Electronic – the response current of a photodiode to an LED emitting light at different intensities

This multivibrator circuit of yours has only two states. In order to get what you ask for, you need more states. You likely will have to build it using several monostable multvibrators, and have each one trigger the next, in a loop.

Edit: Since you posted a photo, I see it is a purchased item. Nonetheless, you can modify it to be monostable, and you can buy several boards to achieve your goal.

The standard answer is to use a microcontroller to do this instead, but I generally believe in answering the original question and letting you decide.

Details: With this two transistor circuit, there are three ways to wire it:

• Bistable: Whichever light is on, stays on. You have to input something (like push a switch) to make the other light come on instead. Then it stays that way. It can be flipped the other way with another switch. This behavior is like the switch that turns on the lights in the room.
• Monostable: Using a switch, you can "kick" it to the other state, where it sits for a given time, then flips back to where it was, and stays that way. This is comparable to a 3-minute egg timer.
• Astable: It stays in one state for a time, then flips to the other. The other state is timed, and flips back to the first state. It never stops. This is commonly called a "flasher".

In this circuit, either the left transistor is on, or the right one is. When the left side turns off, the right side turns on. The corresponding LED's indicate this. It's really flashing one LED, and the second one can be added, but it will always be in the opposite state from the first. These things happen together, so it isn't possible to have both off at the same time. That would require third state--and more parts.

Let's say you could build a three-state circuit. You'd have these states:

1. Left LED on, right LED off.
2. Right LED on, left LED off.
3. Both LED's off.

Question: When both lights are off, which one is going to come on next? Your desired circuit has two places where one of the LED's go off without the other coming on. So, really, you need four states to produce this effect:

1. Left LED on, right LED off.
2. Both LED's off, with right one next.
3. Right LED on, left LED off.
4. Both LED's off, with left one next.

You can use a 555 in place of the multivibrator. And it can be wired to be astable, bistable or monostable. But, you still would need four of them. Both kinds of circuits can provide a pulse to trigger the next circuit, and both have an input where the trigger can be fed in.

I suggest looking at the diagram in the Wikipedia article on multivibrators (or maybe you got a schematic with the board?), and find all the like parts on your board. If you didn't get a schematic, you should try to draw one.

I can make a couple of guesses about what might be going on.

1. You aren't actually measuring the LED current, but instead you are assuming \$I_{LED} = (V_{in} - V_F)/220\$ and assuming \$V_F\$ is a constant. You are operating in a regime where \$V_F\$ will vary significantly (100's of mV), and neglecting this would explain (qualitatively, anyway) the shape of your graph.

2. The LED has some parasitic conductive path through it that does not cause light emission. For low currents, this path is taking more of the current and resulting in reduced light emissions until some threshold is reached and the proper path begins to dominate.

3. At very low currents, the optical emission pattern of the LED is changing, causing less of the light to reach the detector and more to be lost into the package.

4. The way you are hooking up the detector is causing funny behavior. Hooking up the base but not the emitter is not a common way to use a phototransistor (AFAIK), and is not what the device designers would be designing for. Typical phototransistor circuits are shown in an App Note from Sharp starting on page 13.