Jump to content
Electronics-Lab.com Community


  • Posts

  • Joined

  • Last visited


Posts posted by nickagian

  1. Hi all!

    I am confused a little bit with something related with the IC 74LS07.

    This IC has buffers, isn`t that correct? So let`s choose the first gate and apply +5V at the input (1A). I expect to get +5V at the output of the gate (1Y), right? So why is that not working with my chips? ??? (I have already checked two of them).

    I cannot understand what is my mistake... :(

  2. Hi defaced!

    Well this is what I have learned in my university:

    1)At a half-rectifier circuit (look at the first two circuits at the picture posted by Ante) the output voltage amplitude is given by the formula Vout(amp)=E(rms)xsqrt(2)-VD0

    [sqrt->square root, E the voltage at the secondary of the transformer and VD0-> the forward voltage of the diode, about 0.7~0.8V]

    Since sqrt(2) is approximately 1.4 you can say that Vout(amp)=1.4E(rms)-VD0
    so roughly Vout(amp)=1.4E(rms).

    This is one of the formulas that you can see in the picture.

    Moreover the DC output voltage (practically the average voltage) can be calculated by the formula

    VDC=1.4E(rms)/pi  -  VD0/2

    and since 1.4/pi is approximately 0.45, you can say that VDC=0.45E(rms)

    (That means probably the second formula at the picture)

    However if you use a simple capacitor for filtering, the output voltage roughly equals the output voltage amplitude, thus 1.4E(rms).

    2)For the full-rectifier circuit (both with the two diodes and the bridge-rectifier) the output voltage amplitude equals

    Vout(amp)=2*1.4E(rms)-VD0 thus approximately 2.8E(rms)

    and the DC level

    VDC=2*1.4E(rms)/pi - VD0 so approximately 0.9E(rms)

    That`s all!  8) ;)

  3. Moreover I want to add that my idea is to use two couples of sensors, (perhaps IR emmiter/receiver is a very good solution, as StathisCy also said) and put both of the couples at the door with some distance between them. In such a way you can also check if somebody is coming in or out..e.g. when couple #1 gives signal first and #2 afterwards, that means that somebody is coming in and when couple#2 gives signal first, that means that somebody is coming out!

    I have  made something like that in the past, but only checked it with some switches instead of the sensors... If you want some help with what IC to use and in general about the hardware you can ask further!

    Just to give you an idea, I had first designed the appropriate ASM diagram and according to that easily found what and which ICs to use. For counting I had used synchronous up/down BCD counters (SN74LS190) and to display the number of people, some 7-segment displayers (SA52-11HWA) with BCD-to-7 segment decoders (DM74LS47), so as to connect them directly to the counter`s outputs. And of course some flip-flops are needed too.

    I hope I have helped you...

  4. I have asked for a ceramic resonator at about 10MHz for my microcontoller project and I was given this one...the only thing that is written on it is "L10.7A",which I guess shows the frequency, 10.7MHz...that is all I know about it! Oh and it has a red dot above one of the side pins..

    It has an input, an output and a ground. The input and output pins can be swapped, and the center pin is probably the ground pin.

    But thanks anyway, because that is what I was looking for... I didn`t know which pin is for the ground...
  5. Hi Electrical_tech_Student!

    When you use the formula Acm=(Aol/CMMR), all the quantities have to be pure numbers without dimensions (not dB), so you cannot use CMMR=76dB and Aol=81000 at the same time! You have to convert 76dB into 10^(76/20)=6300, since CMMR(in dB)=20log10CMMR. Then you can use this number with the above formula!

    Another solution is to convert Aol in dB and use the formula Acm(in dB)=Aol(in dB)-CMMR(in dB) and then you find Acm in dB.

  6. Thanks a lot Zeppelin! That resistor must be my mistake (I had thought of it, but couldn`t find the proper value). The guide is very helpful on that too!

    With a 5V supply you can't drive that many LEDs, tops 3, I guess-you need a higher supply voltage.

    I know that, I intend to use only two LEDs on series and organize all of them on doubles!

  7. OK Nick, I did misunderstand your question!

    It`s ok! Perhaps I didn`t express myself very clearly and I should have posted that picture too!

    Well I simply triggered it by connecting it to the ground or to the positive supply (later at my real circuit this will be connected to the output port of a PIC, but I guess it is the same thing!)

    I don`t know, I may have made a mistake... I will check it once more!

  8. Hi Nick,

    Why not use the LM317 or LM350?

    Well... I haven`t understand what you mean Ante... :-[ I`m sorry...

    How can the LM317 be used as a current sink?

    Actually I mean something like the following circuit (except that instead of 1 I want to use more LEDs), where the ON or OFF state of the LED can be changed by the voltage applied at the base of the transistor.

    But I obviously make a mistake because it simply doesn`t work at all!


  9. Hi everybody!

    I want to design a project with some LEDs, something like a signboard and I need to use a current sink, which can drive up to approximately 500-600mA. I have thought of using a NPN transistor (for example 2N2222),  but I do not know what the exact circuit should look like. I have simply connected the E to the negative supply (ground), the C to the negative lead of the LED (of course the + of the LED is connected through a resistor to the positive supply) and the B to ground, but it doesn`t work at all!

    Can anybody help me? Perhaps is there another solution, apart from using a transistor? (an IC maybe?)

  • Create New...