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39

First determine the coil current when the coil is on. This is the current that will flow through the diode when the coil is switched off. In your relay, the coil current is shown as 79.4 mA. Specify a diode for at least 79.4 mA current. In your case, a 1N4001 current rating far exceeds the requirement. The diode reverse voltage rating should be at least ...


27

Power output and energy per switching cycle An output of 265 volts at 5 mA is a power of 1.325 watts and this means that the energy that needs to be transfered each switching cycle is 1.325 W divided by the switching frequency. Hence, the energy released by the flyback transformer is 2.65 µJ. Accounting for losses, you should probably bump that up to around ...


23

Consider the operation of the circuit. When the transistor is on current is flowing in the coil from top to bottom as the circuit is drawn we now switch the transistor off. The current in the coil still wants to flow. For the circuit on the left this current can now flow back to Vcc via the diode the voltage across the coil has reversed direction and is ...


21

There are many reasons: - Cost: This is probably the biggest reason. The lighting market is quite competitive. There are lots of Chinese players offering aggressively cheap prices. This forces the other brands (even the bigger European ones like Osram, Philips, etc.) to use "cheaper" components to keep themselves in the game. As an employee for one of those ...


17

A real inductor looks like this (shown below is an inductor with 4 coils) there is a small amount (usually in the pF-fF range) of capacitance between each coil. Each piece of wire also has some resistance associate with it. Because each coil in an inductor has resistance (or each section of wire if you consider one coil) this impedes the current and ...


15

This has nothing to do with a relay, other than its coil acts like a inductor. What you are really asking is how to chose the flyback diode across a inductor. There are three main parameters to look at: Voltage rating. This is the maximum voltage the diode can take across it backwards, and still block current and not get damaged. This must be at least ...


15

You need to be aware that the professor in the video is skipping over a few things. Note that at about 22 minutes, he writes the equation for the current through the resistor $$i = I_0 e^{\frac{R}{L}t}$$ but conveniently fails to evaluate $$ I_0= iR $$ In other words, for his proposed 10,000 ohm resistor, a 1 amp current will provide a 10 kV voltage spike. ...


15

Use a cascade. There are two main choices a) Villiard aka Cockcroft-Walton aka Greinacher multiplier Each stage is stacked on the previous stage, so capacitors and diodes only need to be rated at the input voltage, not the output voltage. However the output impedance increases as the number of stages squared. b) Dickson multiplier Each stage returns its ...


14

Voltage required is the nominal coil voltage, since that is what will be applied. Give it a factor of 2 for safety. Current requirement is the nominal coil current. Speed is probably not a consideration for relay coils, since they are not turned on/off very often, as compared to, for instance, a PWM motor drive. In your case, a 1N4001 will probably work ...


14

Just a bit of preliminary theory. As you probably know, without any flyback diode, be it a rectifier or a Zener, you'll have a (theoretically infinite) kickback voltage from the inductor (valve coil, relay winding or whatever) whenever you try to interrupt its current abruptly. In reality the kickback won't be infinite because the spike will trigger any ...


13

Unlike a forward-topology transformer (where the primary and secondary windings are conducting at the same time), the flyback transformer must store energy during the primary switch on-time, delivering it to the load during the primary switch off-time. A forward-topology transformer doesn't need any gap since the peak flux density is a function of the ...


12

The diodes serve two distinct purposes. Under regenerative braking, they return the generated voltage to the power supply (where with suitable electronics, it can be used to recharge the battery). Note that unless the motor is being run above its normal speed, the generated voltage will be no more than the supply voltage, so it is within the voltage rating ...


12

From the question below: A diode is put in parallel with a relay coil (with opposite polarity) to prevent damage to other components when the relay is turned off. This is called a flyback diode. See also How to choose a flyback diode for a relay? See also the comment of Kevin White (in my own words): The diode prevents current spikes The problem ...


11

The key point here is that without an air gap an inductor will saturate if you try to put any current through it so inductance will fall and you can't store any energy. The term "Flyback Transformer" is a little misleading and its more useful to consider it as coupled inductors rather than a transformer because the action is quite different with a ...


11

Things aren't always as simple as they seem, though in the case of relays it's highly application dependent. While the diode provides a safe discharge path that preserves your switching transistor and power supply, it can cause a few issues in certain applications. Relays on closure can form a small weld at the contacts, and by placing the diode there you ...


11

I'm wondering how one would choose between one or a combination of these options? It's easy, if you understand how inductors work. I think the problem most people have is that they hear words like "inductive voltage spike" or "back-EMF" and reasonably conclude something like So, when an inductor is switched, it's for an instant like a 1000V battery. ...


11

I would still add diodes between the output of the opamp and both supply rails. Such diodes are already present anyway inside the opamp (for ESD protection) but you do not want to damage these diodes as then you'd have to replace the opamp. You're right that under normal working conditions the current through the coil is not interrupted (which is the cause ...


11

Your gate drive voltage is too low. That MOSFET needs 10V to turn on completely. 5V just barely clears the 4V threshold when the MOSFET just barely starts to conduct. DO NOT use the Vgsth if you intend to use your MOSFET at a switch. That is the voltage it just barely starts to conduct at. Use a Vgs at least as high as the one used to obtain the given ...


11

does it physically matter at what position my flyback diode for my motor is ? Ideally, the diode should be as close to the motor as possible in order to minimize the circuit-loop in which the flyback current flows. This is to reduce, as far as is possible, EMI into other circuits. Other than that, there are no special considerations unless you were ...


10

I have not personally worked with solid state relays before, but from my understanding they function similar to an opto-isolator, with an LED triggering a phototransistor (or photodiode) into order to close or open the "switch". As such, the triggering side is isolated from the output side so I don't think you need a flyback diode. However, if you are ...


10

Try researching Geiger counter power supplies. They don't produce 1500 volts but they do lend themselves for modification such as this design: - Picture from MAXIM website. As you can see the output stage is a cockcroft walton multiplier so you can add more stages and get more output voltage. Alternatively you build two of these (with fewer added CW stages)...


10

If the motor is producing power, the net power into the motor must be positive, so the net current out of the batteries must be in the direction that drains them, so you are fine. If the motor is being regeneratively braked, then power can flow out of the motor and can push the supply voltage up and charge the batteries (this is used to advantage in ...


10

I'd wager that the problem isn't an oscillation, it's just the initial inrush of current into the MOSFET that's killing it. When you connect your super-caps to the circuit, it will charge up the parasitic MOSFET capacitors \$\mathrm{C_{oss}}\$ and \$\mathrm{C_{rss}}\$. According to the datasheet, \$\mathrm{C_{oss}}\$ is only about 781pF and \$\mathrm{C_{rss}}...


10

The design of this \$RCD\$ clamp requires the knowledge of the maximum peak current your controller can set up as well as the maximum voltage you tolerate across the MOSFET. I have presented all these equations in a seminar I taught at APEC in 2011 and entitled The Dark side of the Flyback Converter. The equations to determine the components values are there:...


9

The control side of solid state relays is usually just a LED, sometimes two LEDs back to back, and sometimes with integrated resistor. In any case, there is nothing inductive there, so no inductive kickback to protect against. If the relay is on the same board as whatever is driving it, then no inductive kickback diode is needed. It's no different than ...


9

There is no hard limit to the output power from a flyback topology. It's a matter of which is best for a given situation. One could create a 1kW flyback, but it would not likely be economical. This is a business where they have blood-on-the-carpet meetings over 3-cent diodes and recognize that it is cheaper to hire another full-time engineer than to put an ...


9

In the question you say you want 4 A at 7 V (28 W), but in a comment 5 V at 10 A (50 W). That rules out cheap and available POE (power over Ethernet) transformers, which I often use for such things. Flyback transformers at the power levels you want are more scarce. In any case, here are roughly the steps: Pick a transformer. Start with the power ...


9

No diode is needed, but you should put both some bulk capacitance and high frequency bypass capacitance across the power leads. Keep in mind that this is a "hobby servo" motor. You are not driving the motor directly. You are sending signals to a controller in the same package as the motor, and that controller drives the motor. Therefore, you should try ...


9

At face value it certainly appears that the LED's would be damaged by back EMF as generated by the relays on turn-off. However if you take a look at the datasheet for the ULN2803 you see it is designed specifically for driving inductive loads such as relays, and has integrated clamp diodes between each output and common. The common point is connected back to ...


9

A multiple output flyback converter (with only one controller, i.e. PWM circuit/feedback loop) can only precisely control one voltage. If you need multiple precisely controlled voltages, you will need multiple independent controllers. That being said, in theory the output voltage of a flyback converter is dependent ONLY on the duty cycle and transformer ...


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