I need to make a device that counts light impulses, basicly a light detector counter, i need help with my counter cp, how to implement a photodiode(can't find phototransistors in the stores) to do clocking, do i need a IC 555 timer, or can i just use an common transistor and use photodiode as base current of the transistor? im using a 7493 counter a 7447 decoder and BPW 34 photodiode.
Making a clock cp for counter, using a photodiode
clockcounterdevicedigital-logicphotodiode
Related Solutions
At first glance, this appears to be a power problem. It appears that your "5 V" power supply droops to less than 4 V in operation.
Perhaps the power supply is inadequate. So power your circuit from some other wall-wart or other power supply that can supply at least 10 times as much power as you think you need (at least 10 times as much current at 5V). (Another possible explanation: perhaps the wires and connectors and traces between your power supply and the chip's GND and power pins have too much resistance; so replace them with thicker wires and less oxidized connectors).
No matter how good the wall-wart is, it can't do anything about the inductance in the wires between the wall-wart and your PCB. That inductance creates an impedance significantly larger than measuring the resistance of those wires might suggest. You need a "bulk capacitance" capacitor on your board. There are ways to calculate how much you need, but many people just throw on a 1000 uF capacitor on their first prototype -- usually overkill, but not worth trying to find out how much less we actually "need". "Logic Supply Decoupling Techniques", "General PCB Design Guide", "Decoupling Capacitance Design Rules", "Power Distribution System Design Methodology and Capacitor Selection for Modern CMOS Technology", etc.
No matter how good a bulk capacitor is, it can't do anything about the inductance in the traces between that capacitor and each IC. That inductance creates an impedance significantly larger than measuring the resistance of those traces might suggest. Put a 100nF decoupling capacitor across the GND and power pin of every IC. "Decoupling Capacitors Explained -or- If You Like It Then You Shouldn’t Let It Ring Like That", "What is a decoupling capacitor and how do I know if I need one?", "Decoupling by Example", etc.
Unfortunately, these 3 problems bite extremely often, and cause all kinds of strange and not-quite-working-right behaviors.
(How do we add "decoupling and bulk capacitance" to the Electrical Engineering site FAQ ?)
I'll try to cover the parts of your question separately.
Multiplexing
Is the multiplexer represented in the schematic where it says "buffers"?
Nope! The multiplexer is the select digit part of the circuit. As you said, a multiplexer is an electrical switch: if I have \$n\$ "selector" inputs, I can choose from \$2^n\$ outputs. In your case, you have a two bit counter (because 1 bit isn't enough to count up to 2) which is connected to the "select" part of the multiplexer. The mux then sets one of its four outputs high, depending on what the counter is. If you make your counter reset as soon as it hits 3, then your multiplexer will set 0 high, then 1, then 2, and repeat this loop forever.
If the ultimate goal is to display the output on two additional displays, why couldn't we just wire the all the a's, b's, c's, etc. together?
When the mux has set digit 0 high, we only want to light up display 0 (and likewise for 1 and 2). If you wire the displays together, you can't control them all with different digits.
High End Digit Driver
The purpose of the transistor drivers are to supply/sink current to/from the LEDs, but it's unclear to what it's connected to electrically in terms of the emitter, collector, and base.
Look at a single digit driver. When its mux output is high, you want current to flow from your power supply into the LEDs; when the mux is low, you want to block that current. That means your which digit? output is probably connected to the base of the transistor, and setting it high allows current to flow from the collector to the emitter. Is that enough of a step in the right direction?
If the drivers are simply sourcing or sinking current, why are they connected to the multiplexer?
You'll have three drivers. You only want to turn on one at a time, and the multiplexer picks which one. They aren't "simply sourcing current", I guess - they're current sources that you can selectively turn on and off.
Counter Circuit
However, what is considered a desirable output here?
You want to count
0, 1, 2, 0, 1, 2, ...
so you'll need two wires (like I mentioned above). What parts do you have access to? What kind of counter circuits have you seen before?
As Nedd mentioned, you have a good oscillator set up - that'll be the input to your counter. Flip-flops would be a standard approach from there.
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Best Answer
Try this
simulate this circuit – Schematic created using CircuitLab
The 7414 is a Schmitt trigger, and will handle even very slow transitions of light level. If you just use an ordinary gate, and the input changes slowly, you may get multiple triggers as the input responds to noise right at the threshold. Conversely, if your light pulse has a very fast turn-on AND turn-off, you may not need the inverter at all.
R1 is used to set sensitivity. But note that a "real" 7414 needs 1.2 mA of photodiode current, plus the resistor current to operate. If your light source is not intense enough, the photodiode won't be able to pull the input low. In that case you can try a 74LS14 which, like all 74LS, only requires 0.4 mA. If that's still too much, go to a 74HC14. It will work just fine driving the original 7400 series.
And, of course, if you need the clock on the falling edge of the light pulse, just add another inverter.