Suppose you require a smaller capacitance with higher voltage rating for your application but you are left with only higher capacitance capacitor with less voltage rating. How could you overcome this problem?

You can follow a small trick. You put two or more capacitors in series to arrive at your designed capacitance of smaller value.Because

i) It divides the voltage over two capacitors and thus has net effect of doubling (in case of two capacitors) the voltage rating.

ii) It reduces overall capacitance by following

C = (C1*C2) / (C1+C2 )

where 'C' is total capacitance of two capacitors in series.

For instance, consider two capacitors 100 pF @500VDC,

Then keeping them in series we get C = 50pF

voltage is now divided over two capacitors so total rating is 50pF @1000V Dc.

Hope this is helpful for designers.

You can follow a small trick. You put two or more capacitors in series to arrive at your designed capacitance of smaller value.Because

i) It divides the voltage over two capacitors and thus has net effect of doubling (in case of two capacitors) the voltage rating.

ii) It reduces overall capacitance by following

C = (C1*C2) / (C1+C2 )

where 'C' is total capacitance of two capacitors in series.

For instance, consider two capacitors 100 pF @500VDC,

Then keeping them in series we get C = 50pF

voltage is now divided over two capacitors so total rating is 50pF @1000V Dc.

Hope this is helpful for designers.