Signal Generator
http://www.electronics-tutorials.ws/waveforms/555_oscillator.html
Using the square wave signal generator.
For the 555 timer, how have they got the times for charging to be t=ln(2)RC
The time for charging through 2 resistors for a capacitor should be Vin/3 / (R1 + R2) = I
Because we are only charging 1/3 of Vin, from 1/3 of Vin to 2/3 of Vin.
The voltage of the capacitor should be 2Vin/3 and put into the equation C=Q/V we get C = 3Q/Vin.
The charge transfered during that time are equal so we sub for Q.
Getting t = 2C(R1+R2), where have they gotten ln(2)(R1+R2) = t for charging time?
Using the square wave signal generator.
For the 555 timer, how have they got the times for charging to be t=ln(2)RC
The time for charging through 2 resistors for a capacitor should be Vin/3 / (R1 + R2) = I
Because we are only charging 1/3 of Vin, from 1/3 of Vin to 2/3 of Vin.
The voltage of the capacitor should be 2Vin/3 and put into the equation C=Q/V we get C = 3Q/Vin.
The charge transfered during that time are equal so we sub for Q.
Getting t = 2C(R1+R2), where have they gotten ln(2)(R1+R2) = t for charging time?
Anyone
Original post by JokesOnYoo
http://www.electronics-tutorials.ws/waveforms/555_oscillator.html
Using the square wave signal generator.
For the 555 timer, how have they got the times for charging to be t=ln(2)RC
The time for charging through 2 resistors for a capacitor should be Vin/3 / (R1 + R2) = I
Because we are only charging 1/3 of Vin, from 1/3 of Vin to 2/3 of Vin.
The voltage of the capacitor should be 2Vin/3 and put into the equation C=Q/V we get C = 3Q/Vin.
The charge transfered during that time are equal so we sub for Q.
Getting t = 2C(R1+R2), where have they gotten ln(2)(R1+R2) = t for charging time?
Using the square wave signal generator.
For the 555 timer, how have they got the times for charging to be t=ln(2)RC
The time for charging through 2 resistors for a capacitor should be Vin/3 / (R1 + R2) = I
Because we are only charging 1/3 of Vin, from 1/3 of Vin to 2/3 of Vin.
The voltage of the capacitor should be 2Vin/3 and put into the equation C=Q/V we get C = 3Q/Vin.
The charge transfered during that time are equal so we sub for Q.
Getting t = 2C(R1+R2), where have they gotten ln(2)(R1+R2) = t for charging time?
during charge the pd across the capacitor goes from 1/3 Vcc to 2/3 Vcc... i.e double it's initial pd
it's being charged by a RC circuit with a pd of Vcc.
2/3 Vcc is halfway between 1/3 Vcc and Vcc
ln(2)RC gives the time for the pd across a capacitor to double - in this case R is R1+R2
during discharge the pd across the capacitor goes from 2/3 Vcc to 1/3 Vcc... i.e. half it's initial pd.
ln(2)RC gives the time for the pd across a capacitor to half - in this case R is only R2
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