[SOLVED] i need to detect rising edge of pulse

[rakesh1987]

Rising edge of pulse and falling edge of pulse should be determined separately using CMOS IC ?

 

[BradtheRad]

An easy concept is shown in these schematics. A comparator or an op amp in comparator mode.



This probably is not what you want.
Do you want a single short pulse when the signal rises, and a single short pulse when it falls?
There is a circuit called a 'one-shot' (or monostable multivibrator) which will give you that.

 

[crutschow]

If you want a short pulse on both the rising and falling edge of a pulse you can use an XOR gate. Connect the signal directly to one of the XOR input pins. Also connect the signal through a series resistance and capacitor to ground to the other input. The XOR will generate a positive pulse at both the rising and falling edges with a width approximately equal to RC.

 

[rakesh1987]

If you want a short pulse on both the rising and falling edge of a pulse you can use an XOR gate. Connect the signal directly to one of the XOR input pins. Also connect the signal through a series resistance and capacitor to ground to the other input. The XOR will generate a positive pulse at both the rising and falling edges with a width approximately equal to RC.

i need to get it separately....

 

[BradtheRad]

Add this circuit to each comparator, post #2.

I believe this is similar to what Crutschow described, post #3.

 

[crutschow]

Add this circuit to each comparator, post #2.

I believe this is similar to what Crutschow described, post #3.

Not quite. The XOR circuit I described generates a positive pulse for both the leading and trailing edges of the pulse.

To use this circuit to catch the negative edge you would invert the diode polarity and replace the ground to the resistor and diode with the plus supply voltage. If you still want a positive pulse from that then you could run this signal through a Schmidt-trigger logic inverter such as the CD40106.

 

[BradtheRad]

Not quite. The XOR circuit I described generates a positive pulse for both the leading and trailing edges of the pulse.

To use this circuit to catch the negative edge you would invert the diode polarity and replace the ground to the resistor and diode with the plus supply voltage. If you still want a positive pulse from that then you could run this signal through a Schmidt-trigger logic inverter such as the CD40106.

Yes, this clarifies matters. And it works fine. Screenshot:
(Let's hope the image gets uploaded okay.)

 

[crutschow]

The circuit looks OK but the 10k ohm resistor to ground is not needed.

 

[BradtheRad]

The circuit looks OK but the 10k ohm resistor to ground is not needed.

I agree, I only added it to show how the circuit is referenced to ground.

It didn't look right without a connection to ground somewhere.

 

[ALERTLINKS]

From page,
**broken link removed**

https://www.freepatentsonline.com/6756808.html

 

[crutschow]

I agree, I only added it to show how the circuit is referenced to ground.

It didn't look right without a connection to ground somewhere.

The connection to ground is through the resistor and the +5V supply. Also the inverter has a connection to ground. So there are plenty of ground connections, they just are invisible on your schematic.

- - - Updated - - -

From page,
**broken link removed**

https://www.freepatentsonline.com/6756808.html

That appears to be a level detector, not an edge detector.

 

[ALERTLINKS]

For microcontroller use, on zero crossing interrupt, if we check rising and falling inputs, its easy to check status of AC wave.
For simple pulses. i simulated this circuit.