Circuit that converts a triangular wave potential waveform

[mohlam12]

I have to propose a circuit that converts a triangular wave potential waveform to a square wave potential waveform, that has a T=10ms and Umax= +4V (the triangular curve has the same T and Umax)
We have a capacitor of 4micro Farad and 3 resistance (1kΩ ; 250Ω ; 500Ω) andyou should use only one resistance.

So, I know it's this one: (attached)
But I have to find what resistance to use.

$$\ V_{exit}= \frac {4RCV_{max}}{T}$$
I need that $$\ V_{exit}$$ (that I don't have) to solve for the R !
Any help please ?!

 

[mohlam12]

Never mind... got the answer :)

 

[kyle8921]

To convert a triangular wave potential waveform to a square wave potential waveform, we can use a comparator circuit. This circuit compares the input voltage with a reference voltage and outputs a high or low voltage depending on which one is larger. In this case, we want the output to be high when the input is above 2V (half of the maximum voltage of 4V) and low when the input is below 2V.

To achieve this, we can use a voltage divider circuit with a 1kΩ resistor and a 250Ω resistor. This will give us a reference voltage of 2V at the junction between the two resistors. Now, we can use a single 500Ω resistor in series with the capacitor to form a low pass filter. This will smooth out the triangular waveform and make it easier for the comparator to detect when the voltage is above or below 2V.

To calculate the value of the resistor needed, we can use the formula:

R = \frac{T}{4CV_{max}}

Substituting the given values, we get:

R = \frac{10ms}{4\times 4\mu F \times 4V} = 0.156k\Omega

Therefore, we can use a 150Ω resistor for R in the circuit. This will give us a square wave output with a period of 10ms and a maximum voltage of 4V.