Potential Difference Across Sensitive Instrument: 7.5V

In summary, the potential difference across the sensitive instrument in this circuit would be 7.5v. However, the diagram shown in the conversation is incorrect and incomplete as it does not account for a resistance between the source voltage and the zener diode. This causes a potential drop between them, which is why the load resistance does not cause a potential drop. The purpose of using a greater value of resistance in series is to protect the zener diode from damage and to avoid wasting power. The zener diode maintains a constant voltage by utilizing its reverse breakdown characteristic, which causes a small increase in voltage to result in a relatively large increase in current. This excess voltage is then dropped across the resistance in the circuit.
  • #36
gracy said:
Why?When Iload*Rs drops the voltage below the zener voltage ,does zener offer a large amount of resistance so that at current dividing point no current goes in zener?
Bystander said:
Yes
So ,where do all resistance go when breakdown voltage is achieved?
 
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  • #37
It doesn't "go anywhere." This is a characteristic of this type of component, a very non-linear current-voltage curve, and at least two different conduction mechanisms that produce/explain it, or result in zener behavior, neither of which I can explain to myself right at the moment, let alone to someone learning the topic.
 
  • #38
Take a look at this video from time 1:32 to 2:50

in forward bias why diode behaves as short circuit?And in reverse bias why it behaves as open circuit?
 
  • #39
The circled "~" at the left side of the schematic indicates that an AC signal is being applied to the circuit; for the first half of the cycle, the voltage is positive, and for the second half it is negative.
 
  • #40
Bystander said:
The circled "~" at the left side of the schematic indicates that an AC signal is being applied to the circuit; for the first half of the cycle, the voltage is positive, and for the second half it is negative.
Yes,but in forward bias why diode behaves as short circuit?And in reverse bias why it behaves as open circuit?
 
  • #41
It's a regular diode, and when forward biased it passes current with very little resistance. I can't get the video to reset to show me what the load is, but there is a complete circuit, and the "lower" conductor in the circuit diagram is arbitrarily assigned a potential of zero volts. The upper conductor is alternating between positive and negative potentials/emfs.
 
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  • #42
In normal diode forward bias there is a lot more current than in reverse bias.That's why in video forward bias is regarded s short circuit and reverse bias as open circuit.
2000px-Pn_Junction_IV_Characteristic.svg.png
 
  • #43
Yes.
 
  • #44
What is special about zener diode?I mean if we will connect any other load in place of zener, the current will not drop the voltage across Rs by same amount by which the voltage has increased so that the voltage across that load remain constant?
 
  • #45
The special thing about the zener diode is its I-V characteristic curve. This curve results from exploiting certain features of semiconductor physics, and in particular, how a semiconductor junction that is reverse biased can break down by way of certain modes of "failure". Careful engineering of the chemistry of the materials and structure of the junction encourage and "tune" these features to particular "knee" voltages and make the "failure" survivable for the component.

In theory you could replace a zener diode with a handful of other parts that mimic its behavior (an equivalent circuit model). But having a single inexpensive component that delivers this functionality is very practical.
 
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  • #46
I got really good marks in my class test of zener diode.Thanks gneill ,bystander and lightgrav.Can you please explain what is current gain in transistors?
 
  • #47
gracy said:
I got really good marks in my class test of zener diode.Thanks gneill ,bystander and lightgrav.
You're welcome! Glad to hear you did well on your test.
Can you please explain what is current gain in transistors?
That's a new topic, so you need to start a new thread . Be sure to include the current understanding you have from your class and reading your text.
 
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  • #48
gneill said:
That's a new topic, so you need to start a new thread
Ok.But please help me in the new thread also.
 

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