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short circuit protection in inverter

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--BawA--

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I have made a SMPS based sine wave inverter. i just want to incorporate short circuit feature in my inverter. i have thought to accomplish this task using FUSE (fast blow). If i connect a FUSE between my DC-DC stage and DC-AC stage, would it be able to protect the inverter against short circuit events?
Any suggestions will be highly appreciated.
 

--BawA--

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Are you talking about MCBs ? if yes, then i can't use MCB because it will overshoot the estimated cost of my design , i want to implement some cheap hardware .will fuse not do any good in this case?
 

andre_teprom

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Depending on the type of load expected on output, and the allowable cost incresing, a fast circuit breaker would be a reliable option. By the way, additionaly are you already adopting some overcurrent loopback control on inverter driver circuit ?



+++
 

--BawA--

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andre_teprom said:
Depending on the type of load expected on output, and the allowable cost incresing, a fast circuit breaker would be a reliable option. By the way, additionaly are you already adopting some overcurrent loopback control on inverter driver circuit ?
yes, there is a provision of overload protection in my inverter , for that i have used a low side current sensing shunt mechanism. and the expected load is around 800VA / 220V.
there are still two confusion in my mind
1) will fast blow fuse between dc-dc and dc-ac stage will provide protection against short circuit ?
2)can you give me any link where i can read about fast circuit breaker ?
 

andre_teprom

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1) will fast blow fuse between dc-dc and dc-ac stage will provide protection against short circuit ?
The question is, protect what ? Short circuit event on inverter output tends to instantanly get the entire charge of the output capacitor, so that a fast circuit breaker could act faster than inner circuitry would sense.

2)can you give me any link where i can read about fast circuit breaker ?
We used the brand Siemens, and even on their Website you can find a lot of informations.
 

--BawA--

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In most commercial inverter, i didn't see any MCB , so what they are using for short circuit protection? and by protection i mean protecting inverter components from failure.

- - - Updated - - -

As i already mentioned that i am using a low side current sensing mechanism for overload protection , so if i SHORT the ac output terminal of inverter , will that overload protection circuit will protect the inverter from damages?
 

andre_teprom

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I presume you are sensing the DC current flowing on switching elements, just before transformer. This is the usual closed-loop procedure purposed to limit an overload current able to burn these components, but I´m not sure if this would suffice to protect remaining devices against surge current on output stage of transformer.

What is the circuit topology used on this equipment ?
 

FvM

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Safety regulations require a fuse for the battery input of the DC/DC converter, to prevent the device from catching fire in case of an internal short.

An internal DC bus fuse, as considered in post #1 can help the inverter to handle output shorts safely, but only if the output transistors have sufficient safe operation area margin to withstand the initial short circuit current without damage. I believe that an electronical overcurrent shut down would be the better solution.
 

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FvM,one more thing i wanna ask , I have designed a low Battery cutoff circuit using lm358 (comparator),i set the critical voltage as 10.2, now when my battery reaches 10.2v the relay(mechanical) starts oscillating very fast , How to overcome this problem , The circuit is open loop ? if i connect a high value resistance between the output pin and non inverting pin of the IC , will it solve my purpose?
 

R

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You need to put a hysteresis network on the IN+ of LM358 to have two switching points (i.e. 10.2V to cut off, 10.5V to reconnect the battery).
 

--BawA--

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red alert , could you please elaborate about hysteresis network? how to implement it ??
 

chuckey

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What is happening is when the battery falls to 10.2V, the relay opens, the battery gains a little voltage, say 10.25 V, the comparator then switches the battery back into circuit. You need to adjust the comparator so its does not pull the relay in until the battery voltage reaches some more substantial, 12V? One easy way is to use a second set of contacts on the relay to "re-adjust" the reference voltage, so the comparator will not pull in until the battery is 12V. The other way iis to employ positive feed back around your opamp. Could be just a high value resistor between the output and the positive input. This is like the relay contact when the opamps output is low, the resistor is pulling the voltage on the positive input down to earth and once the output is high the resistor is pulling the positive input towards the + Vcc rail. Try :- http://en.wikipedia.org/wiki/Schmitt_trigger .
Frank
 
R

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You need to put one resistor between IN+ and OUT and one (much smaller) in front of IN+, like in the diagram below:

hysteresis circuit.gif

If you could tell me what hysteresis do you need (the cut off/reconnect voltages) I'll do the calculation for you (it's pretty simple anyway).

You could check this link, too: adding hysteresis to comparators.

The hysteresis (the difference between cut off/reconnect voltages) is: Vh = (Vout1 - Vout2) * R1 / (R1 + R2), where

Vout1 = comparator output voltage in HIGH state (almost Vcc)

Vout1 = comparator output voltage in LOW state (almost 0)

Anyway, you could approximate Vh = Vcc * R1 / (R1 + R2) or even Vh = Vcc * R1 / R2 (if R2 >> R1).
 
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