# Increasing current output in capacitor charging circuit

#### lesneypark

Joined Dec 31, 2015
47
Hi,

I have a circuit which is powered by an AC supply and a capacitor.

The AC supply is available for part of each cycle during which the capacitor is charged then when the AC supply is removed for the remainder of the cycle the capacitor discharges and therefore ensures the downstream circuit has a supply throughout the full cycle.

The capacitor has been sized to provide the required current to the downstream circuit but in certain scenarios the AC supply is not present for a long enough period during the cycle to adequately charge the capacitor, therefore there is insufficient power available to support the downstream circuit for the full cycle.

Could I simply add a second capacitor of the same value and properties in parallel with the existing capacitor therefore the AC source will charge both capacitors at the same rate and therefore when the AC supply is not available there will be double the power to support the downstream circuit.

Thanks.

#### Dyslexicbloke

Joined Sep 4, 2010
566
I take it you are talking about some cycle, of an application, other than cycles of an AC waveform.?
When you talk about AC source do you mean some sort of power supply that is connected to an AC supply but produces DC to supply your circuit?

If so then yes a bigger cap could work but you would have to ensure that your power supply was big enough to charge it in thew available trime.
You would also probably need to limit the charge current to protect the supply.

A battery may be a better choice but without knowing what you are doing, voltage, current and duty cycle it is not possible to give a definative answer.

Al

#### lesneypark

Joined Dec 31, 2015
47

I am talking about cycles of an AC waveform.

The AC waveform is converted to DC via a full wave rectifier circuit which is then converted to 12 Vdc via a DC/DC buck convertor (LNK304D). The capacitor is on the output side of the LNK304D and is used to try to hold up the supply.

#### Dyslexicbloke

Joined Sep 4, 2010
566
Hold up the supply for how long?

Firstly a full wave rectifier will not produce DC, at least not smooth DC. What you get is the negative half of the AC sine wave shifted above the 0 point, less the forward voltage of the diodes that is.
Typically, in fact almost always, at least 1 smoothing capacitor is added to the output of the rectifier which absorbs some of the bumps and fills in the gaps giving you an output that is mostly DC with a ripple voltage imposed upon it.
Generally speaking the larger the capacitor the smoother the DC will be and this will be affected by the impedance of the AC supply and the applied DC load.
There are of course other techniques for smoothing lumpy DC but they are many, varied and generally only applied when strictly necessarry.

In your case you should be considdering energy as opposed to voltage or current.

Your buck converter probably will not like a lumpy supply because it will be attempting to reach a steady state, some specific duty cycle, that forwards sufficiant energy to maintain the DC load at a particular level. If the DC input voltage is continually changing then the duty cycle will also have to change at the same rate even if the load on the output is constant.

Having said all that the switching freequency of the converter will be much higher than the AC supply and the unit may cope with the variation without much instibility being evident at its output.

Your buck converter will have a minimum supply voltage that it requires to deliver the 12V output. What it is will depend on the specific IC/Circuit but it is likely to be between 1.5 and 2 V. It will also have a maximum supply voltage which again is IC/Circuit dependent but could easily be 30V or so.

If you think about how a capacitor works, storing a charge at a given potential, it is easy to see that increasing the capacity, more charge, or increasing the potential, more voltage will store more energy. However if you are charging the capacitor with your 12V output then you cant utilise any of that stored energy untill the output has fallen below the voltage at which the capacitor was charged, thus allowing it to discharge some of the stored energy. You may have a huge capacitor and a tiny load resulting on only a slow decreese in voltage but it will drop, it must.

Now considder putting the capacitor on the input of the buck converter... you can increase the smoothed DC to charge the cap to a level well above the 12V you require. if you went to 26V for example the energy stored in the capacitor would be more than twice that at 12V and the top 50% of that energy would be available to the buck circuit to reduce to 12V at your output.
If the supply were interupted the converter would still run discharging the capacitor to 12V + whatever its minimum threshold, input to output, is whilst maintaining the 12V output.
Of course at some point you would run out of stored energy if the supply wasnt restored and simply adding bigger and bigger caps is not an option unless you address the supply impedance and handle the charge current but still it is going to be a better option than adding capacitence to the load side.

Basically you are building a UPS, and conventional wisdom suggests that the prefered option is a battery, charged by the AC supply and delivering a stable voltage to your load via your buck circuit.

If you used two buck converters, one to maintain a reasonable float voltage on your battery, and the second to take that float and buck it down to 12V for your load you would be good. However if the required float was 13.6 and the buck needed 2V over thre 12 to maintain 12 you would be stuck... One option is to buck to 6.8, to float a 6V battery, and then boost to 12V

If you better define the problem, you will get a better answer/s. Its all about the application.

Al

#### lesneypark

Joined Dec 31, 2015
47
Al, that was a very helpful answer and has given me a few things to think about....thanks.

I already have a large capacitor on the output side of the full wave rectifier which is being used to smooth out the ripples as you recommended.

I like the idea of switching the capacitor from the output side of the buck converter and amending it to store more energy with the buck converter being use to provide a stable 12V.

I also don't fully understand your suggestion of using two buck converters, with the first charging a 6V battery and the second boosting to 12V. I don't want to use a battery, presumably the same logic applies to capacitors? Could you draw a rough sketch?