2
\$\begingroup\$

In the classic 555 astable circuit the external capacitor charges to 2/3 of the Vcc (through R1 and R2) and then is discharged to 1/3 of Vcc (through R2) and then it continues to oscillate between 1/3 and 2/3 of Vcc.

I would like to avoid the initial delay while the capacitor is being charged from 0 volts. The duration of the very first square wave produced by the circuit is almost twice as long as the period of all the subsequent square waves and it is this extended period (for just the first square wave) that I'd like to eliminate.

The problem also occurs with square waves produced with op amps ICs (as opposed to 555s).

Somehow maintaining the capacitor at 1/3 Vcc before the circuit is powered up is not an option. Immediately before the circuit is powered up, there will be zero volts across the cap.

I would like the circuit to behave as though the capacitor had a lower capacitance for just the first cycle, and thus charges from zero volts to 2/3 Vcc though the same R1 + R2 as fast as the capacitor normally charges from 1/3 Vcc to 2/3 Vcc in all the subsequent cycles.

Is there a standard method of accomplishing this?

UPDATE:

I could not get @transistor's solution to work, at least not in a simulator. What I did instead was add a separate R-C kick starter circuit:

R-C kick start circuit

If you fiddle around with C3 and R4, you can get the voltage across C1 to 1/3 Vcc very quickly. When you do this, the 555's first square wave is the same duration as all the subsequent square waves.

Overall, I'd have to agree with @Dave Tweed's comment that this is starting to defy the spirit of using a 555 for its simplicity. Plus, it's sort of a clunky solution, not having a comparator to shut off Q1 when C1 reaches 1/3 of Vcc.

\$\endgroup\$
  • 1
    \$\begingroup\$ Go to a digital method of generating the square wave OR have a buffer circuit that nulls the output until the 555 is fully started up. Look at the linear.com/product/LTC6992-1 \$\endgroup\$ – Voltage Spike Jun 1 '16 at 19:23
  • \$\begingroup\$ Once you add the circuitry required to accomplish what you want, you've lost the point (simplicity) of using the 555 in the first place. There are plenty of cheap, low pin count microcontrollers that can be programmed to do exactly what you need. \$\endgroup\$ – Dave Tweed Jun 1 '16 at 20:33
3
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Flying start 555 oscillator.

I've only used a 555 twice in the last few decades but this may do the trick for you.

  • C3 is connected between V+ and the timing capacitor. On power up C3 and C1 form a capacitive potential divider. If the voltage rises smartly the junction of C3 and C1 will rise to \$V_+ \frac {C_3}{C_3+C_1} \$ thus priming the capacitor for you. So, to aim for 1/2 supply set C3 = C1. Bear in mind that R1 and R2 will be discharging C3 during power up so you may have some calculations or experiments to do.
  • Once the oscillator starts up the timing will depend in C1 in parallel with C3 so use value \$C = {C_1+C_3} \$ in your timing calculations. The fact that C3 is connected to V+ rather than GND won't matter.
\$\endgroup\$
  • 1
    \$\begingroup\$ Good idea. Probably want C1 = 2*C3 to get 1/3 V+ \$\endgroup\$ – Spehro Pefhany Jun 1 '16 at 21:47
0
\$\begingroup\$

The startup delay is a 'feature' of the 555 timer.

Instead of the 555 try using a microcontroller such as an ATTiny which is about the same price as a 555 IC. A microcontroller can be programmed to start either high or low and will pretty much straight away. (Alternatively you could use an Arduino if you don't have an ISP programmer.)

Yes, a microcontroller may be considered as overkill but it is fit for the purpose you are after.

\$\endgroup\$
0
\$\begingroup\$

I hope this isn't a I need this yesterday and I only have 555s on hand question. My idea is to place a small MOSFET here:Hopefully the inrush is enough to charge the capacitor.

\$\endgroup\$
  • \$\begingroup\$ Mind providing a few more details about how this works? Also, this isn't going to be well controlled as Vgs(th) is not tightly controlled in discrete MOSFET parts \$\endgroup\$ – ThreePhaseEel Apr 27 '17 at 23:32

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.