Voltage calc across Cap

[Suman Dhua]

I am using one battery 3.6V as a power source . Output is fixed at 3 V using DC converter IC. 300uF cap in input of converter( parallel) maximum output current 25 mA for will draw for a small time duration t.
My question is how i will calculate the capacitor dipping voltage at lower input voltage to the DC converter.\][/b]

 

[keith1200rs]

There is not enough information. You are converting 3.6V to 3V with a DC-DC converter, but is it a SEPIC converter or step down (buck) regulator? What is the minimum input votage befroe the regulator will stop working? What is the converter efficiency? (you need to know the input current for 25mA output current).

When you know that you can use the rate of discharge of the input capacitor to work out dV/dT = i/C. So, if C=300uF and say input current is 30mA, dV/dT = 100V/s. So in 1ms it will drop 100mV. If you can tolerate a 100mV drop on the input then the 300uF will keep the supply going for 1ms. If you can tolerate a 0.5V drop on the input before ther egualtor stops working, the capacitor will hold up for 5ms.

Keith.

 

[Suman Dhua]

yah i got ur idea..here i am using one buck boost regulator min Vin 1.8 V. required Vout=3V.and Io 30 mA
then i think i have to back calculate the input current input current (Ireq) taking regulator efficiency.
then , for particular time T
required Q= T* Ireq
voltage dipping across cap =Qreq/C
thats all i think.....is it curret??

 

[keith1200rs]

The voltage dip will be (i/C * time).

Keith.