Digital Electronics - Multivibrators and 555 Timer - Discussion
Discussion Forum : Multivibrators and 555 Timer - General Questions (Q.No. 15)
15.
If a diode is connected across resistor RB (positive end up) in the given figure, what is the new duty cycle of the output waveform?
Discussion:
7 comments Page 1 of 1.
Kiran Borude said:
9 years ago
Duty cycle = Ra + Rb/Ra + 2 * Rb.
(3.3 + 12) * 10^6/(3.3 + 2 * 12) * 10^6.
(3.3 + 12)/(3.3 + 2 * 12).
15.3/27.3 = .56 = 56%.
(3.3 + 12) * 10^6/(3.3 + 2 * 12) * 10^6.
(3.3 + 12)/(3.3 + 2 * 12).
15.3/27.3 = .56 = 56%.
(1)
Kiran Borude said:
9 years ago
3.3/(3.3 + 12) * 100 = 21.56%.
(1)
Aimal said:
1 decade ago
If you want less then 50% duty cycle then Rb is bypassed by a diode. This left out 56% out of box. The duty cycle can now be calculated (in this configuration) as:
D.C = [Ra/Ra+Rb]*100.
D.C = [Ra/Ra+Rb]*100.
Madhukar said:
1 decade ago
When we connect a diode, the main purpose is to get the duty cycle < 0.5. Before connecting the diode the D = (RA+RB)/(RA+2RB). By the time we connect diode, the D becomes RA/ (RA+RB).
A banerjee said:
1 decade ago
Ees ths is correct. If we want 50% duty cycle this is perfect ckt..only one 10k pot'll be connected betwn Ra & Rb., because for 50% duty cycle Ra must be equal zero.
If we directly connected pin 7 to Vcc ckt may be damaged. (down transistor damaged/)
If we directly connected pin 7 to Vcc ckt may be damaged. (down transistor damaged/)
Manju said:
1 decade ago
Tm=0.7*R1*C, Ts=0.7*R2*C
duty cycle= R1/(R1+R2) or Tm/(Tm+Ts)
duty cycle= R1/(R1+R2) or Tm/(Tm+Ts)
(1)
Amita said:
1 decade ago
What is duty cycle and how to solve this numerical? give full calculation.
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