i want to know the function of the schematic attached and how does it work?
it was in an exam , i can't solve it

 

Hello,

I think it is a power failure monitor.

When the power fails the optocoupler will activate the alarm.

Greetings,
Bertus

 

thanks
but can u explain how does it work ?

 

Hello,

I think it is a power failure monitor.

When the power fails the optocoupler will activate the alarm.

Greetings,
Bertus

Yep...that's what I see too.

 

i'm confused about the thyristor, when it's on and when it's off

 

i'm confused about the thyristor, when it's on and when it's off

The thyristor allows the alarm to stay on even when there is no AC signal input, until it is reset. The momentary switch shorts out the current flow in the thyristor which turns it off if the gate is no longer being driven. EDIT: ignore this, the following post gets it right.

This circuit may be a power detector, but my guess would be a sound detector.

 

Assuming the low-current dynamic resistance of the buzzer (including the LED and its bias resistor) is much smaller than R2 (it should be), then the thyristor will be turned on the moment that Q1 is turned off, which will be when AC power is applied to the source on the left. When the AC source drops to zero, Q1 is turned on, and as the thyristor is still latched the buzzer is energised. R5 provides enough bias current to keep the thyristor latched when the gate current is removed.

When the AC source is present again, then Q1 is turned off again, and the buzzer is de-energised. The thyristor remains latched until the battery B1 goes flat, or the push-switch is pressed.

The thyristor is there so the buzzer can be silenced by pressing the push-switch next to B1. When the buzzer is on, there is no gate current to the thyristor, but it is still latched on from the earlier sequence. Shorting the anode to the cathode commutates the thyristor, and it is now turned off until triggered afresh.

 

I have never worked with or studied thyristors but this is what I think the circuit does.

At power loss the npn part of U2 is open so Q1 can turn on.
The gate (?) of U1 is pulled low through R1 and Q1.
This momentary action turns on U1 if the anode is positive in respect to the cathode.
So BUZ1 has a path on it's right side to the battery positive through U1,
and a path to the battery negative through Q1. So does the LED.

If power is restored Q1 will turn off and BUZ1 will shutup. The LED turns off.

The momentary switch shorts U1 to reset it.
Not sure why a thyristor is used if it requires a manual reset.
If power is not restored a reset will not stop the noise.
The gate will still be low and U1 will turn on again.

Guess I'm missing something.

 

I agree with those who have thus far identified this circuit as one that activates its audible and visual alarms when the applied AC signal drops below the designed limit.

Once the circuit is triggered, the push-button is provided to reset the thyristor since it latches once triggered. The button robs the thyristor of current and that is what cause it to reset.

hgmjr

 

thanks alot , i got it

 

Doesn't the AC drop below the designed limit once every half-cycle? Since there are no capacitors in the circuit to filter the DC voltage (C1 does not filter the DC voltage), I think this would cause Q1 to turn off momentary each time the AC passes through zero.

Q1 off, U1 on, buzzer and LED on, once every half cycle. Hmmmm...

 

DonQ has a very good point, the buzzer will indeed be modulated by the AC waveform. Nicely spotted. If the opto isn't being under-biased then the buzzer should be on most of the time, but there will indeed be brief dropouts, 2 per cycle.