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Posts posted by HarryA

  1. There is  information here on Visual Servoing that maybe helpful;  particularly  the numerous references at the end of the article.


    If you look at a bmp file of a smd image like this one (I could not find an image of the bottom of a smd).

    You can see the white background as FF FF FF and one scan line across the component highlighted.  So in theory one could  find the four corners and calculated the rotation and any x y offset. In bmp files the image is inverted.

    If you only need the rotation then only part of one edge would be required to calculate it I believe.

    smd compoint.png



    smd compoint.bmp

  2. I am confused by the circuit. If -v is grounded at connector CN1 then it is grounded at connector CN2 also. If so then capacitors C18,C19, C20, and  C21 are grounded on both ends?  Thus there would be  no 1/2 voltage and +v would be at full voltage off the bridge rectifier?

    I must be reading the schematic wrong.


  3. the circuit below works in the LTspice simulator and as a prototype:


    Traces from the simulator:


    The green trace is the input from the sensor, the yellow trace is at pin 2, and the red trace is the output at pin 3. Note the positive going pulse when the input drops back to zero. That is why the Zener diode is required to limit it. Here it is  a 15v Zener diode.

    The output from prototype the circuit: Here I am using 9 volts as I only have 12 volt Zeners.


    Here the blue trace is the trigger signal at pin 2 and the yellow trace is the output at pin 3 across a 150 ohm resistor for a relay load. The diode across the relay coil can be any general purpose diode like a 1N007 for example.  The output is much wider then I would expect for R4 at 1Megohms and C3 at 1 microfarad. Expected 1 sec.  not 1.4   There  are numerous online calculators for calculating these values.  

    The values for the resistors are not critical except what you pick for R4:

    • R1: 1000 to 1500 ohms
    • R2; 20k   to  30K   "
    • R3: 30K  to  50K   "

    all work in the simulator.



  4. The problem with the 555 timer is that it is not edge triggered and it is retriggerable. You would like a one-shot that is rising edge triggered. Retriggerible means as long as  a moose stands in front of the PIR the 555 would output a pulse; having inverted the output with a transistor. Thus the relay would stay energized. That is okay  if you want to  make a movie of a moose but not good for one  photo.

    The 74121  is a nonretriggerable and is edge triggerable, that would be a better fit.

    I will look at your circuit to see if it be used.  I am thinking perhaps one could invert the PIR output with a transistor and discharge a capacitor to momentarily  pull trigger pin 2 low.

    What is the resistance of your relay if you already have one?  One needs to know how much current it would draw.



  5. You need to create a pulse that goes from low to high and back again to low. Like the first pulse B:


    If you connect a resistor ( say 10k perhaps ) from the pulse in  line to ground that will hold  it low. Then connect your switch from the pulse in line to the Vcc the supply voltage. When you switch on the switch you create the rising edge like pulse B above. See line 4 below. When you open the switch you get the trailing edge of pulse B above. That is the transition for line 5 below.  The function table may differ from this one for the one-shot you use.  Remember if your switch is closed longer than the output pulse you create be sure to use a nonretriggerable chip. Analog makes chips of this type also: the LTC6993 family of one-shots.




  6. The circuit below uses edge triggering so you get an output on the rising edge and the falling edge of a pulse. The SN74LVCIG123 is a surface mount component but there is a throw hole chip; the 74123. The 74123 is a retriggerable monostable multivibrator (one shot).  Retriggerable means  that  the output persist as long as the input is applied so if you switch closure  is longer than the output pulse it still remains high.

    The 74121  is a nonretriggerable version. 

    If you use a two position selector switch that would  work; selecting ground or high. Else you   may have to use retriggerable monostable to get a real pulse if you use a simple push button; as input to the circuit.


  7. Doing a prototype of the circuit as shown below using a voltage regulator as I have no 5 volt Zeners:

    I used 1N007 diodes and a LM7805 regulator. Transformer output 13.85 Vac 17.3 Vdc at the capacitors.


    What I got for ripple voltage at the capacitor was:

    • 536 mvpp with 1523 mfd cap.  140 mv ac with multimeter
    • 288 mvpp with 3005 mfd cap     71 mv
    •  200 mvpp with 4454 mfd cap.   48 mv
    • 140 mvpp  with 6050 mfd cap    36 mv





  8. The 12 volt transformer supplies a peak voltage of:

    Vpeak = 12 * square root of 2 = 12 * 1.4 = 16.8  

    With 16.8 volts into the bridge we subtract 0.7 volts for each diode. As the current
    passes through 2 diodes for 1.4 volts.

    Vdc = 16.8 - 1.4 = 15.4 without a load.

    The actual loaded voltage will be more likely around 14 volts as we will see later
    with a capacitor and load. I am using 20ma LED here.

    For the capacitor:
    ripple voltage Vr:

    Vr = (0.0024) * I/C  = 100 mv

    C = 0.0024 * I / Vr  =  0.0024 * (0.120A / 0.10V)  = 0.00288 F

    0r: C = 2880 micro-farad

    from Schertz and Monk "Practial Electronics for Inventors" section 11.6

    For a Zener diode
    Lets try the 5.1 volt 1N4733A. As 5.0 volt Zeners are hard to find.

    For the 1N4733A Zener diode the value of Vz is 5.1V and Pz is 500 milliWatts
    now with a Vdc (dc supply voltage) of 14V the value of R2 will be:

    Rs = (Vdc - Vz)/Iz

    Rs = (14 - 5.1)/Iz

    Iz = Pz/Vz = 500mW / 5.1V = 98mA max.

    Thus Rs = (14 - 5.1)/98ma = 90.8 ohms

    Rs = 91 ohms (approx)  91 ohms is a standard.

    The power/watts into R2 will be:
    Pr2 =  V^2/Rs = ((14 - 5.1)^2)/Rs = (8.9 * 8.9)/91 =  0.87 watts or 2 watts in practice.

    So with a load of 120 ma the voltage drop across Rs:

    Vrs = 91 * 0.12 =  10.92 volts ........oops! Looks we only have 14 - 10.92 = 3.08 for
    the load and the  Zener is out of business!

    Lets try a 1 watt Zener and 5.1 volts:

    Iz = 1.0  watt / 5.1 volts = 0.196 A or 196 ma.

    Rs = (14 - 5.1)/ 0.196A =  45.4 ohms or 47 ohms standard

    Voltage drop across Rs at full load:   Vrs = 47 * 0.12 = 5.4  volts;

    Or 14v - 5.4v =  8.6 volts for the Zener to chew on.

    The power/watts into R2 will be:
    Pr2 =  V^2/Rs = ((14 - 5.1)^2)/Rs = (8.9 * 8.9)/47 =  1.685 watts or 2 watts or better in practice.

    Next we can run this through the simulator and see what it looks like. (to be continued).

  9. Are you  interesting in designing a circuit using mathematical equations or are you interested in building a power supply using proportional engineering (as a 6.3 volt filament transformer should work)? Proportional engineering is the  way I build most things.   Are you looking for help here?

  10. Two things from your circuit are the 20k ohm potentiometer and the 10k resistor at pin 2.

    The potentiometer going from 0 to 20k ohms can produce a wide range of outputs in time/widths. So the RC circuit must charge above 4 volts fairly fast to accommodate the shortest output and while charging through the 10k resistor. A 1.0 ufd would require 20 ms, perhaps to slow. So trying a 0.1 ufd gives 400 microsecond which would seems fast enough(?).

    To discharge the capacitor before the next switch closure the  parallel resistor must set the voltage at pin 2 above the required 4 volts while the switch is closed. A 50k sets the the voltage at: 50K/(50k+10K) * 12volts or 10 volts a bit higher then necessary. A 10k resistor would set the voltage at 10k/(10k+10K) * 12 volts or 6 volts which would be adequate. Discharging from  6 volts to 4 volts in 2 ms after the switch is opened.

  11. In the photo the yellow trace is pin 2 (trigger) and the blue trace is pin 3 (output). Starting on the left pin 2 is at the Vcc voltage of 12 volts (0 is at the bottom of the screen) . Pin 3 is at 0 volts. On switch closure pin 2 drops to  near 0. And pin 3 rises to near 12 volts. The switch is held on until the later part of the trace on the right most side of the screen. DSCF3681.JPG.1f6d97a0c484b521e759b3b7f7f4205c.JPG

  12. Try putting a resistor and capacitor in parallel after the switch. The capacitor will pull pin 2 down until it charges through the upper  10k resistor. The resistor in parallel   with the capacitor will discharge the capacitor during the period the switch is open.

    Do you need help with the calculations? As a quick test make the new resistor 50k and the capacitor perhaps 0.1ufd - I am just guessing. You want the capacitor to discharge before the output ends.


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