What happens to the frequency of an astable multivibrator if the value of ONLY the timing capacitor is doubled?
The frequency is reduced by half.
An astable multivibrator has no stable state; it continuously oscillates between HIGH and LOW, generating a square wave without any external trigger. Its timing is set by charging and discharging a timing capacitor (C) through resistors, so the switching period is directly governed by the RC time constant.
For the common 555-timer astable configuration, the output frequency is:
f = 1.44 / ((RA + 2RB) × C)
The key point is that frequency is inversely proportional to the capacitance. A larger capacitor takes longer to charge to the upper threshold and discharge to the lower threshold, so each cycle takes more time (lower frequency).
Doubling only the capacitor (C → 2C), while keeping RA and RB the same:
fnew = 1.44 / ((RA + 2RB) × 2C) = (1/2) × foriginal
So the frequency is reduced by half. (Equivalently, the period T = 1/f doubles.)
The choice claiming the frequency doubles confuses the direct relationship of period with the inverse relationship of frequency — since T grows, f must fall. The fourfold increase would only apply if C were involved as a squared term, which it is not. And the frequency cannot remain the same, because C is an active part of the timing expression and changing it necessarily changes the timing.
Important Questions from Multivibrators
Multi-vibrators can be used to produce which type of signals?
In astable multivibrator using IC555 timer design, the duty cycle D is given by _______.
A monostable multivibrator has which of the following state(s)?
I. One stable state
II. One quasi-stable state
Which of the following methods can result in a square waveform?
- A monostable multivibrator (MMV) is frequently used
