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need help... urgent!! plzzzzz


Ann Margareth
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Hi Ann Margareth:

1 uF at 15 or 25 volts is a very common capacitor value that you should be able to find at Radio Shack, any electronic supply store, electronic surplus store, Digikey, Mouser, or on scrap circuit boards.  But you can also substitute with little difficulty.

No identification is shown for IC1, but it looks like a classic 555 timer application.

The capacitor on pin 7 of IC1 is part of the time constant setting circuit that determines the on and off times of the output to the LED.  I don't know what the capacitor and resistor values are, but you can calculate the design duty cycle and period by reference to the 555 spec sheet.  Google "555" and you will find many tutorials showing how to calculate period and duty cycle for the recommended components.  If you don't stray too far from the 1 uF design value you can compensate for a different value of capacitor by selecting different values of resistors in the string.  Thus, if you use a 0.5 uF capacitor at IC1, double the values of the two resistors above it in the schematic.

The second capacitor is not at all critical, as it is only setting a response time for the LED drivers.  A larger capacitor here will make the LEDs slower to respond to the input pulses and conversely.  I'd bet that you could vary the value of the second capacitor by +/- 50% without noticing any change in performance.  I wouldn't try to compensate for any modest changes in the value of the second capacitor.  The value of this capacitor does not have to match the value of the capacitor at IC1, since it does not have any "tuning" or frequency determining function.

Have fun

awright

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Hi Awright,
Welcome to our forum. ;D
At the top of this page is a row of buttons. The one labeled "Projects" links to our projects section. The Park-aid project that Ann Margareth is building can be found there and its link is here: http://www.electronics-lab.com/projects/automotive/005/index.html .

We have discussed this project before on the forum like other projects. A discussion can be found by using the "Search Site" button at the bottom of this page. ;D

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You want to use the correct capacitors. It does make a difference. Besides, this is not a hard cap to find. You should get used to keeping a good selection of parts if you are going to be buiding projects.
If the comparators do not function as intended, then your sensor will be unreliable.
The second cap which is parallel with a resistor is also a tuned network.

Beginning hobbyists should purchase cap kits and resistor kits which have a few of many values so that they do not run into the probems of trying to substitute the wrong parts. This often results in asking others why the circuit does not work.

MP

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I agree just use a 1uF capacitor they're no expensive and you can get them from scrap items, if you're that desperate go down to youe local dump and pickup an old radio or TV and you're bound to find one on the PCB.


The second cap which is parallel with a resistor is also a tuned network.

How?
Aren't tuned circuits normally made of inductors and capacitors, I know RC filters can be used as phase shift networks but they're normally connected in series not paralell and are found in filters and oscilators.

I notice IC2A doesn't have any negitive feedback as there's no resistor in series with the - pin, isn't this bad? What's the internal impedance of D2? If it's fairly high then this wouldn't be a problem but if it's low the op-amp will have it's open-loop gain and become unstable.
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Hi Alun,
C2 has a very small value because it is used to correct the frequency response of the opamp IC2A. The 1M negative feedback resistor R4 and the high impedance of the non-conducting photodiode causes a high frequency peaking effect due to stray capacitance at the inverting input of the opamp to ground. The RC time constant of R4 and C2 is chosen to equal or exceed the time constant of the stray RC to ground at the inverting input of the opamp to result in a flat high frequency response or drooping one.

The photodiode in this circuit is used in its photo-voltaic mode, where it generates a current like a solar cell. The current is very small so the opamp has a high value for its negative feedback resistor R4 for lots of gain.

At first I thought the opamp was incorrectly biased. But it is a "single-supply" opamp that works fine when its inputs are at its negative supply (ground, in this circuit). So the output of the opamp will be near ground without light and rise positively when the photodiode detects light. ;D

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