Here is circuit from Horowitz's "Art of electronics" book:
The problem is that I can't get expected signal from R2 resistor in OrCAD:
Here is diode pass both rising and falling edges.Why?
EDIT:
Here is correct settings of voltage source:
Electronic – Get rising edge using diode
diodesschematics
Related Solutions
Look at the Zener diode curve. You will see that the device breaks down at the Zener voltage when reverse-biased, and conducts. That property will fix the output voltage at the breakdown voltage, over a range of output currents, when used with a resistor, with relatively small voltage changes. It will also stabilise the output against changes in the input voltage.
Strictly speaking, Zener diodes are low-voltage devices (up to about 5V6). Higher-voltage ones have a different mode of operation and are called avalanche diodes. Both types are commonly referred to as Zeners, though.
No current is right, because your circuit is interrupted, and you need a closed circuit to get current flowing.
And when there's no current through the resistor there won't be a voltage drop across it neither, per Ohm's Law: Voltage = Current x Resistance. So 0 V across the resistor means the full 10 V is across the diode (not 5 V!).
If the diode would fail shorted, there wouldn't be a voltage drop across the diode, and the full 10 V would be across the resistor, giving (again due to Ohm) 10 V/ 1 kΩ = 10 mA.
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Best Answer
In short, parasitics.
A parasitic model of a diode unfortunately contains a capacitive component. The art of electronics was describing what would occur with ideal components. SPICE attempts to model actual components.
A 1n5817 has a total capacitance of 110pF
If the circuit is simulated with an IDEAL diode (with a 10V pulse with 1ns risetime) the resultant waveform is:
If a capacitor is placed in parallel with the diode:
The waveform that is produced is in line with your simulation