HarryA

Most likely it is the manufacturer's part number. Perhaps you can find a schematic diagram?

When the switch is closed in the debouncing circuit the input to the inverter is a logic 0 and the output a logic 1. Current flows from the battery supply of the inverter to its output. But the absolute maximum current rating for the output of a 74HC14 is only 24 ma. Do you really need a debouncing circuit? I wonder if you could replace the mosfet and debouncing circuuit with a 3 position switch; An On-Off-On. If not you may need a transistor between the inverter and the load. But you get a lost of 0.5 or 0.7 volts. Also the 100mfd capacitors are quite large for you circuit; charging the one through 2000 ohms will take awhile. Perhaps 1.0 mfd?

If you search for Arduino function or signal generator you will find some useful information. Like: https://circuitdigest.com/microcontroller-projects/arduino-waveform-generator

The URL is: https://www.youtube.com/watch?v=bIz9ektLunE Following the link from youtube does not lead to anything useful. The code has numinous errors. It would never run, for spuddo: This is a snapshot of the serial monitor displaying the output of the code below. This is the circuit that I used. The duty control on the left. The two resistors limit the range of the output. You do not want zero duty cycle for example. The period control on the right. The resistor limits the period from going to zero for example. The code: // // A rework of the code from the video: https://www.youtube.com/watch?v=bIz9ektLunE // void virtual_Scope(); int iDutyCyclePin = 2; //set input pin for the left pot = duty cycle int iPeriodPin = 1; //set input pin for the right pot = peroid int iLedPin = 13; //ser pin for output to LED //float fFrequency; //frequency = how many full periods per second or per 1000 millisedonds float fPeriod = 10; //value from the left pot float fDuty = 1; long int liPeriod, liDuty; int iDutyPercent; long int liHigh_Time, liLow_Time; void setup() { pinMode(iLedPin, OUTPUT); //declare ledpin as OUTPIN //Serial Port begin Serial.begin (9600); } void loop() { fPeriod = analogRead(iPeriodPin); //read the value from the right pot = period fDuty = analogRead(iDutyCyclePin); //read the value from the left pot = duty cycle //limit the range of the duty cycle to 20 to 80 percent of the period because that is what works! fDuty = (fDuty*9/10000) * fPeriod ; //with 5 volt supply liDuty = int(fDuty); //for printouts below liPeriod = int(fPeriod); //iDutyPercent = 100*liDuty/1024; //cal duty cycle as percent for serial monitor ???? iDutyPercent = 100*liDuty/liPeriod; //cal duty cycle as percent for serial monitor // fFrequency = (1000/float(liPeriod)); //cal frequency in units of Hertz for serial monitor ??? // liHigh_Time = liDuty*liPeriod/1024; //cal high = on time ??? liHigh_Time = liDuty; // high = on time liLow_Time = liPeriod - liHigh_Time; //cal low = off time Serial.println(); //new line Serial.print("period "); Serial.print(liPeriod); Serial.println(" milliseconds"); Serial.print("duty "); Serial.print(liDuty); Serial.println(" milliseconds"); Serial.print("on time "); Serial.print(liHigh_Time); Serial.println(" milliseconds"); Serial.print("off time "); Serial.print(liLow_Time); Serial.println(" milliseconds"); Serial.print(" duty cycle "); Serial.print(iDutyPercent); Serial.println(" % "); // Serial.print("frequency "); //what is frequency here? // Serial.print(fFrequency); // Serial.println(" Hertz"); digitalWrite(iLedPin, HIGH); //set the ledPin on delay(liHigh_Time); //pause the program for high_time in microseconds digitalWrite(iLedPin, LOW); //set the ledPin off delay(liLow_Time); //pause the program for low_time in microseconds virtual_Scope(); //display trace delay(4000); //slow down the scrolling } void virtual_Scope() { //display a line of pulses int iHighDashes, iLowDashes; int iCount; //let one dash = 30 ms iHighDashes = liHigh_Time/30; iLowDashes = liLow_Time/30; iCount = 60/(iHighDashes+iLowDashes); //try to make the lengths similar // Serial.println(String(iCount) + " "+ String(iLowDashes) + " " + String(iHighDashes) ); //new line for space at top Serial.println(); //space at top while (iCount--) { for (int i=0; i <iLowDashes; i++) { Serial.print("_"); } for (int i=0; i <iHighDashes; i++) { Serial.print("-"); } } Serial.println(); //space at bottom }

Given what he shows in the video one can emulate the code easily. If that is what you would like.

I gather one would need a microcomputer to interface one with the other. I take it the output one is not compatible with the input of the other?

or cleaner: String DisplayString = " "; //a blank string int percent1 = 0; int OldPercent = 0; void setup() { // Serial.begin(9600); //testing } void loop() { if( OldPercent != percent1 ) //do not scroll needlessly { //previous string - set the cursor to second line LCD.setCursor(0,2); LCD.print(DisplayString); //move previous line down //set the cursor to for new first line LCD.setCursor(0,1); if (percent1 > 50) { //new string DisplayString = "Moisture S1 : " + String(percent1) + "%" + " Irrigaion sistem ON"; } else { //new string DisplayString = "Moisture S1 : " + String(percent1) + "%" + " Irrigaion sistem OFF"; } //set the cursor for new first line LCD.setCursor(0,1); LCD.print(DisplayString); //new line to top of the LCD OldPercent = percent1; //set new percent } } //end loop the first "set the cursor for new first line" code is redundant; I can not see how to edit the code

Perhaps: String DisplayString = " "; //a blank string int percent1 = 0; int OldPercent = 0; void setup() { // Serial.begin(9600); //testing } void loop() { if( OldPercent != percent1 ) //do not scroll needlessly { //previous string - set the cursor to second line LCD.setCursor(0,2); LCD.print(DisplayString); //moves previous line down if (percent1 > 50) { //new string DisplayString = "Moisture S1 : " + String(percent1) + "%" + " Irrigaion sistem ON"; //set the cursor to for new first line LCD.setCursor(0,1); LCD.print(DisplayString); //new line to top of the LCD OldPercent = percent1; //set new percent } else { //new string DisplayString = "Moisture S1 : " + String(percent1) + "%" + " Irrigaion sistem OFF"; //set the cursor to new first line LCD.setCursor(0,1); LCD.print(DisplayString); //new line to top of the LCD OldPercent = percent1; //set new percent } } } //end loop

"Could a capacitor smooth the frequency? " If one does not know what is wrong with the system then one can not know what the affects of a capacitor would be. you have a 12 5kw diesel generator the generator outputs 120 or 240 volts at 50 or 60 hertz. the generator supplies the mains that has numerous outlets on the boat. the generator output is stable until you turn on the ac unit. with the ac running the voltage at the various outputs are no longer 50 or 60 hertz but varying in frequency 6: adding any load to an output corrects the problem. Is that correct?

48 volts is above the absolute voltage for this voltage regulator: 35 volts maximum:

In that type of air conditioner the input ac voltage is converted into dc via a power supply. The dc voltage is converted (inverted) into variable frequency ac. The frequency depends on the demand on the unit. A microcontroller samples the ambient air temperature and adjusts the speed of the compressor motor via the frequency of the ac applied to the compressor motor. Invert versus convert is interesting: Inverter - dc to ac converter - ac to ac (transformer, etc) converter - dc to dc (potentiometer, etc) power supply - ac to dc

see: http://starseedsportal.org/tag/stephen-dickens-magnesium-water-copper-battery/ plus other sites; search on "stephen dickens water battery" Also see Voltaic pile at: https://en.wikipedia.org/wiki/Voltaic_pile

I have not used varistors but not letting ignorance slow me down. The V20E14AUTO is a 16 volt MOV. Voltage Rating DC: 16vdc Varistor Voltage: 22 volts this is where the varistor starts to conduct at 1 ma. Varistors draw some small current always Clamping Voltage: 43 volts the voltage any spike is truncated to. Peak Surge Current: 20 amperes. Surge Energy Rating: 28 joules. 28 watts-1 second, 56 watts-500 ms, etc Unlike Zener diodes that fail as an open circuit varistors fail as a short circuit so in some cases a fuse is used with them. Most likely on the larger ones.. see: https://www.mouser.com/ProductDetail/Littelfuse/V20E14AUTO?qs=%2Fha2pyFadugE%2FoOLoA7J9TCvi67Mtc%2F%2FOVMAqZDk45RR%2F7GD1QKNkA%3D%3D or: https://www.digikey.com/products/en?mpart=V20E14AUTO&v=18 That is my best guess; perhaps someone else has experience with varistors? "The MagLock Draws 420mA 2 12 Volts, The panels power supply provides 500mA @ 14.3 V " Do you need to reduce the 14.3 volts to 12 volts?

Wikipedia has a well written article on operational amplifier applications. see: https://en.wikipedia.org/wiki/Operational_amplifier_applications#Non-inverting_amplifier

There is information here on Visual Servoing that maybe helpful; particularly the numerous references at the end of the article. https://en.wikipedia.org/wiki/Visual_servoing 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.bmp

You may find some useful information here. If you need more help come back here. https://github.com/openpnp/openpnp/wiki/Setup-and-Calibration%3A-Bottom-Camera-Setup I worked for a company that made PnP and Through-hole machines but I only worked on the Through-hole machines.

what is the replacement for the d13005ed? Perhaps the MJE13005G ? The data sheet is online. The MJE13005G from China see: https://www.ebay.com/itm/5PCS-X-UTC-MJE13005G-C-TM3-T-TO-251-I-PAK-NPN-Vceo-700V-Ic-4A-Transistors/333530195913? hash=item4da7f2e7c9:g:9MgAAOSwr9ReWXce The D23005Ed 20 of them! "20 x D13005ED Transistor TO-220" from China. https://www.ebay.com/itm/20-x-D13005ED-Transistor-TO-220/184022252029?hash=item2ad894b9fd:g:cLQAAOSwOFZdwBIz also see for large list of similar transistors: https://alltransistors.com/crsearch.php?at=Si&struct=NPN&pc=65&ucb=700&uce=400&ueb=9&ic=4&tj=150&ft=4&hfe=15&caps=TO220

If it is worth 21$ to you. You can always sell it on ebay when you are done with it 😉 https://www.amazon.com/Service-Technical-Information-Compact-Digital/dp/B00KV315WI It's 29.99$ on ebay https://www.ebay.com/itm/Sharp-dx-110-service-manual-original-repair-book-stereo-cd-player/123711246781?hash=item1ccdc3d1bd:g:8RoAAOSwk~NZ2OdD

If you open the unit up and find out what is wrong with it that will give you insight into what the problem is/was. If the problem is on the AC input side it may well be, as you say, a spike from the power source.

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.

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.

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.

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?

How fast is an MLX90614 IR temperature sensor? https://www.electronics-lab.com/project/infrared-thermometer-arduino-mlx90614-temperature-sensor/

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. .