# "a factor of"

Discussion in 'Math' started by Life617, Aug 22, 2012.

1. ### Life617 Thread Starter Member

May 23, 2012
30
0
I am having trouble trying to understand this capacitance ratio equation for varactor diodes. I think I have figured some of it out like the $C_2_0$ means the capacitance of the varactor diode at 20 reverse voltage. Now am now trying to figure out CR (Capacitance ratio)... for each volt increase how much exactly does the capacitance decrease.

$C_2_0=\frac{C_2}{CR}=\frac{22pF}{5}$

I interpret this equation as What is the varactor capacitance at 20 reverse voltage if the capacitance at 2 reverse voltage is equal to 22 pico farads using a varactor capacitance ratio of 5.

....

$\frac{22pF}{4.4pF}=\frac{C_2_0}{C_2}$

$\frac{20V}{4.4pF}=\frac{2V}{22pF}$

$\frac{22/22}{4.4/22}=\frac{1}{.2}=5$

This does not add up in a linear fashion, how would I go about creating the equation for the graph of capacitance vs reverse voltage.

Last edited: Aug 22, 2012
2. ### Life617 Thread Starter Member

May 23, 2012
30
0
Sorry about the tittle, I forgot to change it...

3. ### walk on New Member

Jul 12, 2012
11
0
Im going to look at this from a math POV, (very little knowledge of the components involved). If i made an error, which is likely, please point it out. I think the first part of the string of equations you listed seems incorrect. From the first formula you gave, I see that ...
C20 = 22pF / 5
Which simplifies to
C20 = 4.4pF
In that case,
C20 / C2 = 4.4pF / 22pF
or
C20 / C2 = 1/5

Now on to your question, we were given: C20 = 22pF and we found C2 = 4.4pF
So first find the slope, aka the change in capacitance divided by the change in reverse voltage
(C20 - C2) / (20-2) = slope
****Then slope = 17.6 / 18 = .977777.......pF/V****

We almost have everything we need to get the equation, but now need to find out what the capacitance is when voltage is zero. So using the formula for slope, and using the given capacitance of C20, we see that:
(C20 - C0) / (20 - 0) = slope = .97777....
So, (22pF - C0) / 20 = .9777....
Then, 22pF - C0 = (20 * .9777...) = 19.5555....
Or, 22pF - 19.5555.... = C0
****Therefore, C0 = 2.4444...pF ****

Using the equation, y = mx + b where "y" represents Capacitance, and "x" represents reverse voltage, and "b" represents C0,
The equation is now Capacitance = ( 17.6 / 18 ) * reverse voltage
*****Or, C = (.97777 * RV) + 2.4444******
(forgive me for dropping the units throughout the process)
Once again, this was just taken from a math POV so i dont have the "logical" check to rely on, but if any steps were incorrect, point them out please.