# [SOLVED]Calculate 555 astable high time accurately with diode

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#### d123 Hi,

I did some calculations that were wrong, I believe, and I've just looked at a few webs and none so far show an accurate way of calculating t1 (output high, charging) for the 555 when the duty cycle is modified for <50% with a diode. They omit the diode resistance or how it affects the standard astable calculation for high output. Normal astable and astable with diode formulas are in the attached schematic/jpg.

Ra = 10k
Rb = 1M
C = 7.05nF

t1 = 0.693*10k*7.05nF = 488uS Wrong.
t2 = 0.693*(2M + 10k)*7.05nF = 9.82mS Right.

If I make t1 = 0.693*20k*7.05nF = 97.7uS Right.

Frequency counter gave 101Hz in a real circuit, and:
1.44/((10k + 2M)*7.05nF) = 100.8Hz Right/matches real circuit.

I checked 1n4148 resistance at 630uA (6.3V/10k) on a website, and it is far from 20k, it's about 100R, and doing a parallel calculation for 100R and 1M is not 20k. Not sure what I'm getting wrong.

Any ideas of how to factor in the diode resistance to the formula for t1, or how to make calculations match reality? Or point out what I have got wrong in my calculation for t1 with the 10k?

Thanks. #### std_match The diode is a non-linear device. It has more of a constant voltage drop than a constant resistance value.
If the diode voltage is exactly constant (which it isn't), it will give a new VCC-dependent value for the 0.693 factor in the t1 calculation.

• d123

### d123

points: 2

#### d123 Hi,

Thank you.

The diode is a non-linear device. It has more of a constant voltage drop than a constant resistance value.
That I can understand. And your definition matches my understanding of the graphs I saw on the linked website.

If the diode voltage is exactly constant (which it isn't), it will give a new VCC-dependent value for the 0.693 factor in the t1 calculation.
That sentence I don't fully understand. Do you mean that the 0.693 value would need to be modified to calculate t1 correctly?

[Why I chose a hobby I love but I'm so bad at, especially as I always found understanding 'high school' mathematics akin to attempting to read Sumerian tablets, is mystifying to me.]

It would explain why no-one bothers to provide highly accurate formulas for the calculation, especially as it's only a 555.

Thanks, again.

#### std_match The time t1 is for charging C3 from 1/3 VCC to 2/3 VCC. With a diode with constant drop Vdiode and the 1/3 VCC as the reference, the charging (=final) voltage is (VCC - Vdiode - 1/3 VCC) = (2/3 VCC - Vdiode)
Again, with the 1/3 VCC as a reference, we want to charge C3 to (2/3 VCC - 1/3 VCC) = 1/3 VCC
The equation to solve for t1 is then
1/3 VCC = (2/3 VCC - Vdiode) * (1 - e^(-t1 / (RC)))

If you set Vdiode to zero, the equation solves to t1 = 0.693 * RC
If Vdiode isn't zero, t1 depends on VCC (and the nonlinear behaviour of the diode).

• d123

### d123

points: 2

#### d123 Hi,

That's so nice of you. Thank you for taking the time to show how to calculate it accurately. I really appreciate it. I wouldn't even have known where to begin... The formula will be added to my notebook of useful formulas, and used more than once, I'm certain of that.

Many, many thanks.

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