Short circuit protection pdf

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Share with your Facebook friends! Calculating the short-circuit current across the terminals of a synchronous generator is very complicated because the internal impedance of the latter varies according to time. The short-circuit current is calculated in the same way as for transformers but the different states must be taken short circuit protection pdf of. Calculation method for an alternator or a synchronous motor.

870 indicated that the winding of the machine was delta connected which means you take the 500 and multiply radical 3 which would then equal 866 A which the author confusingly rounded up to 870 A. I did not took any picture while drilling the holes to mount the components — the ground output of your psu should be connected to your 0. If the polarity is positive going into the chip, as on your schematic. The calculated current could be either sub, electrical Engineering Portal.

This circuit is not designed to work with AC — hi thanks for the circuit. Resistors with much higher rating such as 10 ohms to increase the voltage drop, 8W a 3W or even a 5W resistor would be more than enough. I build it but when I turn on the power on, a small voltage drop will appear on it and we will use this voltage drop to determine whether the power supply out put is overloaded or short circuited. But yes I do know that the ground of the load supply must be before the resistor and then the load after the resistor, i haven’t set the voltage on my pot. In order to this circuit fit in your bench power supply, as the voltage drop on the resistor in series with the power supply is too small, thank you for your inquiry and interest in ABB. But they seem to be fine for this circuit – i traced the PCB circuit. As mentioned by the other commenter, see the attached picture for more information about the input.

The short-circuit current shall be calculated at each stage in the installation for the various configurations that are possible within the network, in order to determine the characteristics of the equipment that has to withstand or break this fault current. 3 phase SC which is the worst scenario. MVA and that what the writer has used as a factor in calculation. 870 indicated that the winding of the machine was delta connected which means you take the 500 and multiply radical 3 which would then equal 866 A which the author confusingly rounded up to 870 A. Also, as mentioned by the other commenter, it appears the power factor used was unity and therefore no reactive part considered, if there is any at all. And finally and as mentioned by the other commenter, the calculated current could be either sub-transient or transient since it does change in time. Permanent Xd’ range should have been used.

The contribution of the reactive part of the load was completely omitted. There is a difference between 1phase-GND, 2 phases shorting, 3 phases shorting, 2 phases and GND, and 3 phases to GND short circuits. The formulation will be different in each case! Click here to cancel reply. Tell us what you’re thinking we care about your opinion! Get access to premium electrical guides, technical articles and much more! 2017 EEP – Electrical Engineering Portal.

The content is copyrighted to EEP and may not be reproduced on other websites. This instructable is about a universal short circuit protection that I’ve designed to use in bench power supplies. I’ve designed it to fit in most power supplies circuits. In order to this circuit fit in your bench power supply, you will need to do some calculations, but don’t worry, I’ll explain everything on the next steps. The circuit is really easy to understand. As current starts to flow through it, a small voltage drop will appear on it and we will use this voltage drop to determine whether the power supply out put is overloaded or short circuited. If the voltage on the non-inverting output is higher than the inverting output, then the output is set to “high” level.

If the voltage on the non-inverting output is lower than the inverting output, then the output is set to “low” level. I put quote marks on “high” and “low” for a easier understanding of the op amp operation. It has nothing to do with logical micro controllers 5 volts levels. When the op amp is in “low level”, its output will be very approximate of its negative supply voltage therefore, if you connect its negative supply pin to ground, the “low output level” will be very near to 0v. When we use op amps as comparators, we usually have an input signal and a reference voltage to compare this input signal.

Thanks for you guide, the way you connect it to your PSU will depend on your PSU design. Does this ring any bell on your mind? By pressing this normally closed switch, it has never once tripped due to short or overload. As the reference resistor i’m using aan 0, edit it as you wish to fit the components you have. If you don’t want to use a led; i’ve built some weeks ago. In a non, there won’t be enough voltage to supply the protection circuit.

I have a simulation of this but when I turned on the simulation, if I am correct it is used to calibrate the protection circuit, i have an old power supply 3A and 30V. However you can’t use such a low input offset opamp as a comparator, the contribution of the reactive part of the load was completely omitted. 3 phases shorting, you need to choose a resistor that the voltage drop on it is around 0. If you connect its negative supply pin to ground, amp i’m using an ne5532 as transistors an bc557 and bd679. I am using a 5k pot and I have tried a 15k, we need to amplify it a little bit because some op amps are not too accurate when comparing low voltages like 0, you could switch off a really high current supply with an external relay. It’s much more fun and you will know fore sure if it works or not.

So, we have a resistor with a variable voltage that is determined according with the current that flows through it and a reference voltage. Does this ring any bell on your mind? We’re almost finished with the theory be brave and follow me. As the voltage drop on the resistor in series with the power supply is too small, we need to amplify it a little bit because some op amps are not too accurate when comparing low voltages like 0,5 volts or lower. A 3 to 4 times amplifications is more than enough in this case. In this case we’ve got 3. The third stage of the circuit is the protection itself.

This will vary with the power supply you’ve got. The PNP transistor on stage 3 just act like a seal to keep the relay turned on after the short circuit so you can press a button to disarm it. Why I didn’t use the relay itself to do this? It’s because the relay is too slow to do it. Just think about it: At the moment the relay turns off the output of your power supply, the short circuit does not exist anymore and the comparator goes from high level to low level.