Comment summary #50
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Comment 491 ... 500
| date | article | author | comment |
| 24 feb 2014 12:17:31 | Theory and Definitions | Masello | Can some1 ple help me to get power and rms value of a signal: 10 cos ( 100t + pi/3) |
| 17 feb 2014 23:04:41 | Fast Lux meter | steve | Thanks a lot; I really appreciate your time and patience. I will keep you updated with the result of the work... |
| 17 feb 2014 10:26:53 | Fast Lux meter | Freddy | That's the ground symbol. It has no connection to anything. Or, in case the circuit is build in a metal casing: you could make a connection to the case at that point. |
| 17 feb 2014 00:29:24 | Fast Lux meter | steve | Thanks for your patience with me... the image clarify the connections very well. It is very clear now. Just one Last question...the very last question... Which part of the power supply is the tick black small block of pin 3 connected to? Please, If my question is not clear, you can use this link: http://kenfreetutorial.blogspot.com/2010/03/tutorial-world.html I mean the side that is marked with red X in the above Link. Which part of the power supply is it connected? |
| 16 feb 2014 23:47:08 | Fast Lux meter | Freddy | I hope this image will clearify the connections. |
| 16 feb 2014 22:54:04 | Fast Lux meter | steve | Thanks... Finally, if am using the original circuit in figure 6 and connecting my power supply to pin 7 (+) and pin 4 (-); should I use the same +12v for pin 7 ? OR should I refer to your previous reply and I Quote At 10 feb 2014, 02:27:26 wrote Freddy... "The zero volt power supply is not the same as the ground. It is only the reference for the positive power supply. If your using the ground as power supply zero reference, than you can connect pin 7 to +5V and pin 4 to -5V." Based on this previous reply of yours, Should I connect pin 7 to +5V and pin 4 to -5V OR pin 7 to +12V ? |
| 16 feb 2014 22:00:29 | Fast Lux meter | Freddy | Both circuits will not work. Keep in mind that the OP37 can only operate correctly if both inputs (pin 2 and 3) are at least 4 volts higher than the voltage on pin 4 and at least 4 volts lower than the voltage on pin 7. The output voltage on pin 6 can only become 3 volts higher than the voltage on pin 4 and 3 volts lower than the voltage on pin 7. With other words: The input and output voltage range is limitted. This is the reason why R2 and 3 with the capacitors are there. They provide the circuit with a third voltage that lays halfway the supply voltage. Use the original circuit in figure 6 and connect your power supply to pin 7 (+) and pin 4 (-). Measure the voltage with a multimeter between pin 3 and 6. |
| 16 feb 2014 21:19:00 | Fast Lux meter | steve | I really appreciate your reply. The calculation is clear. Please, can you look at the second question... I have re-drawn the circuit diagram based on your previous explanation on how to connect a Voltmeter. Which of the two diagrams is correct? I cannot upload the diagram in this post; please, use this link: http://kenfreetutorial.blogspot.com/2010/03/tutorial-world.html Thanks for your good work... |
| 16 feb 2014 00:03:16 | Fast Lux meter | Freddy | Hi Steve. At 1000 Lux the photo diode outputs a current of 10 µA, see figure 2. Because this relation is linear it is 10 nA/Lux. Both opamp inputs are kep on 0 volts and all the current flows through R1, the equation is a simple Ohm's law. The circuit of figure 2 outputs 500 µV/Lux. So we can calculate R1=U/I = 500*10^-6/10*10^-9 = 50000 ohms. The same as given in the circuit. |
| 15 feb 2014 02:47:49 | Fast Lux meter | steve | Well done Freddy. You told me to Multiply the read voltage by 2000 to get the Lux value. Can you give me the equation that shows how this 2000 is being used? OR Is the 2000 a constanst; please explain. I have re-drawn the circuit diagram based on your last explanation on how to connect a Voltmeter. Which of the two diagrams is correct? I cannot upload the diagram in this post; please, use this link: http://kenfreetutorial.blogspot.com/2010/03/tutorial-world.html |