Kevin Weddle Posted April 3, 2004 Report Share Posted April 3, 2004 I am refering to the project found on your home page. What is going on with the batteries? Please explain the relative voltages surrounding the photodetector, which is forward biased. Are the two batteries working together?Project link: http://www.electronics-lab.com/projects/pc/023/index.html Quote Link to comment Share on other sites More sharing options...
serge_saati Posted April 15, 2004 Report Share Posted April 15, 2004 The batteries are used to convert the infrared signal into electric signal. Of course, the electric signal is alternative. The filter lets pass only the good frequency. But it has other ways to do that. We can use only the voltage source of the computer's power supply. We have to use the red wire (+5V) and the black wire (GND). But we must add to other logic gates. Those must be of series 74AS and type TTL. Remember, all the logic gates of the circuit must be powered with the 5V power supply of the computer. So the +5V must be connected to the Vcc pin and the ground must be connected to the GND pin. I send you the new schematic circuit possible for the receptor of the desktop computer. Quote Link to comment Share on other sites More sharing options...
Kevin Weddle Posted April 17, 2004 Author Report Share Posted April 17, 2004 I don't think the outputs are differential like they are supposed to be. The outputs are complementary. The purpose of the differential is to treat the inputs as two signals. Sometimes this is better than using ground because the ground can contain many signals. Quote Link to comment Share on other sites More sharing options...
Kevin Weddle Posted April 22, 2004 Author Report Share Posted April 22, 2004 I am perplexed. Why is the photodetector forward biased? The gate must me CMOS. Am I right? I have another question. The original circuit is what I am concerned with. It has a battery that works in conjunction with the power supply, but without a common ground. The use of devices without a common ground is circumspect, but it still doesn't appear to have the correct voltages. Quote Link to comment Share on other sites More sharing options...
audioguru Posted April 28, 2004 Report Share Posted April 28, 2004 I agree with you, Kevin:1) Why is the IR photodiode FORWARD biased? Most photodiodes are REVERSE biased, so that there is a very small leakage current when dark, but IR radiation causes much more leakage current. An IR photodiode project is here:http://www.comdev.ca/esq/Lightwavecommunicator.pdfThe photodiode leakage current modulation causes a voltage signal across it. The amplifier following the photodiode is used only to make the circuit extremely sensitive for long-range pickup. Our project doesn't need long-range and therefore doesn't need that amplifier.2) Why is the function of the rectifier diode and why is it forward biased? Since it is in the middle of a voltage divider then it will not rectify. Quote Link to comment Share on other sites More sharing options...
serge_saati Posted April 29, 2004 Report Share Posted April 29, 2004 The infrared diode is not biased. When no infrared signal come on it, the voltage at the input of the filter is 5V. When infrared signal come on it, the voltage at the input of the filter is -5V. When the infrared signal oscillates, because the data transmission is an oscillating signal (up to 100 MHz), the voltage at the input of the filter oscillates between -5V to 5V. So there is an alternating current (peak voltage is 5V). The filter is used to let pass the good frequency. The digital amplifier is used to smooth the signal. It's mean to convert analog signal (caused by the limited response time of the infrared diode) into direct current digital signal. The network card receive this signal. We can't use CMOS gates because they are too slow. The TTL logic gates of the series AS (74ASxx) have short response time (less than 1.7 ns) so thy can response to frequency higher than 200 MHz. The 9 volt battery of the first circuit is only used to convert direct current into alternating current. The 5V source is used to supply all of the digital components and works with the 9V battery to produce the alternating current. Because the ground of the power supply is alway connected to the ground of the logic gate, the gate can receive the negative polarity of the 9V battery if its positive polarity is connected to the positive supply of the 5V source. Quote Link to comment Share on other sites More sharing options...
audioguru Posted April 29, 2004 Report Share Posted April 29, 2004 I disagree with you, Serge:In all of your circuits, of course the IR diode is foward-biased, with its anode shown connected to +9V, +4.5V or +5V, and the cathode connected through a resistor to negative or ground. The diode is shown conducting forward current, all the time.Please check your working circuits and you will find that the IR diode is connected in reverse. For example, in your 3rd circuit, the anode of the IR diode is shown connected to +5V. Since its cathode connects to ground through a resistor, then you must measure about +4.3V at the inputs of the logic gates without IR radiation, since the diode will be forward-biased. But when you measure it, I think that you will actually measure about 0V, since the diode is actually reversed and will not be conducting. I do agree that the 9V battery and resistor in the 1st and 2nd circuits and the +5V and resistor in the 3rd circuit will convert DC to AC during modulated IR radiation, but only if the IR diode is reverse-biased.Again, I ask what is the function of the rectifier diode and why is it forward biased? Are you using the voltage divider that includes the rectifier simply to bias-up the input of the schmitt trigger and therefore make it more sensitive to low-level signals? If so, then the rectifier is not needed. Quote Link to comment Share on other sites More sharing options...
MP Posted April 30, 2004 Report Share Posted April 30, 2004 Serge, I am assuming that you have built this circuit and it is working good, yes?MP Quote Link to comment Share on other sites More sharing options...
audioguru Posted April 30, 2004 Report Share Posted April 30, 2004 Which circuit works OK? Why was it completely re-designed? Quote Link to comment Share on other sites More sharing options...
Kevin Weddle Posted April 30, 2004 Author Report Share Posted April 30, 2004 I don't think the voltages are respective. You have to connect the 9v with the 5v somehow. What about the signal at the 9v negative terminal. If you have a signal there, you will have a signal on the other side. You have the 5v connected to the 9v through the photodetector and the 5v connected to the 9v through a 1kohm resistor. What is the voltage at the 9v negative terminal? Is it 5v? That would put the 9v positive terminal at 14v. Quote Link to comment Share on other sites More sharing options...
audioguru Posted May 1, 2004 Report Share Posted May 1, 2004 Kevin,In the 1st and 2nd circuits that use a 9V battery, the IR signal is taken across the 1K resistor. The +5V is usually bypassed with a capacitor and therefore is at signal ground.It doesn't matter what the voltage is for the - terminal of the 9V battery, since it only affects the 1pF capacitor, which surely can withstand a few volts. If the IR diode is forward-biased as shown and has about 1.5V across it, then there will be about 7.5V across the 1K resistor. So the - terminal of the battery will be negative 7.5V from +5V which is -2.5V with respect to ground. If the IR diode is reverse-biased then the 1K resistor will have 9V across it when dark, but signal with IR radiation. So when dark, the - terminal of the battery will be -4V with respect to ground. Quote Link to comment Share on other sites More sharing options...
Kevin Weddle Posted May 1, 2004 Author Report Share Posted May 1, 2004 Is the signal fed through the capacitor and inductor? This makes for a changing voltage at the negative terminal. To make it work for me, I would ground the 9v negative terminal to start. Quote Link to comment Share on other sites More sharing options...
stuee Posted June 16, 2004 Report Share Posted June 16, 2004 I cant find any replacemnts for the 74hs over in this part of the world (AUS) can anyone help me out, i really want to build this..the best place for bits if from here www.rsaustralia.com Quote Link to comment Share on other sites More sharing options...
audioguru Posted June 16, 2004 Report Share Posted June 16, 2004 Stuee,RS seems to sell only stuff that I haven't used for about 30 years, like old-fashioned standard TTL.Don't you have good English suppliers such as Mappin or Farnell over there?Any members from "down under" to help him out? Quote Link to comment Share on other sites More sharing options...
Kevin Weddle Posted June 21, 2004 Author Report Share Posted June 21, 2004 Referring to the original circuit found on the home page. I can see that you are using the 5volts respective to 9volts through the infrared diode. I don't think this is a reliable situation. I could go into detail as to why, but I will leave it at that unless you are interested. The signal, by the way, may not care that the DC voltages have run amuck because of referencing a voltage through an infrared diode. Quote Link to comment Share on other sites More sharing options...
stuee Posted July 17, 2004 Report Share Posted July 17, 2004 Yet again my neck of the woods prevents me from making things like the rest of the world, Could someone please spent 5 mins and tell me what i can use in place of the hs series used, i seen on my supplier (only one available for 1000s of ks) is the LS or HC series?Please help before my wife tips over the network cable to my laptop in the lounge and breaks her neck. Quote Link to comment Share on other sites More sharing options...
cristopher Posted August 3, 2004 Report Share Posted August 3, 2004 serge,what are the advantages of this project? and the limitations? Quote Link to comment Share on other sites More sharing options...
harsh Posted August 3, 2004 Report Share Posted August 3, 2004 stuee,The HS/HST series are fabricated on a 2 micron advancedCMOS process which enables AC switching performancecomparable to the standard HC/HCT logic families. However,HS/HST series have reduced output drive whichyields significantly lower dynamic power consumption thanHC/HCT.i got this off of http://www.fairchildsemi.com/ms/MS/MS-503.pdf ...page 4...i would say yes ,you can definatly use the HC part as a replacement for the HS part... Quote Link to comment Share on other sites More sharing options...
audioguru Posted August 3, 2004 Report Share Posted August 3, 2004 Why aren't modern parts available in Aus?Just dig a channel in the dirt floor of your cave and bury that dangerous cable! (Get kangaroos to do the digging) Quote Link to comment Share on other sites More sharing options...
serge_saati Posted September 13, 2004 Report Share Posted September 13, 2004 Hello, the advantage is the wireless anand the limitation is the distance range. Quote Link to comment Share on other sites More sharing options...
ali_abbass85 Posted April 21, 2005 Report Share Posted April 21, 2005 hello everybody can anyone tell me what is going on i cant find where did u use the 9v in the first circuit ,and what is the filter in the circuit (is it the circle around the diode),and i only found the 74hc04 and 74hc08 ics. Quote Link to comment Share on other sites More sharing options...
audioguru Posted April 21, 2005 Report Share Posted April 21, 2005 Hi Ali,Welcome to our forum. ;DThis is a confusing project, isn't it? ??? Quote Link to comment Share on other sites More sharing options...
ali_abbass85 Posted April 21, 2005 Report Share Posted April 21, 2005 :)thx audioguru :) ,ya it is confusing ??? but i really want to make this circuit i need some help from u guyz 1- i found the 74hc series in my country ,is it ok to us them .2- what is the range of this device .3- i cant find the 200mhz crystall so i will try to use the frequency multipier with 50mhz.thx for ur help Quote Link to comment Share on other sites More sharing options...
audioguru Posted April 22, 2005 Report Share Posted April 22, 2005 Hi Ali,The circuit has changed a lot since last year. It had a 9V battery.1) The author says it won't work with 74HC series because they are too slow. Other people say 74HC is fine.2) I don't think it will work. Its photodetector diode is backwards.3) I have never heard of crystals that operate at such a high frequency, even using overtones. Maybe it doesn't need a crystal oscillator and the unsyncronized gate. Quote Link to comment Share on other sites More sharing options...
ali_abbass85 Posted April 22, 2005 Report Share Posted April 22, 2005 dear audioguru if it is not a crystall then wat is the 200mhz in the transmitting part of the circuitplz relpy fast Quote Link to comment Share on other sites More sharing options...
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