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This question is ambiguous, which is perhaps why it has not been answered. A spark ignitor is designed to produce spark when the power is applied, which is accomplished by generating sufficiently high voltage to break down the dialectric between a spark gap - every time. A spark igniter that did not spontaneously spark would be a failure for the purpose of ...

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That is a long shot, but the high ripple voltage you are seeing is because of the electrolytic capacitor $C_4$ you picked. According to the its datasheet it has an $ESR=160m\Omega @ 100kHz$ The capacitor used in the reference design is a tantalum one, with an equivalent $ESR=18m\Omega @ 100kHz$ Although you are switching at a lower frequency ($\... 1 It seems that it is actually based on a "typical" application, as you stated. Searching this in WE site, these results came out: https://www.we-online.com/catalog/en/search?search=750343140 The first one implies that this transformer (750343140) is a "MID-OLTI Offline Flyback Transformer for Texas Instruments". By clicking on the second ... 2 You are right about everything as far as I can see. If you have 100 Ohms in parallel with an inductor, and there is 1A flowing through the inductor, then you open the switch that was supplying the current, then you will have 100V across the resistor and inductor (briefly) because the only path available to the inductor current is to loop through the resistor ... 3 Without a component across the inductor, at transistor switch off, the voltage across the inductor will rise to be V = Ldi/dt. With a resistor in place across the inductor, at transistor switch off, the voltage across the inductor (and resistor) will rise to be V = iR. The resistor would usually be sized to make iR less than Ldi/dt without the resistor. That ... 0 Consider the mechanical analog of flowing water. If you have a large mass of water, equivalent to a large current, but at a low speed, and you put an obstacle in it's way (e.g., a steep gradient or open switch), the water will rise to a much higher level, equivalent to an increase in voltage. This abrupt change with time is$V=L\frac{dI}{dt}\\$, where dI is ...

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To me, keeping the current flowing exactly as it was means the current is the same. The inductor "tries" to keep the current constant, or "does everything in its power" to keep the current constant. That doesn't mean that it actually keeps the current constant. Similarly, a resistor "resists" the flow of current. That doesn't ...

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You should always measure within the oscilloscopes input tolerances. However if your scope is CAT rated it should have some tolerances to handle spikes over the rating depending on if it's CAT I CAT II, III and IV and so on. That is why different voltage levels for example may be displayed on a multimeter in accordance like CAT II 1000V and CAT III 600V, as ...

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Note that if you are seeing 400V with a x1 probe then changing to say x10 will usually increase the probe's load impedance, and so the spike may be correspondingly higher. Best to start with a higher-voltage probe, say x100, and work down to the more sensitive ones as required.

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Are there any better ways of doing this? You use a step up transformer. If the step-up ratio is (say) 50:1 and the output voltage is (say) 10 kV before it sparked, then the primary voltage back emf is: - $$\dfrac{10,000}{50} = 200\text{ volts}$$ Note that when the secondary output produces a 10 kV spark it causes a primary back-emf of 10 kV divided by the ...

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How can I use the voltage generated from inductive kickback? The current flowing through an inductor, upon closure of switch contacts, would result in energy being stored in its magnetic field. Disruption of the current, caused by the switch contacts opening, would lead to a collapse of the magnetic field and consequent induction of a high voltage that ...

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Yes it can damage the scope or the probe or both. A scope that is rated to handle 400V without damage should not be used to measure voltages over 400V. A 10x probe will divide the voltage by ten, but then the probe must be rated to handle over 400V as well, or it might be damaged. And if the probe gets damaged, it can pass the full measured voltage to scope ...

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The video shows 400V spikes. If max input on your scope is 400V you should be ok with a x10 probe.

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The optocoupler 1st-order model is a simple current-controlled current source whose output is decoupled by a parasitic capacitor. This is what has been represented in the post. This capacitance needs to be characterized and accounted for when adopting the compensation strategy. The below circuit shows how to extract this parasitic element. More details can ...

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I have a 1N4007 diode parallel to the solenoid. No you don't - not according to your schematic - the anode is connected to ground: - The flyback diode needs to be in parallel with the solenoid. Look where you have it connected - it is ineffective how you have connected it. If you want to get rid of the capacitor fix how you connected it - i.e. place it ...

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Should the flyback diode have a "Maximum average forward rectified current" above 2A, 200mA or 12A? If PWM is applied using a single FET (not 'half-bridge') the diode recirculates current through the motor during the 'OFF' portion of the PWM cycle. Therefore it has to handle a peak current equal to the motor current. But how much will that be, and ...

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200 mA, 2A or 12 A? When the motor open drain/collector turns off, the motor coasts to a stop and the current at the time of switching is all that needs to be sustained for a brief time constant T=L/Rdiode thus V^2/Rd=Pd is minimal and <=1A rated will do. This of course must include start/stopping really fast at NEAR 0 RPM, which is unlikely but then &...

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At turnoff the flyback diode will momentarily carry the same current as your motor is consuming. This current will then reduce quickly depending on the inductance and resistance of the motor. You will find a rating in the diode data sheets for non repetitive surge current normally about a 10ms rating. The inductive spike at turnoff should not last anywhere ...

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