That's kind of like asking "how long is a piece of string".

If your light source is a fluorescent light that flickers 100 or 120 times per second, you'll need to add significant capacitance (e.g. 0.1 uF or more) to prevent multiple clocking. That capacitance works by delaying the change in the phototransistor's collector voltage. Therefore it will delay the initial detection of the light significantly as well.

It also depends on the resistance of the pot in the phototransistor's collector circuit.

Perhaps I can explain how the circuit works, and how the capacitor will affect it.

The phototransistor is a variable current sink. It conducts a certain amount of current from its collector to its emitter. That amount of current is roughly proportional to the instantaneous amount of light it receives.

The potentiometer between the collector and VCC converts this current into a voltage. If the phototransistor is dark, and is sinking no current, there is no current through the pot, and therefore no voltage across it. Therefore the collector, and the 555's input, is at the VCC voltage (or nearly).

As the amount of light falling on the phototransistor increases, the phototransistor starts to sink current. This current flows through the pot, and causes the voltage across the pot to increase. This is another way of saying that the collector voltage decreases.

At a certain point, the voltage drops below the 1/3 VCC threshold of the 555's input, and the 555's output goes high. The 555 has a Schmitt trigger action that means that its output will stay high until its input rises above the higher threshold voltage, which is 2/3 VCC.

Now if you add a capacitor across the optocoupler, the capacitor will slow down the voltage change. Its behaviour is: the rate of change of the voltage across it is proportional to the current flowing into it.

It's a bit like a rechargeable battery - the more current you force into it, the quicker it charges, and the more current you draw out of it, the quicker it discharges. Except with a capacitor, everything is linear and smooth, unlike rechargeable batteries which only have a limited usable voltage range.

The formula for charge/discharge in a capacitor is:

dV / dT = I / C where

dV is the change in voltage, in volts;

dT is the time period, in seconds;

I is the current, in amps;

C is the capacitance, in farads. (Be careful here - 1 uF is only 1x10^-6 farads.)

This formula shows that applying or drawing more current to/from the capacitor will cause the voltage across the capacitor to increase or decrease at a faster rate.

It also shows that for a given current, a smaller capacitance will allow the voltage across it to change more quickly than a larger capacitance. This is why when you put a capacitor across part of a circuit to slow things down, a high capacitance has more effect than a low capacitance.

So let's connect a capacitor across the phototransistor, and assume there's no light on the phototransistor, so its collector voltage is roughly VCC. Now shine a light on the phototransistor.

The phototransistor will sink current. Some of this will come from the potentiometer, but some will also come from the capacitor. If we ignore the potentiometer, the collector voltage will drop at a rate that is determined by two factors:

(a) how much current the phototransistor is drawing (more current makes the voltage fall faster), and

(b) the amount of capacitance you've added (more capacitance makes the voltage fall more slowly).

At a certain point, the phototransistor's current and the current through the pot will be equal and the collector voltage will stabilise. But before they do, you will see a delay, due to the capacitor's presence.

It's convenient that a higher current (corresponding to a brighter light) will cause the capacitor to discharge more quickly. This means that the delay in responding to a bright light will be shorter than the delay in responding to only a small increase in light level.

If the illumination on the phototransistor decreases, the phototransistor will sink less current. There will now be a positive current flowing into the capacitor, because the potentiometer is trying to pull the collector voltage up towards VCC. Again this increasing voltage will be slowed down by the capacitor, and this will delay the detection of "dark". But it seems that in your case you are mainly concerned about the delay in detecting a change from dark to light, right?

So I'll plug some approximate numbers in, to make it a bit more concrete.

1. Let's say the circuit's supply voltage is 12V. This means the 555's input thresholds are at 8V and 4V.

2. Let's assume that the "brightness threshold set" potentiometer, from the phototransistor's collector to VCC, has been set to 20 kilohms. We have tested with our light source and found that to get the 555 to report "light present" (output high) at the illumination level that we want to initially detect as being "light" requires the trimpot to be set to 20 kilohms.

3. Initially, there is no capacitor.

We can calculate the phototransistor current that corresponds to "light". The threshold voltage for detection in this direction is 4V, so at that time, the pot has 8V across it. Use Ohm's Law, I = V / R, with V = 8 and R = 20k. The current is 400 uA (microamps).

Now let's add a 10 nF capacitor across the phototransistor and shine the same amount of light onto it. In this calculation I'm going to neglect the effect of the current drawn by the potentiometer, to simplify the calculations.

So initially the collector voltage is 12V (darkness). We shine a light on the phototransistor and it starts sinking 400 uA.

Ignoring the potentiometer, and just considering the capacitor, the collector voltage will fall at a certain rate, which can be calculated from dV / dT = I / C.

I = 400e-6

C = 10e-9

so dV / dT = 40,000V/s.

So to calculate the delay, we need to work out how long it will take for the collector voltage to drop from +12V to +4V, i.e. to drop by 8V.

40,000 = dV / dT rearranges to dT = dV / 40,000 which is 8 / 40,000 which is 200 us.

So adding a 10 nF capacitor delays detection of light by about 1/5000th of a second.

Those calculations are simplified because I ignored the effect of the current supplied by the potentiometer. I did that because the dV/dT=I/C equation can only be used in that simple form if the current is constant. This is roughly true of the phototransistor, but not true of the pot - the pot's current depends on the voltage across it, which varies significantly as the collector voltage falls during the time period I'm trying to calculate.

I hope this is interesting to you - or somebody