# math/circuit help

#### ecjohnny

Joined Jul 16, 2005
142
hey guys. Need some help on my project-based learning problem.

It about displaying television picture(flickering). And i will quote the part where i need help and hopefully i can get some help from you guys. Here

To prevent flickering,the time needed to perform this "painting" must be completed within 1/60s

"One method of generating such a waveform is to repeatedly charged a capacitor with a large time constant , and repeatedly discharges it with a very small time constant . The sawtooth generation circuit and its waveform is shown in Fig.2."

Q2. With reference to the sawtooth waveform in fig2,calculate the value of R1 in order to produce the charging voltage shown.

how to find resistor,R1? i need help here. Thanks a lot.

#### Papabravo

Joined Feb 24, 2006
15,562
1. Do you know the expression, derived from the differential equation, for the voltage on the capacitor, of a series RC circuit?

Hint: It is an exponential function.

2. What is 1/60 s minus 1333 microseconds? That is how long the charging cycle lasts.

3. What voltage would be on the capacitor if there was no switch and time was very large, like approaching infinity?

#### ecjohnny

Joined Jul 16, 2005
142
1. Do you know the expression, derived from the differential equation, for the voltage on the capacitor, of a series RC circuit?

Hint: It is an exponential function.

2. What is 1/60 s minus 1333 microseconds? That is how long the charging cycle lasts.

3. What voltage would be on the capacitor if there was no switch and time was very large, like approaching infinity?

1. is it V(1-e^-t/RC)???
2. ans = 15.33ms??
3.v(0) = 0V??

t=RC , therefore R = t/c = 15.33e-3/0.47e-6 which is 32.61Kohms??
Am i right? and can you explain a little more about the circuit for me? and the steady state(5tau)...etc....
thanks for the guidelines...

#### Papabravo

Joined Feb 24, 2006
15,562
Answer 1: v(t) = 20*(1 - e-(t/RC))
Why is the 20 there? Because that is what the voltage will be at t = infinity. The voltage will approach that value asymptotically. That means it will never quite get there.

Rich (BB code):
10 Volts = 20 * (1 - e-(15.33e-3/(R*0.47e-6)))
I get just over 47K Ohms

Answer 3: v(inf) = 20 Volts

#### ecjohnny

Joined Jul 16, 2005
142
Answer 1: v(t) = 20*(1 - e-(t/RC))
Why is the 20 there? Because that is what the voltage will be at t = infinity. The voltage will approach that value asymptotically. That means it will never quite get there.

Rich (BB code):
10 Volts = 20 * (1 - e-(15.33e-3/(R*0.47e-6)))
I get just over 47K Ohms

Answer 3: v(inf) = 20 Volts

thanks Papa. Can you explain to me how you get the "10 volts" 10 volts = 20 *(1 - ......???

Vc(t) = 63% of V?? which is 12.6V???

Also my mathematics is quite poor. how do i calculate the value R? i mean can you break it into steps? like R = something over something or multiply..etc..

#### Papabravo

Joined Feb 24, 2006
15,562
10 Volts is the value from the graph at t=15.33 msec = (16.67 ms - 1.333 ms). I don't know why you are trying to compute 63% of something it has no relevance in this problem.

Here are the steps
Rich (BB code):
10 Volts = 20*[1 - e(-(15.33e-3)/(R*0.47e-6))] ; Divide both sides by 20
10/20 = [1 - e(-(15.33e-3)/(R*0.47e-6))] ; Subtract one from both sides
1/2 - 1 = -e(-(15.33e-3)/(R*0.47e-6))    ; Multiply both sides by -1
1/2 = e(-(15.33e-3)/(R*0.47e-6))    ; Take the natural logarithm of both sides
-0.693 = -(15.33e-3)/(R*0.47e-6)    ; Multiply both sides by R
-0.693*R = -(15.33e-3)/(0.47e-6)    ; Divide both sides by -0.693
R = (15.33e-3)/((0.693)*(0.47e-6))  ; Evaluate the right hand side
R = 47067
Was this little bit of algebra really that much of a challenge?

#### MarkYeap

Joined Dec 14, 2011
3
How to find i(t) when I am using the formula: iR2 + Vc(t) = 0?

#### Papabravo

Joined Feb 24, 2006
15,562
It makes absolutely no sense to say that a current plus a voltage equals 0. Where did you get that quaint notion from?

When the switch is closed the current in R2 is 0 and the voltage is 10V. The current will increase and the voltage will decrease according to the exponential rule.
Rich (BB code):
0.1 = 10 * [e^(-1333e-6 / R2*0.47e-6)]  ; we cant use 0 because it will never get there
0.01 = e^(-1333e-6 / R2 * 0.46e-6)
ln(0.01) = -1333e-6 / (R2 * 0.47e-6)
-4.605 = -2836.17 / R2
R2 = 615 Ohms
Because of Ohms law you have the instantaneous current through R2 by dividing the voltage equation by 615
Rich (BB code):
i(t) = V(t) / 615 for the discharge cycle through R2

Last edited:

#### MarkYeap

Joined Dec 14, 2011
3
Sorry, I forgot about the value of resistor R2. I tell you again.

How to calculate i(t) when I am using the formula: iR2 + Vc(t) = 0 when R2 = 470ohms?

#### MarkYeap

Joined Dec 14, 2011
3
I got that from my worksheet.