NEED HELP WITH ELECTRICAL ENGINEERING (WBE) ASSIGNMENT?

rewiizzByrewiizz

Hello,

I'm in my first semester studying electrical engineering and I'm stuck with this task and urgently need help!!!

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Thegendarmerie
Thegendarmerie
1 year ago

The object is a circuit with a diode and a Zener diode-like voltage source which serves to limit the output voltage. The Zener voltage source is set to 5 volts, and the diode has a forward voltage of 0.7 volts. This circuit ensures that when the input voltage which is a sine wave rises above this value of 5.7 volts (i.e. the sum of the Zener voltage and the forward voltage of the diode), the diode becomes conductive and the output voltage is limited to 5.7 volts.

So what happens exactly? The input voltage, which is a sine wave, fluctuates over time. When this sine wave reaches a value above 5.7 volts, the circuit will intervene and limit the voltage to precisely this value. This means that parts of the sine wave, which would go beyond 5.7 volts, are simply cut off and instead a straight line is produced at 5.7 volts. Below 5.7 volts, the output voltage will simply follow the input voltage without the circuit engaging.

To display the graphic, you would draw a normal sinusoidal curve, but whenever the curve would rise above 5.7 volts, you will instead draw a horizontal line at the height of 5.7 volts. The remainder of the sine curve, which is below this value, remains unchanged.

The end result would be a sinusoidal curve which is cut off at the tips and forms a straight line at 5.7 volts. This straight line represents the limited output voltage of the circuit.

1
Kelec
Kelec
1 year ago
Reply to  Thegendarmerie

If the input voltage >5V is blocking the diode and is not conductive.

The diode is then polarized in the flow direction when the input voltage is less than 5V.

So this doesn’t correspond to a Zener diode.

1
paprikaw22
paprikaw22
1 year ago
Reply to  Thegendarmerie

The end result would be a sinusoidal curve which is cut off at the tips and forms a straight line at 5.7 volts.

actually at 4.3V

0
Kelec
Kelec
1 year ago

Generally calculate the voltage at the diode.

You can replace the diode here as a 0.7V voltage source with the voltage in the flow direction and an ideal diode.

Now if the voltage at the ideal diode <=0V then locks it and you can leave the branch.

If at the ideal diode more than 0V, the ideal diode is replaced by a short circuit.

For these two cases you calculate the output voltage quite normal.

In order to get your transfer function, you must, of course, reset to the input voltage at 0V at the ideal diode and then you will get a function of the shape

If Uin>U then Uout=f1(Uin) otherwise Uout=f2(Uin)

1
paprikaw22
paprikaw22
1 year ago

When does the diode begin to run?

You set the stitch equation

For this area, you replace the diode with a voltage source – 0.7V .

So…

Otherwise,

In summary:

4