Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

increasing inductance/capacitance by soldering more tin onto two terminal of ind/cap

Status
Not open for further replies.

caoquyen164

Junior Member level 2
Joined
Aug 2, 2015
Messages
21
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Activity points
180
hi, as the tittle, in some circuit, I figure out that we can change the value of inductor or capacitor by soldering more tin. so why is that ? tks alot.
 

The capacitor plates would need to be very tiny, if a bit of solder can make a difference in Farad value. Is that what your question is about?
 

There is a clever way to fabricate your own 'gimmick' capacitor, by twisting two wires together to create the same effect as capacitive plates. You can change the Farad value by changing the lengths of wires.

https://en.wikipedia.org/wiki/Gimmick_capacitor

-------------------------------

Recently I made my own capacitor by cutting several square inches of kitchen aluminum foil, making 2 of these 'plates', and wrapping them between sheets of paper. I put it in an oscillating circuit, then watched the frequency change as I squeezed and released the homebrew capacitor. Its value is in the area of 1nF.

In comparison a plain ceramic capacitor 1nF is tiny. The dielectric must be extremely thin compared to my bulky homemade capacitor.

All this suggests that adding a layer of solder is not likely to alter the Farad value very much.
 

The value of cap or inductance changes by soldering or removing tin.
But to what extent? The variation will be very very small . Unless the orig value of the component is very small , the effect cannot be noticed.

In practice this have no effect in normal circuit.
If it is RF of very High Freq ,then you will notice the difference.

The eqn C=εA/d will give the reason if you take into account all materials and frequency of interest.
 

Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top