supply is given to reset bar pin of the microcontroller through the circuit i have shown in the attachment but the problem is why we apply the diode in reverse bias in circuit, is there any need to apply this diode in circuit?

 

That diode D1 helps capacitor C1 to discharge quickly when Vcc powers down.

This kind of power on reset circuit is needed when you have a slow power supply ramp up.

Good Luck

 

That diode D1 helps capacitor C1 to discharge quickly when Vcc powers down.

This kind of power on reset circuit is needed when you have a slow power supply ramp up.

Good Luck

But what is the requirement of capacitor and diode we can directly connect the voltage supply through the resistor.

 

As I already said this kind of power on reset circuit is needed when you have a slow power supply ramp up i.e.. suppose your power supply voltage is 5V DC,now at the moment you switch on the supply it doesn’t gets to 5V ,it slowly increase from 0V to 5V then your power supply becomes stable at 5V,so the time when the power is unstable your mcu will not work properly.

Now here is the trick, when you switch on the power the capacitor starts to charge through R1 slowly, the reset pin gets a voltage lower than the supply voltage Vcc(logic low), so it keeps the mcu in a reset condition. After some time when the capacitor gets fully charged ,the reset pin comes to Vcc (logic high) through R1.And as the reset pin of the mcu gets the logic level high i.e.. Vcc ,your mcu can start function properly.

The circuit is just a RC delay, so that after you switch on the power the mcu starts a bit later.

Good Luck

 

we can directly connect the voltage supply through the resistor.

Yes,I do it most of the time but only when my power supply do not power up slowly.

 

thank you very much debijit to solve my problem and giving me such a wonderful answer.