overload protection circuit in inverter

I have seen that in sg3525 based power inverter , the overload protection module incorporates two opamps, why is this so?
i mean we simple monitor the current and compare it with a reference , using one opamp and connect the output of opamp to the pin 10 of sg3525. but many schematics contains 2 opamps in cascade , why?

can you post the schematic?

please see the schematics in the attached file in the below thread https://www.edaboard.com/threads/227240/

The first opamp is for overcurrent protection and the second one is for battery undervoltage protection.

i have made a overload protection circuit using opamp , but when , overload event occurs, the oscillations starts , i.e the bulb starts blinking very fast. how to remove these oscilations?
i connected the output of opamp to the pin 10 of sg3525.

You need a hysteresis circuit for your opamp (one resistor in front of IN+ and another one between IN+ and OUT).

BTW, I answered same questions in your old thread.

https://www.edaboard.com/threads/320591/#post1376661

thanks redalert, my overload protection circuit is working now , but still there is a problem , after overload condition occurs, the circuit doesn't reset itself, it goes into shutdown mode . how to overcome this problem?

If current returns to zero (after shutdown) the opamp output should change, too (to disable the shutdown command).

Anyway, you have to insert a delay circuit to avoid shutdown oscillations (if overcurrent/overload persist).

- - - Updated - - -

That's an example of shutdown (reset) delay:



Actually, the circuit reacts very quick to overcurrent events but it requires some more time to deactivate the shutdown command (you could choose a resonable delay - few seconds, by example).

when overload event occurs the system shuts down , but when i remove the load , the system doesn't restart automatically , i have to restart it manually , i want like ,during overload condition the system shuts down and after 5 seconds , the system must try to restart itself . What changes should i do? Do i need to use 555 or opamp is sufficient for this task?

If you're using exactly the circuit from the pdf you did mention in the post #3, there is a diode (D7) across the OUT and IN+ of the IC2B opamp that keeps its output high indefinitely after any alarm triggering event.

Just remove that diode D7, put another diode across R21 resistor and change that resistor (R21) value to 1M or something.

That way, the C13 is charging fast (through the new diode) and is discharging slowly through the R21 (new value), R4 and Q17.

Or you can put D8 across R21 (still modifying the resistor value) and put a wire instead of (former) D8.

i tried this , but it's still oscillating

Could you post the exact schematic you're using?

If that's the one from the pdf, none of those opamps have hysteresis so they could oscillate (excepting the second one, if D7 it's used).

- - - Updated - - -

They did not use hysteresis because after first output pulse (overcurrent event) of the IC2A opamp, the second opamp (IC2B) is latching its output so any further osciallations of the IC2A output have no effect.

VREF is adjusted by pot
capacitor is of 10uf
R2 = 1M
R1 = 3K

during overload event , the oscillations start..

You have to add the following resistors (at least R3 and R4) to avoid opamp output oscillations:



Those resistors actually change the IN+ voltage depending on opamp output.

By example, to trigger high the opamp output you have to apply an OVERCURRENT_SENSE voltage of VREF + 1V but to trigger it low you have to apply an OVERCURRENT_SENSE voltage of VREF - 1V.

With your current circuit, the opamp triggers its output for the same IN+ voltage (VREF) so the oscillations begin.

The R3/R4 ratio set the voltage difference between positive and negative triggering thresholds.

On current implementation, the capacitor is discharging through the SG3525 pin 10, too (not through R5 only). The pin 10 could sink up to 1mA (reading the datasheet) so it's discharging the capacitor much faster than the R5.

The shutdown voltage is also decreasing during capacitor discharging process.

For a more accurate and steady shutdown signal, I STRONGLY RECOMMEND YOU to put another opamp in a similar configuration (hysteresis comparator) after the current one. This way, you could supply a stable and constant shutdown voltage and a controlled shutdown reset delay.

If you want, I'll post a complete schematic of this circuit.

- - - Updated - - -

That's the complete schematic:

Thanks red alert ..
Could you please tell me how did you calculate the resistors value in the schematics that youvhave posted..?

This is how the hysteresis circuit works:

The two resistors represent a voltage divider. The voltage at the cross point (between the two resistors) is Vin+ = Vref + k * Vr, where k = R1/(R1+R2) and Vr is the voltage across both resistors.

When the opamp output is HIGH then Vr = Vcc - Vref thus Vin+ = Vref + k * (Vcc - Vref).
When the opamp output is LOW then Vr = Vref - Vgnd so Vin+ = Vref + k * (Vgnd - Vref).

For example, if k = 0.1, Vref = 5V, Vcc = 12V and Vgnd = 0V we have the following thresholds:

Vtrig+ = 5 + 0.1 * (12 - 5) = 5.7V
Vtrig- = 5 + 0.1 * (0 - 5) = 4.5V

Thus the opamp will trigger its output in HIGH state if CURRENT_SENSE is greater than 5.7V and will trigger its output LOW when CURRENT_SENSE is smaller than 4.5V.

You have to adjust R1, R2 to get the desired thresholds voltage difference.

If you tell me the Vref value, I could suggest a k factor for the first opamp stage.

For the second opamp, you could choose a Vref of half the Vcc and a large thresholds difference (R1 = 5 * R2), to allow a wide range of voltage across the capacitor (between charged and discharged states).

For example, if Vcc = 12V, you could choose Vref = 6V, Vtrig+ = 10V and Vtrig- = 2V.
Thus the opamp will trigger its output HIGH when voltage across capacitor is greater than 10V then wait for the capacitor do discharge bellow 2V to trigger LOW its output (reseting SHUTDOWN signal).

- - - Updated - - -

Please post the CURRENT_SENSE and Vcc voltages to help you choose the right components values.

Thanks alot Red_alert for your support ,, now things are getting much clear
by the way , in present scenario
CURRENT_SENSE = 0.8V
Vref = 0.5 V (when the current sense is greater than 0.5V the sg3525 should shutdown)
VCC = 12V

redalert please calculate the new values for 2 opamps,
when the current sense voltage is greater than 0.5V, the sg3525 should shut down and after 2 secs it should automatically restart.
VCC is 12V

Sorry for the delay. I had to redraw the whole schematic for better results.

In the previous schematics, the second opamp hysteresis resistors have been contribute to capacitor (dis)charge so I had to isolate them by switching IN+/IN- for both opamps.

That's the new schematic:



A short description:

When OVERCURRENT voltage rise over 0.5V, the first opamp output turns LOW, quickly charging the C1 capacitor through D1 diode.

When voltage across capacitor reach 9V (the IN- of the second opamp reach 3V) the second opamp output switches HIGH, activating the SHUTDOWN signal.

If the OVERCURRENT voltage drops bellow 0.2V, the output of the first opamp switches to HIGH and the capacitor starts discharging through R4. When the voltage across it reaches 3V (the IN- of the second opamp reaches 9V) the second opamp switches its output to LOW, resetting the SHUTDOWN signal.

Thus the first opamp triggering thresholds are Vtrig- = 0.5V and Vtrig+ = 0.2V and the second opamp thresholds are Vtrig- = 9V and Vtrig+ = 3V.

Please check if the current R4 value (1M) give you the expected delay (3 - 5 seconds).

Thanks a ton Red_Alert ,, the circuit is working perfectly

one more thing to ask , I want to incorporate a low battery shutdown feature also , the system should shut down when battery voltage reaches 10.2 and it should reconnect when battery voltage reaches 10.8 , Is it possible using a single opamp with hysterisis or should i use 2 opamps in this case also?

- - - Updated - - -

red_alert , in post#19 , the voltage 6V should be regulated or not , i am thinking of voltage divider network for generating this voltage.
am i right or should i use any regulator for 6v generation?
and Vcc of ic will be direct from 12V battery.