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I am trying to build a 555 timer circuit and am stuck on understanding how exactly the discharge pin works. I am going off of these block diagram from the datasheet

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RA, RB and C form an RC network in which C will charge and discharge. I very well could be wrong as I am still learning, but it seems that the discharge pin input acts as Vcc to the transistor's collector, and the base input to the same transistor is the SR flip-flop Q' output. Furthermore, RA and RB form a voltage divider network (this is where I believe I am most confused) that divides the voltage that has formed over the capacitor. The discharge input, which is Vcc to the transistor, takes the voltage over Rb.

The capacitor will discharge when the transistor is closed, meaning Vcc > base voltage on the transistor and the charge built up on the capacitor will be sunk to ground.

I don't understand exactly how RA and RB interplay with the discharge pin. Can someone explain this?

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    \$\begingroup\$ If the transistor is inactive, then the collector is high impedance. So capacitor C charges through RA and RB from Vcc until it reaches 2/3 of Vcc. Then the transistor becomes active and the collector is very low impedance and close to ground. So now the capacitor discharges through RB until it (voltage on C) reaches 1/3 of VCC, when it (transistor) turns back off. (While active, the transistor will be sinking both the discharge current and also the current from Vcc through RA.) \$\endgroup\$
    – periblepsis
    Commented Jun 20, 2023 at 2:32
  • \$\begingroup\$ Forget all about Ra and Rb acting as a voltage divider. That's only true(ish) part of the time, and not relevant to the 555 circuit operation all of the time. \$\endgroup\$
    – brhans
    Commented Jun 20, 2023 at 3:02
  • \$\begingroup\$ DISCH is just an output, doing nothing when open and draining current when closed, in most simple words. Does this help you understand? \$\endgroup\$
    – the busybee
    Commented Jun 20, 2023 at 5:43

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When output is high that is pin 3 is 1 , the capacitor charges, through RA and RB. path is Vcc-->RA-->RB-->C. The transistor is off , so negligible current flow through the transistor , hence discharge pin.

On the other hand , when output is low pin 3 is 0, the transistor is on , hence current flow through discharge pin. The capacitor is discharged through RB. Path is V(capacitor)-->RB-->Transistor. Current also flow through RA but path is VCC-->RA-->Transistor.

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