Smoothing dc question ?

[tron87]

Bit of a beginner question but suppose you had a load attached across a battery the load is switch on and off every 0.25 sec the voltage across the battery drops slightly each time its turned on and rises back to near starting voltage when load is disconnected.

My question is how would you calculate the required capacitor value to place across the power supply to remove the loaded voltage drop during the time the load is connected?

 

[ronv]

Ripple

I think you can use the formula Vr=Vp/Rl*C x ΔT.
V ripple = V peak / R load x Capacitance X time between peaks.
Often however the drop from a battery is due to internal resistance (if the load is high) and the capacitor series resistance will make it less efficient.

 

[MikeMl]

You cannot afford, or have room for, a capacitor big enough.

 

[MrAl]

Hello there,

What size cap you need and just how effective this method will be depends largely on what your circuit parameters are. For example, what your batteries internal resistance is, what your load resistance is, what your nominal voltage is.

Also, this method is only so so effective anyway in that if your load without capacitance loads the voltage down by 20 percent then the best you can achieve using capacitance alone will be 10 percent loading. In other words, if you have a 10 volt battery and it loads down to 8v with the load (and no cap) then using even a very large cap will only improve the voltage by 1v, up to 9v, so that the nominal output will now be 9v instead of 8v which is an improvement but still not maintaining the 10v original battery voltage. The main reason for this problem is the duty cycle which for your app is 50 percent. For smaller duty cycles we can see much better improvement but with 50 percent that's it.

A series inductor improves things a bit too, but the size and cost depends again on your circuit parameters. Perhaps you can tell us what they are (nominal voltage, battery series resistance, load resistance).

Again depending on your circuit parameters, there may be another better solution such as some type of regulator.