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| General Electronics Chat This forum is for general chat about electronics, eg: Dont know what a part does? Dont know how to read a circuit? Want to get an opinion? |
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28th January 2007, 06:12 PM
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 Photometer I need help in figuring out the amperage of the 5V supply. I also want to know what the output voltage would be like and what the transistor is doing there. Thanks URGENT | |
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28th January 2007, 06:24 PM
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 You would only need a low current supply, a 78L05 should be more than enough - on a quick glance the transistor looks to be a common base amplifier?. Presumably the text explains it all?. | |
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29th January 2007, 01:53 AM
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| The transistor (Q1) allows a single supply to be used. The op amp input and output are at the same voltage level, only the pulse from the opto is amplified. If the 510 ohm resistor were directly connected to the op amp, the output DC would be very difficult to control.
__________________ see my website: www.geocities.com/russlk | |
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29th January 2007, 05:47 PM
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| I have a 400mAh, 4.8V . would this do? | |
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29th January 2007, 06:04 PM
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| Should be fine!. | |
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29th January 2007, 09:47 PM
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| What should the output be like. between 1-3 volts? i need something more presicise | |
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30th January 2007, 07:10 PM
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| If i were to measure the variation in potential difference as a function of intensity, where would i have to plug the volt meter. | |
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30th January 2007, 07:31 PM
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| What are you trying to do?, this circuit merely amplifys the AC signal received by the photodiode, it's not measuring the intensity. | |
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30th January 2007, 08:28 PM
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| I'm trying to measure the intensity of incident light on the photodiode. Why can't i do this | |
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30th January 2007, 08:33 PM
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| how much of bias voltage should i apply and how will it affect my ouput | |
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30th January 2007, 08:33 PM
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| The application note you posted does nothing like that, you need a DC coupled circuit. | |
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30th January 2007, 08:45 PM
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| could you help me couple it? please. its urgent. i have a week to finish this. | |
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30th January 2007, 08:53 PM
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| what i meant was like this circuit will have a different vout as the current on the photodiode varies right. so i can, measure the voltage recorded under different sources of light cant i?otherwise wht youre saying is that this will always give a constant voltage indepedant of the the intensity? | |
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30th January 2007, 11:29 PM
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| You need to turn the light off and on or put a rotating wheel between the light source and photodiode to chop the light.
__________________ see my website: www.geocities.com/russlk | |
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31st January 2007, 04:02 AM
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| Ok , may i know why im doing that please? | |
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Linear Mode
