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Maximum voltage in Zener-diode circuit

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Teszla

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What is the maximum voltage v from this circuit? The Zener-diodes demands a forward voltage of 0.6V.

00.png

Where do you start when calculating this?
 

Assuming you have enough current flowing (determined by R) to get beyond the zener "knee", the voltage is simply 7.4 + .6=8V
 

Can't the forward voltage of the diodes be illustrated as a ideal diode in series with a voltage source with V=0.6?

I.e. like this:

00.png

Why is this not correct?
 

1) Your bottom diode is not forward biased.
2) Your diodes are zener diodes, not ideal diodes. An ideal diode won't conduct in the reverse direction..
 

1) Why not? Shouldn't the forward biased voltage be in the direction of the diode?

What is keeping the voltage V from being higher than 8 V? If we have a very large voltage source and a small resistance, why wouldn't the voltage be higher?
 
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If we have a very large voltage source and a small resistance, why wouldn't the voltage be higher?
You are asking for an exact answer without providing a specification, that can't work.

We would generally assume that the components are operated within their maximum ratings. So the actual zener diode voltage might be slightly above it's nominal value, but not much. For an exact solution for various input currents, you have to refer to the z-diode characteristic, e.g. from a datasheet.
 

Can't the forward voltage of the diodes be illustrated as a ideal diode in series with a voltage source with V=0.6?

I.e. like this:

View attachment 97299

Why is this not correct?
Your equivalency is not correct because the 0.6 volts only appears when the diode is forward biased, and the Zener voltage only appears when the diode is reversed biased beyond the Zener voltage. At other voltages the Zener diode appears essentially like an open circuit.
 
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    Teszla

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Okay, for the Zener diode of 3.4 V we can say that

* When the voltage is above 0.6 V, it's like a short circuit in the forward direction but an open circuit in the backwards direction.
* When the voltage is between -3.4 up to 0.6V, it's like an open circuit in both directions.
* When the voltage is below -3.4V, it's like a short circuit in the backwards direction but an open circuit in the forward direction.

I know in reality diodes are not linear, but when doing a model of a Zener diode, would the above be correct?
 

Observing a certain voltage across a diode and stating it's operated in forward/backward direction is just the same thing.

E.g. if you apply 0.6V, the diode is biased in forward direction, it can't be "open circuit in the backwards direction" at the same time. In so far your statements make no sense.
 

No the above statements are not all correct.
When the voltage is above 0.6 V, it's like a short circuit in the forward direction but an open circuit in the backwards direction.
1)It acts as a short circuit when the applied voltage is greater than 0.6 V but the diode always act as circuit element like resistor with the only difference being that it always drops a constant voltage of 0.6 V.It is open circuit if it is a normal diode,but this is a zener diode so the the diode conducts in the reverse direction only when reverse voltage applied is greater than 3.4 V in the reverse direction.
2)Between -3.4 V<Vdiode<0.6 V, the diode does not conduct and acts as a open circuit.

From the above circuit,
In the forward direction, the diode conducts when properly biased : Vcir=0.6 V(D1 Forward Biased)+7.4 V(D2 Reverse Biased)=8.0 V
In the reverse direction,(i.e.) if Voltage is negative, then the voltage drop is :Vcir=3.4 V(D1 Reverse Biased)+ 0.6 V(D2 Forward Biased)=4.0 V

The circuit will work only if the current flowing through the circuit(diode) is properly biased to get a drop of 0.6 V.
WORKING:
1)Positive Voltage it acts as a limiter with the maximum voltage drop being 8.0 V for input of 8.0 V and beyond.
2)Negative Voltage it acts as a limiter with the maximum voltage drop being 4.0 V for input of 4.0 V and beyond.

Hope this helps and clarifies your doubt.
 
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