I have been using a 555 timer to modulate an audio signal and it works as expected, but I wonder how.

Async 555
(source: ibin.co)

If I am right, when the control voltage pin is positive, the "trigger points" get to CV and CV/2, but what happens when \$ CV \leq 0\$, as it does with an audio signal ?

Are potentials below zero simply clipped because of the comparators ?

  • \$\begingroup\$ A capacitor is not the same thing as a wire \$\endgroup\$ Mar 8, 2017 at 12:02

2 Answers 2


CV is tied to the inverting input of the top comparator, therefore it is biased at 2/3 Vcc due to the internal resistor network.

555 block diagram

  • \$\begingroup\$ Doesn't it change since the control pin is used ? \$\endgroup\$
    – Skami
    Mar 8, 2017 at 1:06
  • \$\begingroup\$ Do you understand what "biased" means? \$\endgroup\$ Mar 8, 2017 at 1:06
  • \$\begingroup\$ Do you mean biased as in default value ? \$\endgroup\$
    – Skami
    Mar 8, 2017 at 1:10
  • \$\begingroup\$ en.wikipedia.org/wiki/Biasing \$\endgroup\$ Mar 8, 2017 at 1:27
  • 1
    \$\begingroup\$ The capacitor in the schematic means that the voltage is never applied to it; it is decoupled by the capacitor which means that the audio signal swings around the bias voltage. This is one of the basic tenets of audio electronics. \$\endgroup\$ Mar 8, 2017 at 1:49

The frequency of the 555 astable is controlled by the capacitor being charged between 1/3rd and 2/3rds of the supply voltage. These reference voltages are set by three internal (trimmed)resistors - each 5k (hence the name, 555).

To understand how we produce frequency modulation we only need to consider what is happening to these (comparator) reference voltages when we vary the control voltage.

enter image description here

Circuit (1) shows the input circuit as we would normally think about it. The control voltage in this case is an audio signal. We add a capacitor (C) to block any DC component. With no AC or just DC input (blocked by the capacitor,C) the internal reference voltages will be 2/3rd (Vhigh) and 1/3 (Vlow).

Circuit 2: From an AC point of view the supply lines are connected together so R1 will be in parallel with R2 + R3. This gives an input resistance of about 3k3 (5k//10k). This allows us to calculate a suitable value for C. The circuit formed is a high pass filter so C must be big enough to pass the lowest frequencies required. Let's choose 20Hz. This gives a value of around 2u2.

Circuit 3. If the high and low references simply moved up and down by the same amount the charge and discharge times would remain the same so the frequency would not vary.

Looking at the (ac input) circuit the high reference voltage changes by the full control input but the low voltage reference only changes by half the control signal.

The result is that the voltage between the two references changes by half the amplitude of the audio input.

As the voltage difference between high/low references increases the output frequency decreases (more time to charge/discharge), as the voltage difference decreases the output frequency increases (less time to charge/discharge).

This is the key to understanding how the 555 produces an fm signal (frequency varying with the audio input).


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