I have a technical question about selecting voltage ratings for a capacitor.
e.g. 1 Let's say I need to use a 2uF cap in a 5V circuit.
As I have understood it is a common practice to use twice the supply voltage as for the cap voltage rating for safety (to avoid blowing up the cap).
That means for my 5V circuit I should use 2uF cap with 10V rated voltage.
With the cap equation, Q = C x V , where Q = charge (in Coulomb), C = capacitance (Farads), V = voltage (Volts)
1) 2uF cap with 5V ratings will have maximum charge of,
Q = 2*10^(-6) * 5 = 10 * 10^(-6) Coulombs
2) 2uF cap with 10V ratings will have maximum charge of,
Q = 2*10^(-6) * 10 = 20 * 10^(-6) Coulombs
Now if I use 2uF, 10V cap in the 5 V circuit it will have maximum capacitance of
C = Q/V = 20* 10^(-6) / 5 = 4 * 10^(-6) = 4 uF. Now capacitance value has been change from 2uF to 4uF.
In this case, if I want to double the voltage rating (for safety) of a cap, should I halve the capacitance in order to have the same capacitance (2uF)?
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jfdo
Hi All,
I have a technical question about selecting voltage ratings for a capacitor.
e.g. 1 Let's say I need to use a 2uF cap in a 5V circuit.
As I have understood it is a common practice to use twice the supply voltage as for the cap voltage rating for safety (to avoid blowing up the cap).
That means for my 5V circuit I should use 2uF cap with 10V rated voltage.
With the cap equation, Q = C x V , where Q = charge (in Coulomb), C = capacitance (Farads), V = voltage (Volts)
1) 2uF cap with 5V ratings will have maximum charge of,
Q = 2*10^(-6) * 5 = 10 * 10^(-6) Coulombs
2) 2uF cap with 10V ratings will have maximum charge of,
Q = 2*10^(-6) * 10 = 20 * 10^(-6) Coulombs
Now if I use 2uF, 10V cap in the 5 V circuit it will have maximum capacitance of
C = Q/V = 20* 10^(-6) / 5 = 4 * 10^(-6) = 4 uF. Now capacitance value has been change from 2uF to 4uF.
In this case, if I want to double the voltage rating (for safety) of a cap, should I halve the capacitance in order to have the same capacitance (2uF)?
Thanks for any comments.
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