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Voltage controlled oscillator using 555
Short answer
Control voltage Vc at pin 5 sets thresholds Vc and Vc/2. Charging time depends on Vc; discharging time tL = 0.693 RB C. Frequency f = 1/{ (RA+RB)C ln[(VCC–Vc/2)/(VCC–Vc)] + 0.693 RB C }. Increasing Vc lowers frequency.
Draw voltage controlled oscillator circuit using IC 555 and derive expression for frequency of oscillation.
A voltage controlled oscillator (VCO) gives an output frequency that depends on an applied control voltage. In the 555 timer, the control voltage is applied at pin 5 (CONTROL), which sets the comparator thresholds.
Circuit
+VCC
+------+------+
| | |
[RA] (8)(4) |
| |
+----(7) |
|
[RB] (3) -----> output
|
+----(6)
+----(2)
|
[C]
|
GND
Vc o----(5) control voltage input (e.g. from a pot or source)
Same as an astable multivibrator, but pin 5 is driven by an external voltage Vc (instead of the 0.01 µF capacitor to ground).
Derivation
With control voltage Vc at pin 5: the upper threshold = Vc and the lower threshold = Vc/2 (the internal divider of equal resistors).
Charging (HIGH): The capacitor charges toward VCC through (RA + RB) from Vc/2 to Vc:
Vc = VCC − (VCC − Vc/2) e^(−tH/(RA+RB)C)
tH = (RA + RB) C ln[(VCC − Vc/2)/(VCC − Vc)]
Discharging (LOW): The capacitor discharges through RB from Vc to Vc/2:
Vc/2 = Vc e^(−tL/(RB C))
tL = RB C ln 2 = 0.693 RB C
Frequency:
f = 1/(tH + tL) = 1 / { (RA + RB) C ln[(VCC − Vc/2)/(VCC − Vc)] + 0.693 RB C }
When Vc = 2VCC/3 (pin 5 unconnected): (VCC − VCC/3)/(VCC − 2VCC/3) = 2, so tH = 0.693 (RA + RB) C and f = 1.44/((RA + 2RB)C), the normal astable result.
Behaviour: Increasing Vc raises both thresholds, so charging takes longer and the frequency decreases. Reducing Vc raises the frequency. The frequency is therefore controlled by Vc, which makes a VCO. Applications: FM generation, tone generators, FSK, PLLs.
✓ Answer checked by Claude Opus 5.5 (high effort)
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