Zener Diode Equivalent Circuit Analysis: Finding the Operation Point

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The discussion focuses on determining the operating point (OP) of a Zener diode through graphical analysis. The measured OP is 15.35mA and 3.24V, but the calculated equivalent circuit graph does not intersect at these values. Key parameters for the equivalent circuit, including a no-load voltage of 3.827V and a short-circuit current of 18mA, yield an incorrect OP of approximately 5mA and 2.8V. The Zener diode appears to behave like a forward-biased diode under the given conditions, allowing about 0.7V across it. The impact of the 2700-ohm resistor on the circuit's performance is questioned, indicating its potential influence on the overall behavior of the diode.
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Hello, I've got a problem I simply can't get my head around. Basically I need to determine the OP of the Z-Diode through graphical solution. Thus I need the graph of the Z-Diode (which we drew by hand with help of some measured values) and the graph of the equivalent circuit.

Well, I also measured the OP for the Z-Diode with the given circuit, which is 15,35mA and 3,24V.

With my calculated equivalent circuit graph, there is no interception at those values, not even close.

One basically needs to find the no-load voltage and the short-circuit current to draw a characteristic line of the equivalent circuit.
I got 3,827V for the voltage source and 18mA for the current. With these values I would get an OP for the Z-Diode of ~5mA and 2,8V which is bullocks.
 
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If the voltage is indeed as shown for Vg (18 volts with the positive terminal downward) then the zener diode is acting like a forward biased normal diode and will allow about .7V across it, so Vba will be .7V (or you could say Vab is -.7V). What effect do you reckon the 2700 ohm resistor has in the circuit under those conditions?
 

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