Generating a delayed pulse of 0,5sec with one 555 timer upon powering up the circuit

Question

In the link below, this circuit I found is what I needed. Only the length of the pulse and the delay are not what I want.

When I want just 0.5sec before pin 3 is up and the pulse duration I want is also 0.5sec, can someone tell me how big my resistors and capacitors must be? Or can someone tell me how I can calculate this?

My source power is 16 Volts I have already found 555 timers that can operate with 16 Volts, so thats not the problem.

When someone has another idea with two 555-timers, or something else that does the same, it's also OK.

Thanks

Circuit I need with wrong delays

Answer 1

Assuming you're talking about akhmed's circuit:

The width of the output pulse is calculated in the normal way:

$$T_{ON} = 1.1\cdot R4 \cdot C4$$

The initial trigger delay uses the same formula:

$$T_{DELAY} = 1.1\cdot R1 \cdot C1$$

For both of these, you might use 470 kΩ and 1 µF, giving a nominal value of 0.517 seconds for each.

The only "trick" required is that R3+C3 (via D2) must pull the trigger voltage above 1/3 V_CC sometime after \$T_{DELAY}\$ but before \$T_{DELAY} + T_{ON}\$ (i.e., between 0.6 and 0.9 seconds). You need to keep the ratio of R3:R1 at around 1:3, and adjust C3 to get roughly the delay you need.

This is complicated by the forward drop of D2, but an approximation would be

$$T = -ln(1/3)\cdot R3 \cdot C3 = 0.4 \cdot R3 \cdot C3$$

If you set R3 to 150 kΩ and T to 0.75 s, you can solve for C3, giving 12.5 µF. This isn't a standard value, but 12 µF is, which should be close enough. Some experimentation may be required.