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Reference following circuit:

555 Timer Bistable

which comes from the article Bistable Multivibrator Using 555 Timer.

I am assuming the capacitor leaves pin 5 floating and therefore the comparators into which pins 2 and 6 feed remain at reference volatages of 1/3 and 2/3 of Vcc, but that there is some noise filtering achieved.

However, I would like to change the reference voltages and think I understand that if I connect pin 5 to my desired control voltage (via voltage divider), call it Vx, I should get a reference voltage at the comparator connected to pin 6 of Vx and a voltage of Vx/2 at the comparator connected to pin 2.

(1) Is this correct?

(2) What is the capacitor connected to pin 5 in the schematic as drawn achieving?

(3) Is the capacitor connected at pin 5 still needed if I connect the pin to my required Vx?

EDIT: MODIFIED CIRCUIT AND APPLICATION

With thanks to @GodJihyo for very helpful answer, modified circuit is on the right of the below schematic.

With R3 between pin 5 and ground, V5 control voltage is now set to Vs/3, so second internal comparator of 555 IC references Vs/6. R2 is changed so that 'transient' voltage at pins 2 and 6 is now Vs/4. Resistor R12 added so that C3 charges only to Ve = V3/2. Since TSOP4838 operating voltage is 2.0-5.5V and its output is active low, R4 and R5 achieve close to Vs/2 (internal resistance of 30k) and PNP now replaces the switch in the original circuit.

End result is in response to a burst of IR, LED switches on and stays on, in response to the next burst of IR, LED switches off and stays off, and whole circuit is powered by a single supply. And V3 in high state is (close to) Vs which can supply components requiring Vs.

schematic

simulate this circuit – Schematic created using CircuitLab

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1 Answer 1

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The voltages you get will depend on how 'stiff' your control voltage is, that is to say what the impedance of your CV source is. If you're using a voltage divider to feed pin 5 you need to take the existing divider's resistances into account as you'll have something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

In this case you might expect CV to be 5 V but it will be 6 V, and the voltage to the other comparator will be 3 V.

As for the capacitor, it is as you suspected there to reduce noise, and often is not used. When using a control voltage you can use it or not, as long as you're not having any problem from noise you could leave it out, but it shouldn't hurt to leave it in.

Also note that you can raise the voltages to the comparators by putting a single resistor between Vcc and Pin 5, so it is in parallel with R1, or lower them by putting a resistor between Pin 5 and ground, so it is in parallel with R2 and R3.

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  • \$\begingroup\$ Nice answer, thank you. I'll test it out and respond. \$\endgroup\$
    – RickyBoy
    Apr 15, 2023 at 20:00
  • \$\begingroup\$ Edit noted, thank you \$\endgroup\$
    – RickyBoy
    Apr 15, 2023 at 20:10
  • \$\begingroup\$ Since you are driving those components way over on the left (reference designators - !!!) with the 555 output that has only two possible states that are not adjustable, why do you want to change the internal reference voltages? \$\endgroup\$
    – AnalogKid
    Apr 15, 2023 at 20:27
  • \$\begingroup\$ Thank you very much for this answer. \$\endgroup\$
    – RickyBoy
    Apr 26, 2023 at 21:22

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