Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

ESR of capacitor.

Status
Not open for further replies.

rainman1

New Member
Does tan(lamda) equal [ (1/jwC) / ESR ], where ESR is the ohmic resistance of the capacitor?

According to datasheet, tan(lamda) = 0.25 @ 120Hz,
And it turns out that R = 1 / (2pi*120Hz*10uF*0.25) = 530ohm
And it seems too large to me.

Is it correct please?
 
Last edited:

crutschow

Well-Known Member
Most Helpful Member
It should be Tan(δ) = ESR/(1/jwC).

Thus the ESR = 33.2Ω at 120Hz
 

picasm

Member
According the the web page below, tan δ is the "dissipation factor" and in order to calculate the ESR you need to multiply it by the reactance value in ohms
reactance = 1/2πfC **broken link removed**

The esr for an average 10uF capacitor should be less than 2 ohms. (ESR is usually measured at 100Khz)

The tan δ is related to esr: lower=better
 
Last edited:

Papabravo

Well-Known Member
ESR is just the real part of the complex impedance WHICH IS A FUNCTION OF FREQUENCY. As such, no single value will do. To actually see it you should try to get your hands on a network analyzer. If you can't afford one maybe a friend will let you borrow one -- LOL.

Ponder the following. There is a frequency at which the complex impedance of a capacitor has no imaginary part and it looks like a pure resistance.

Even stranger. There is a frequency at which the capacitor behaves like an inductor. It's complex impedance has a positive imaginary part.
 
Last edited:
Status
Not open for further replies.

Latest threads

Top