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HarryA last won the day on January 5

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About HarryA

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  1. 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.
  2. 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.
  3. HarryA

    Adblock Plus?

    Is there a way to block listings containing Realtek and Ameba using Adblock Plus? I added them to the filter list but it did not work.
  4. That looks good. Good use of the capacitor. If you replaced the DPDT toggle switch with a relay powered of the supply on power failure would it switch to inverter automaticlly?
  5. How fast is an MLX90614 IR temperature sensor? https://www.electronics-lab.com/project/infrared-thermometer-arduino-mlx90614-temperature-sensor/
  6. 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. .
  7. 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.
  8. HarryA


    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
  9. It looks similar to: y[n]=0; for(i=0; i<n; i++) y[n] += x * b; see: http://pdfserv.maximintegrated.com/en/an/AN3386.pdf What language is that? Never seen (long long) before. Perhaps like a double in C ?
  10. HarryA


    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).
  11. HarryA


    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?
  12. Without a diagram of the circuit it is impossible to help you in spite of the good photographs. It reminds me of the game tossing set mouse traps between players 🤪
  13. Help NASA explore the moon and win up to 30,000$. See: https://www.herox.com/crowdsourcing-projects/government
  14. 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.
  15. What you want to know is will the capacitor charge to above 1/3 of Vcc within the period of the 555 pulse; thus it will not get re-fired. Using: You get 405 microseconds, See: http://mustcalculate.com/electronics/capacitorchargeanddischarge.php And will discharge to 4 volts from 12 volts through the 50k resistor in 5.6milliseconds.
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