigger PSU is simpler, easier to design, and likely more reliable. So, this would really only be a viable option if you could not get a bigger PSU, such as if your power was coming from a limited resource, such as solar panels, etc.

Another option would be to use a battery, rather than a capacitor. There are plenty of battery types that might be suitable: lead-acid is affordable but bulky, Li-ion has higher energy density but considerably more cost and complexity in the control electronics. But still, if you can just get a bigger PSU, that's the most economical option.

To oversimplify the other answers: The energy is ultimately coming from the wall outlet, which can supply much more power than the ~240 Watts you require in pulses. (A space heater is 1500 Watts.) Use a voltage regulator that can pass it through from the wall to the load on demand, instead of storing up so much energy in a capacitor or battery. There's no need to smooth out the load on the wall outlet with gigantic local energy storage unless you're concerned about operating through blackouts or overloading very old wiring.