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HarryA

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HarryA last won the day on November 13 2020

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About HarryA

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  • Birthday June 23

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    USA, Pennsylvania, Susquehanna County, Forest Lake Township
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    Canoeing, woodworking, electronics, and gardening.

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  1. Step 1: Are you committed to using to using the TC4420 and 40193? Consider using a circuit like the one below. The direction of current flow through the relay coil depends on which transistor is on. Also if you do not have the relays yet these may simplify your design: https://www.amazon.com/SMAKN-Trigger-Switch-Latching-controlled/dp/B01CN7V0F6/ref=sr_1_20?dchild=1&keywords=latching+relay&qid=1617978673&sr=8-20 What are you using of inputs to the circuit?
  2. There are a couple of videos on Youtube you may find helpful. The first one uses a socket with colored wires while the second one shows the terminals on the back of the switch. In the first one he connects directly to 12 volts while in the second one he claims the LED is 3 to 6 volts and uses a resistor to connect to 12 volts. If you need help connecting to the board I will look at its specifications. https://www.youtube.com/watch?v=nlPj0VL_bGU https://www.youtube.com/watch?v=zd3n8pYFx1g
  3. That is a complex beast! Given: "Each LED output can be off or on (no PWM control), or set at its individual PWM controller value. The LED output driver is programmed to be either open-drain with a 25 mA current sink capability at 5 V or totem pole with a 25mA sink, 10 mA source capability at 5 V." If you are using the PWM control the meter would not show the peak currents to the LEDs. Other then that it is difficult to help troubleshoot something that requires programming.
  4. As the TC4420 accepts logic level inputs and the 40193 output is a logic level why is anything needed between them? Also both only require a single voltage supply (Vdd) for the circuit shown a dual supply would not be required. There are single Vdd supply op amps if you need one. I have a circuit that uses a TC4420 driven directly via a Arduino Uno output.
  5. Are pinball schematics any different than electronic schematics? https://learn.sparkfun.com/tutorials/how-to-read-a-schematic/all
  6. I gather you are looking for a circuit like this one. Do you have any idea what the currents are? Perhaps one could key off the change in voltage at the battery; low voltage turns a transistor on to power a relay.
  7. There is something about pulse transformers dealing with the pulse width. If the input pulse width is two short the current never has time to build up completely do to the lagging current in an inductor/coil. That would look like an impedance mismatch also. If the pulse is to wide current is wasted when the coil is fully energized. For example see https://www.youtube.com/watch?v=vvXbTIqBY4o
  8. Yes you can use a battery as long as it is 6.0 volts. How long it will last will depend on how much current the camera draws. If you are using 4 AA batteries there is good information here: https://www.powerstream.com/AA-tests.htm
  9. Your getting just 3 volts suggest an impedance mismatch between the transformer and the cell. In looking at using automobile ignition coils for high voltage I see a number of videos and articles where they get huge sparks from the coils that I can not get. I feel like you; where is the spark! I am thinking my coil is a dud! See for example: https://www.youtube.com/watch?v=PTt5sM3moqQ There are a number of similar videos there using ignition coils for high voltages.
  10. We know the output voltage is related to the ratio of the number of turns. And the output current related to the inverse of the number of turns. The more voltage gives one more current at the primary so in an ideal transformer the power is the same on both sides of the transformer. But in a pulse transformer is it not the rate of change of current in the primary that gives rise to the change in magnetic flux that produces the voltage? I need to learn more about pulse transformers. I had in mind trying out different frequencies on the automobile coil/transformer but had a problem when
  11. Using the 100 watt bulb and two different ignition coils this is what I got: AC Output(VOM) 1 meg load Input sig. 204 Hz Peak voltage across 7ohms coil type 67.4 v 6.0v 1.84 v std auto ignt. coil 159.8 v 9.68 v 5.44 v 17.4 v 8.0 4.0 - 4.2 v Capacitor discharge coil The std auto coil is a 12 volt
  12. Don't laugh that is an emulation of your circuit. I was thinking that there is not enough load in the circuit with only the coil (the mosfets are acting like switches only off or on) so I started out with 7 watt incandescent lamps but could not get two mosfet to work so I switched to only one; that works well. In the photo at the lower right is an automobile ignition coil and the big black thing is an 80 volt battery (borrowed from my chainsaw). Working up to 13 bulbs (normally 7 watt bulbs) I got 0.571 ma through a 7 ohm resistor. Switching to a 100 watt bulb got 0.594 ma, the IR
  13. I was using the fixed resistors as the simulator dos not have a concept of a potentiometer. Putting the 530 back in an using a combination of two resistors between 30k - 17k to 46k - 1k (where the lower value is to ground) I get good results with 40k - 7k and 30k - 17k but the peak current is 2 amperes for some reason. At 44k - 3k the current is 900ma but the output is down to 50v. I will continue looking at it. If you have a symmetrical pulse waveform say of 1 ampere I gather the rms value of the current is 1 * 0.707 and the rms voltage would be 100 * 0.707 so the equivalent pow
  14. I would think you are not hitting it with enough current. Using a resistor instead of the IRF530 as it is faster to use rather then dorking around with two resistors to simulate a pot. I am using a input pulses that 2.5 ms on and 2.5 ms off at 6.8 volts peak No bifilar coil nor diode in the output: resistor current output capacitor output resistor output voltage 100 1A 1.058 nf 1.16 meg 720 volts 100 1A 0 1.16 1.05 kv 50 2A
  15. New section: For R1(max). Assuming a 8 volt transformer. Vin = 8.0 * 1.4 or 11.2 Taking the min and max to be plus or minus 10%. One would not need a voltage regulator if there where nothing to regulate. Vmin = (Vin - 10%*Vin) - (2 * voltage drop across the two diodes). Typically 0.7v each. And Vmax = Vin + 10%Vin - (2 * voltage drop). R1max = (Vmin - Vz)/(Izmin + Ib) Izmin is found from the zener data sheet; typically 10 ma. What happens at Vmax? R1max = (Vmax - Vz)/(Izmax + Ib) Where Izmax is the only unknown here. Izmax must be within the maximum curr
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