need help in 4N26.

Discussion in 'The Projects Forum' started by skychan1213, Mar 26, 2009.

  1. skychan1213

    Thread Starter New Member

    Jul 10, 2008
    9
    0
    Dear everyone.thank you for open my question. i have a problem in my project. the diagram is shown in the jpeg picture.my problem is about the 4N26. as shown in the picture, when the input of the full wave rectufier is 8VAC single phase, the output of the 4N25 are very low which is not detectable by the hc11 microcontroller. after i increase the input voltage to 16VAC, the output value of the 4N26 are still remain same or just increaly very slightly. when i remove the AC supply to the rectifier, i get the same value at the output 4N25. how could this can happen? since the AC supply already removed? how to solve it? what i want in my project is when the input of the rectifier increase, the output of the 4N26 also will increase. meaning input AC source of VAC should propotional to the output of the 4N26. the 5 V supply is supply from the microcontroller, and the range which the microcontrooler can acceprt from the output 4N26 is form 0 to 5V. any mistake in my circuit? the first ua741 and TL081 are use a peak detector to detect the peak voltage form the full wave rectifier. very thank you for your solution.
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    The transistor in a 4N26 should be connected with the emitter to ground and connected to Vcc through a resistor. The transistor is fairly weenie, so use a value of 47 - 100K for the resistor. The signal at the collector should switch between ground (saturation) and Vcc (off).

    You will need a high impedance input/amplifier to buffer the output of the 4N26.
     
  3. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    There is no discharge path for C1 except for leakage current; once it is charged, it will take a long time to discharge. Consider a 220k Ohm or so resistor to ground there.

    Consider changing the ratio for R5/R6. Right now, U2 is basically a voltage follower.
    Try 50k for R5, and 10k for R6.

    Consider using a better opamp than a 741. LF353, TL082, etc.
     
  4. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    656
    I believe a phototransistor such as that found in the 4N26 will saturate just as well with the load on the emitter as it will with it on the collector.
     
  5. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    656
    The optocoupler needs an antiparallel diode (e.g. 1N4148) across the LED to protect it from reverse breakdown.
    I would like to know what this circuit is supposed to do. If it is intended to be a peak rectifier with an isolated analog output, it will not be predictable.
     
  6. skychan1213

    Thread Starter New Member

    Jul 10, 2008
    9
    0
    Mr. here is thecircuit what is supposed to do. hope you can help me to solve it. thank you
     
  7. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    656
    OK, I looked at your methodology. The big problem I see is that you are passing an analog value through an optocoupler. An opto's transfer function will vary with temperature. I would move the microcontroller into the peak detector block, with the output of the peak detector going directly into the ADC. Use four optos to couple from the μC output to the triac block. You won't need op amp U2. You will need to increase the attenuation between the FW bridge and the peak detector.
    You might want to consider either using a precision FW rectifier (op amps and diodes) instead of a FW bridge and a precision peak detector (as you have now), or just feeding the output of the FW bridge directly to the ADC, with appropriate filtering and attenuation. The bridge, with an added capacitor, is a peak detector.
    Regarding your question about a resistor across C1: C1 will probably self-discharge, and also leak back through D2, but it will be a very slow process. I would change C1 to a nonpolar cap (1uF or 100nF), and add a parallel resistor to control the discharge time constant. The time constant controls the response time to changes in input voltage. With a long time constant, it will take longer to detect a decrease in input voltage. With a short time constant, the response time will be faster, but you will get more ripple into your ADC. A filter can remove the ripple, at the expense of more delay.:(

    I'm curious about what sort of load this circuit is intended to drive. Can you tell us?

    EDIT: How about feeding the attenuated output from the bridge directly to the ADC, and doing the peak detection and filtering in software? Or feeding the transformer output (level shifted and attenuated) into the ADC, and doing the peak detection and filtering in software?
     
    Last edited: Mar 29, 2009
  8. skychan1213

    Thread Starter New Member

    Jul 10, 2008
    9
    0
    To Mr, sorry for late reply. actually the load is a AC varible resist0r which use to connect to the tapping transformer. thank you for your suggestion. may i ask you some more question again? as look for my diagram, if i send a signal to the first moc3041, will the triac turn on and supply voltage 110 to load? i cant run it by harware. thank you.
     
  9. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    656
    According to the datasheet, your input current needs to be at least 15mA, for pulse widths >100us. For narrower pulse widths, the input current has to be higher. See fig. 6 in the datasheet. With 5V in and 300 ohms, as in your schematic, you will only get about 12mA.
     
Loading...