Hello Guys,

Could anyone explain me the purpose of C3, D4, D5 in below circuit ?



Thank you !

 

Is 'no' the correct answer?
It looks as if the cathode of D4 also connects to something else?
What does the red writing around the MOSFET drain mean?

 

Is 'no' the correct answer?
It looks as if the cathode of D4 also connects to something else?
What does the red writing around the MOSFET drain mean?

D4 cathode is connected to source of M1, no other connection.
Red wire is Buzzer .

Trying to drive buzzer at the same time short circuit protection needed.

 

The V6 battery polarity is backwards.

 

The V6 battery polarity is backwards.

But it is set at -5V so the result is a positive voltage!

 

The battery is shown with a negative output and the text says it has a negative output.
The battery output must be positive for this circuit.

 

The battery is shown with a negative output and the text says it has a negative output.
The battery output must be positive for this circuit.

Two negatives make a positive.

 

Without the connection from D4 to the M1 source, C3, D4, and D5 form a charge pump, a low-power type of dc/dc converter. To the right of D4, off the page, there probably is a capacitor to GND to filter the ripple. The circuit is getting double duty out of transistor M1, driving both the buzzer and the charge pump. The buzzer acts as the pull-up for driving the pump. The output of the pump will be something less than twice the input voltage, so 5 V minus two diode Vf drops.

With standard silicon diodes, a charge pump running on 5 V is not very efficient because the diodes eat up a lot of the boosted voltage. Changing D4 and D5 to Shottkey diodes will increase the output voltage by approx. 0.5 V.

ak

 

Two negatives make a positive.
View attachment 260229

Thanks,
V6 is positive voltage to drive buzzer.
Since I made wrong connection so I changed voltage to -5V. So the overall is +5V.

 

Why show its polarity and text its polarity WRONGLY? It is simple to turn the battery symbol around on LTspice.