electrolytic cap ripple

Status
Not open for further replies.
Do you have a brownout specification?
You are asking this question such that I can size my capacitor right? If 12msec is my brownout then c= I*t/V = 0.2*0.012/12= 200uF.
My understanding was wrong earlier. I was thinking the current through cap will be (12-0.3)/0.15 fora input of 0.5Vrms upon 12V. But as cap will be charged to the DC value of 12V the current will not be so huge for case 4 and it will be 0.3/0.15.
With this understanding I think adding a diode is more appropriate than to add a resistor as it is redcuing the power dissipation compared to resistor.
The regulated output of capacitor will be 12V.
 

The problem is, if you don't include a diode then when the battery voltage drops for 12ms it will discharge your capacitor. A battery dropout in a vehicle is not due to disconnection of the power source but a drop in the voltage source so the impedance is low.

I have attached a simulation showing the difference. With no diode the power will drop to zero almost straight away. It also shows that 220uF is not large enough for 12ms brownout at 200mA. You can work that out easily - dv/dt = I/C.

My question about the regulated voltage is - what voltage are you regulating to? 5V for example? You need to ensure that during brownout the voltage doesn't dip so far that you lose regulation.

Keith.
 

I used the same formula to calculate capacitor C= I*dt/dv = 0.2*0.012/12= 200uF. But the mistake is that I took 12 as the dv value which should not be the case.
I understand that the voltage should be atleast 5.5(considering 0.5 dropout) for 5V LDO to function. So I would use 12-5.5= 6.5 and the capacitor becomes 0.2*0.012/6.5=370uF.
By the way what simulator are you using?
Thanks for your time
 

Status
Not open for further replies.

Similar threads

Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…