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Does anyone know how to use the 555 timer control voltage? My thoughts are if -input in the upper ("Threshold") comparator is at 2/3Vcc, then any voltage should act against the 2/3Vcc into the -input. But would it increase the voltage at 1/3Vcc into the +input of the lower ("Trigger") comparator?

Essentially how does the control voltage affect the voltage divider and current through the voltage divider line?

Thanks

555 timer internal block diagram

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  • \$\begingroup\$ Resistor can be connected to change reference voltage \$\endgroup\$ – user257432 Jul 12 at 19:30
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If you establish a certain voltage Vc at the control pin, either with a stiff voltage source or some kind of network of resistors or whatever, then the capacitor charges from that Vc/2 up to Vc and discharges back to Vc/2. Note that the resistor values are not well specified, 5K is a 'typical' value, but they are well-matched to each other.

(Looking into the control pin it looks like a voltage source of Vcc*(2/3) with a source resistance of roughly 5K||10K = 3.3K. Some variants of the original NE555 such as the TLC555 use 100K nominal resistors so it would be more like 67K.)

The consequence of this is that changing Vc by means of external resistors (rather than, say, an op-amp output with feedback) generally will significantly degrade the initial accuracy and the temperature stability of the output frequency since the internal resistors are not very stable with temperature nor necessarily all that close to nominal to begin with.

schematic

simulate this circuit – Schematic created using CircuitLab

If we simulate this with the control voltage swept from 0 to Vcc over 1 second you can see the capacitor voltage extremes change (orange/yellow trace) , and the duty cycle of the output (blue trace) changes.

enter image description here

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The control voltage is directly the the node with 2/3 VCC.

If you externally set the voltage to some value V, then that will be the voltage the upper comparator sees.

And the divided down voltage for the lower comparator will be half of it, V/2.

Currents can be calculated with ohm's law.

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