0
\$\begingroup\$

In a standard astable configuration of the 555 timer, we have two resistors R1 and R2. The capacitor charges through both resistors but only discharges through R2, consequently, the high interval is longer than the low interval.

I am having trouble understanding the purpose of R1 here, won't removing it make the charge and discharge times equal, effectively making it a 50% duty cycle?

Circuit

I tried following this question, but the answers there is not comprehensive enough. Moreover, other resources I came across states that the resistor is essential and can't be removed. Other than changing the ratio between the low and high interval, why is R1 essential?

\$\endgroup\$
  • 2
    \$\begingroup\$ The discharge pin pulls low, to ground. If you connected it directly to Vcc... that would be a "bad thing." \$\endgroup\$ – jonk Dec 9 '18 at 21:31
1
\$\begingroup\$

Pin 7 is the discharge pin. It's an open-collector type, so it can only sink current to ground. It cannot source any current, at all. (It only becomes active when one of the comparators determines that it is time to discharge the capacitor.)

If pin 7 were directly connected to the \$V_\text{CC}\$ power supply then it would be just fine, while pin 7 is inactive. In this case, the capacitor's charging rate would be determined only by \$R_2\$ and \$C\$ (since \$R_1\$ would effectively be \$0\:\Omega\$.)

However, as soon as pin 7 goes active and attempts to sink as much current as possible, the very low resistance of a short-circuit from pin 7 to \$V_\text{CC}\$ would mean that very high currents took place. These would be almost certainly be sufficient to vaporize the aluminum interconnects on the IC die somewhere near where pin 7 is bonded. So a direct connection from pin 7 to \$V_\text{CC}\$ is ill-advised because of the problem that occurs when pin 7 becomes active.

A resistor from pin 7 to \$V_\text{CC}\$ isn't entirely desirable as pin 7 not only has to work on discharging \$C\$ via \$R_2\$, but then pin 7 also must sink any current produced by \$R_1\$, too. However, that resistor is absolutely required in order to limit the current that pin 7 sinks, when active.

Pin 7 will have a specified current limit and this must be more than enough to cover the discharging current from \$C\$ as well as the excess current being supplied via \$R_1\$ when pin 7 is active. So that's something to watch for in the datasheet, too.

In general, directly tying pin 7 to \$V_\text{CC}\$ is a "bad thing" to do.

| improve this answer | |
\$\endgroup\$
2
\$\begingroup\$

If you look at the 555 internal schematic as shown on the TI datasheet, you will see that the Discharge pin is only connected to the collector of a transistor - that transistor can pull the Discharge pin Low, but there is nothing in the chip to pull that pin high, or to provide and charging current to the capacitor. R1 pulls that pin high, and provides the charging current for the capacitor.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ I get the point of @jonk, but I am still confused. During charging, the discharge transistor is in cut-off mode and the discharge pin is no longer short circuited to ground, thus the current goes through both resistors to charge the capacitor, then why do you state "there is nothing to provide the charging current or pull the pin high"? What do you mean by "pull the pin high"? \$\endgroup\$ – Omar Emara Dec 10 '18 at 7:44
  • \$\begingroup\$ @OmarAhmad It's a common phrase in electronics. Resistors are often used as a passive device to "pull" some node towards one supply rail or another. If it is tied to a power supply with relatively positive voltage, it's called "pulling upward." Otherwise, it is probably "pulling downward." In this case, \$R_1\$ pulls upward on pin 7. Since pin 7 isn't actively pulling to ground in the situation Peter was referring to, \$R_1\$ continues to pull upward through \$R_2\$ as well, to the capacitor. (Pin 7 has no effect since the collector is "open" at the time.) \$\endgroup\$ – jonk Dec 10 '18 at 7:56
  • \$\begingroup\$ @OmarAhmad During charging, since pin 7 is inactive at the time, you could eliminate \$R_1\$, shorting it instead. However, as soon as pin 7 goes active, you suddenly find yourself needing a resistor between pin 7 and the Vcc rail. Otherwise, it's just a way to vaporize an IC's aluminum interconnects. So you are stuck with the resistor (or some other means to limit pin 7 current when it goes active.) \$\endgroup\$ – jonk Dec 10 '18 at 8:03
  • \$\begingroup\$ @jonk Thanks for clarifying this. Should your comments be added as an answer? I think it should be the accepted answer. \$\endgroup\$ – Omar Emara Dec 10 '18 at 8:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.