HarryA

Folks that restore old radios say do not power them up until replacing the capacitors. It is difficult to find information on failing due to disuse. Wikipedia has some information related to storage. See "Performance after storage" here: https://en.wikipedia.org/wiki/Electrolytic_capacitor#Operational_characteristics Years ago, okay many years ago when I was in HS, I used to repair radios and bw televisions. I repaired a radio that a mouse had chewed into a wax coated capacitor.

Modern aluminum electrolytic capacitors have longer shelf lives then the old ones, usually around 2 years. Perhaps powering up old equipment a couple of times a year would be sufficient. Once a month is perhaps unnecessary. I have never found any good information on that subject. I have an old audio amplifier I power up a couple of times a year. I keep thinking someday I will use it - perhaps.

There are three things to consider. First the voltage from the charger is higher than the batteries voltage in order to charge the batteries. Second the current supplied by the charger is most likely much smaller than the current supplied to the unit by the batteries during operation. Also there is a var (volt-ampere reactive) which is an back emf from the coil as the field collapses. This maybe more then the charger can handle. Shavers typically draw 10 to 15 watts off the mains and some higher.

I would think that using the devices, as you suggest, once a month would be a good idea. Also you could do a quick test on the capacitors if you have not used the device in some time. I often check the capacitor with a multi-meter using the ohmmeter range without disconnecting it. The reading will start low then rise as the capacitor charges indicates a good capacitor. If this fails I disconnect one lead and try again. If the reading stays low it could be shorted. If no capacitance it will often just show a high reading. see: https://www.wikihow.com/Test-a-Capacitor If you doing many capacitors perhaps a tester would be worth the monies: https://www.amazon.com/HiLetgo-Multifunctional-Capacitor-component-Backlight/dp/B01MYU0QI3/ref=sr_1_10?dchild=1&keywords=tester+electrolytic+capacitor+tester&qid=1619267945&sr=8-10

Are you looking for a circuit like this? This is not a finished circuit it is just an example.

Sorry but you must tell us what you have now to work with and what are your goals. Do you have a sensor or need to find one to buy? What type of air pump or inflator do you have or need? If you need a sensor this company has 1 - 5 volts ones that you could subtract 0.5 volts from as an example. https://www.sensorsone.com/10bar-150psi-0-10v-0-5v-1-5v-pressure-sensors/

"Hi, I need a digital car blower circuit with the ability to adjust the pressure with a pressure sensor 10 times 5 volts with an output of 0.5 to 4.5 volts." Sorry Google does not translate Persian into English well. What is "a pressure sensor 10 times 5 volts" ? با عرض پوزش ، فارسی را به انگلیسی ترجمه نمی کند. "سنسور فشار 10 برابر 5 ولت" چیست؟ You may find this translator helpful: به نظر شما این مترجم مفید است https://translate.google.com/

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?

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

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.

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.