HarryA

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.

Are pinball schematics any different than electronic schematics? https://learn.sparkfun.com/tutorials/how-to-read-a-schematic/all

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. Edited: The circuit needs a momentarily open push button switch between the battery and the relay coil to switch if off.

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

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

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.

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 I connect the second probe to the current monitoring resistor while the first probe was floating on the output of the coil. The 100 watt lamp when full brightness and it when down hill from there. I will continue later. Anyway I had put a voltage divider across the output of the auto. coil (a 1094k and a 9.84k) and got + and - 32.8 volts peaks on the scope at the 9.84k. I calculate the output voltage at +3679 volts and -3679 volts peaks. DUT device under test?

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 and a few amperes coil. While for the CDI coil I am running it via discharging a 1 mfd capacitor charged to 350 volts in my experimental chain saw ignition system. So neither are designed for what you are doing. Speaking of coils/transformers Scherz and Monk have information on transformers in their "Practical Electronics for Inventors" you may check it out the next time you are in a book store. I think using a incandescent lamp (as a ballast resistor) and controlling the current via the input signal is useful but using a mosfet as current regulator maybe useful also. This regulator is from the IRF530 pdf sheet. I use a similar circuit in my ignition system floating a 9 volt battery at 350 volts. The 12 volts at the battery got cut off.

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 IRF530 got up to 32 degrees C. I do not have any IRF580's. The input signal is 6 volts peak. The 50k resistor is a pot. max'ed out. I turn the input up from low to high to prevent any inrush of current into the lamp(s), I do not know if that is necessary or not.

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 power would be: 0.707 * 70.7 = 49.98 watts not the 100v * 1a = 100 watts of a continuous drain. So one may be able to run it above 1 ampere. ref: https://masteringelectronicsdesign.com/how-to-derive-the-rms-value-of-pulse-and-square-waveforms/

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 1.058 nf. 1.16 1.0 kv 50 2A 0 1.16 1.2 kv The current being a symmetrical pulse(?) would be only be half of a continuous draw from the power supply.

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 current for the Zener. If that works out okay use R1max. For C1; using the equation from Scher & Monk's "Practical Electronics for Inventors" section 11.6. Vripple(rms) = 0.0024s * Il /Cf Where Il is the output current and the current through Iz plus Ib. Taking the Vripple = 100 mv or 0.10vrms = 0.0024 * Il / Cf Remember Cf is in farads and Il in amperes. You may expect Cf in the order of 1000uf. Use whatever Vripple you wish. Digital ICs need less than 0.25% * Vout but the zener does regulation also. If you wish I can run whatever values you come up with through the circuit simulator if you do not access to one.

Part one: Starting on the right: the load resistor minimum would be Ro = Vout/Iout. As the transistor is used as an emitter follower the output will be near a voltage gain of one. For the Zener diode and Vout of 6.0v you must allow for the voltage drop Vbe (Vb to Ve) of 0.6 to 0.7 volts. So Vz = Vout + 0.7 volts or a Zener with a voltage at or above wanted Vz. R1 supplies current to the base of the transistor and the Zener. The current input into the base of the transistor can be found using the hFE (the forward current transfer ratio) or current gain of the transistor. Ib * hFE = Io. hFE is in the order of 100 or so. So Ib = Io / hFE; thus Ib adds little to the current through R1. You can calculate the minimum value of R1 given the wattage of the Zener. Iz(max) = watts/Vz. A zener for this type of regulator could be 500 milliwatts. Then R1(min) = (Vin - Vz)/Iz the voltage drop across the resistor divide by the current through it. But what is Vin? Using my rule that the input voltage is 1.5 volts times the Vo or better. Picking a 8 volt transformer (for now). That gives a Vin of no more than 8 * 1.4 or 11.2 volts as the peak voltage. What is there a maximum value for R1? to be continued...

In the simulator (with 100 ohms between the PSU and the coil and 1 ohm resistor in the source lead of the ITF840) the output is about 1.2kv peak with or without the 1.164 meg resistor. With the 1.058nF capacitor it drops to about 750v and with the 5.18nF it drops to about 450v. The current through the secondary coil is +29ma to -42ma peaks. The current through the primary is 0 to 1.8 amperes peak. This is with a 100 ohm resistor between the PSU and the coil. At the 1 ohm resistor in the source lead of the ITF840 it shows 1 ampere peaks. I found an ITF840 spice model. If you put a small resistor in the source lead of the IFR840 you could measure the voltage across it and see if the current is what you expect.

Can you measure the resistance of the cell? Possible it got comtaminated. Also knowing the size of the electrodes and their separation in water one could get a first order approximation of the cell's capacitance.

I must admit the more I look at the circuit the more confused I get. Why the negative rail connects to the TL081 is beyond me. Why it connects to the 30 v rail when it is only rated 18v max is a mystery to me also. Also there are many different up grades to the circuit I gather; for example I have this one I got from somewhere. Note the differences in connections to negative rail and a voltage regulator not the zener diode. Also if you search for "0-30 V PSU not workingy" on the internet you will find this PSU has been beaten to death numerous times on various forums including this one.

What happens when you connect the load? You may try to connect a small capacitor across the output if you have one that has the required voltage rating. The last trace maybe as good as you can get. I will try adding distributed capacitance to the windings to see if that produces similar results in the simulator.

Putting just the front end into the simulator. The labels in large print map back to the schematic. Using a 4.7 volt zener simulator doesn't have a 5.1 zener. In these traces the green is the voltage on the negative rail and the yellow is between D6 and R3. In this one the green trace is between D5 and D6 while the yellow trace is the current through R2. And this one is with C3 disconnected. voltage at the negative rail.

What is a Nilma Radio? Are these band pass filters used in the antenna input used when selection one of two radio bands? If you have both types of inductor cores why not use them? As you are on the upper limit of the T37-2 cores and the T37-6 cores (lower limit) do not need to go below the upper limit of the T37-2 cores for the higher frequencies.

As a first try the circuit below has a transformer with the primary of 1.8mh and 0.1 ohm while the secondary is 720mh and 150 ohms. The inductance is related to the number of turns squared ratio: 25^2 to 500^2 = 625/250000 = 0.0025 thus ( 0.0025 * 720mh) = 1.8mh So your guess is a lot better than mine would be. For input the pulses are 2ms on and 2ms off for a period of 4ms. Amplitude is 8 volts. The out put would make a good TENs device for an elephant.

I set the browser to display the links in large bold fonts. As the first page is all links that did what I wanted to do.

What are you using for input pulses? Any idea what the inductance of the primary and secondary windings of the transformer are? Else does anyone have a good guess what they may be? I have never used inductors enough to get any concept of what transformer inductance like that would be. I can run the circuit in the simulator if you like.

I believe you are correct. Yes they are beefy - see: https://www.ebay.com/itm/OHMITE-RHEOSTAT-100-OHMS-100-WATT-MODEL-K-STOCK-NO-0451/223494235238?hash=item34094b6466:g:MHYAAOSwJHhcwbwy You could consider using a 1 ampere fast blow fuse. Ideally one could put a second power mosfet at the output of the power supply and use that to control the current and control it via a potentiometer. What is 1 ampere when using a pulse amplifier? Is it the peak pulse current or would you have to integrate the sums of the currents? I would think it relates to the power dissipated by the power transistors in the power supply.

The codes most likely only have meaning to the manufacture of the circuit board they are on. Manufactures do not want customers to be able to repair their products rather them buy new ones. Can you find a schematic?

Bear in mind the headroom for the LM388 is 3 volts. The headroom voltage refers to the actual voltage drop across the regulator that must occur during operation. So that means that the most you can get out of the above circuit is 9 volts. Minus the drop across the 0.3 ohm resistor. You maybe better off using your one transistor circuit if you do not need the power/current for extended periods. If so perhaps there is a spice model for the IRFB4110? You could run it in the LTspice simulator and/or I could run it in the TINA-TI simulator. Also there may be a similar transistor that there is a spice model for; for simulation if you decide to go that way.