What is the Current Flowing Through Series Zener Diodes with 8V Input?

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The discussion centers on a circuit with two series Zener diodes, both rated for a reverse breakdown voltage of 10V, while the input voltage is only 8V. Participants clarify that under these conditions, the diodes are technically in reverse bias and should not conduct significant current, but a small reverse leakage current may still flow. Questions arise about the relationship between the minimum currents for each diode and how they affect the overall current in the circuit. The conversation highlights confusion regarding the calculations and assumptions made in analyzing the circuit, particularly concerning the behavior of Zener diodes near their breakdown voltages. Ultimately, the need for accurate application of diode equations and understanding of reverse bias behavior is emphasized.
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Hi. A circuit has two zener diodes connected in series. The voltage reaches the diode's reverse breakdown voltage (VzDiode1 and VZDiode2 are the same = 10V). The Vi of the all circuit is 8 V.

/ /
8v -----|<|----|<|---- Ground
/ /
+ +DZ1- +DZ2- -

Both diodes have: IzminDz1= 1microAmp and IzminDz2= 2 microAmp

My question: What is the current flowing through this circuit?

I understand that these diodes are off and that should be open, but in reality there is current flowing through the circuit?

Thanks in advance for your help...
 
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Antonio G said:
Hi. A circuit has two zener diodes connected in series. The voltage reaches the diode's reverse breakdown voltage (VzDiode1 and VZDiode2 are the same = 10V). The Vi of the all circuit is 8 V.

/ /
8v -----|<|----|<|---- Ground
/ /
+ +DZ1- +DZ2- -

Both diodes have: IzminDz1= 1microAmp and IzminDz2= 2 microAmp

My question: What is the current flowing through this circuit?

I understand that these diodes are off and that should be open, but in reality there is current flowing through the circuit?

Thanks in advance for your help...

There should be a small reverse leakage current up to the point that they start to break down.
 
Thanks for your answer.
This small reverse leakage current is the same to or less than Izmin?
If Iz1min (1mA) is different from Iz2min (2ma). What is the current series (Iz serie) in the circuit. The diode 1 or 2 ?. Which predominates?
Thanks again !
 
Antonio G said:
Thanks for your answer.
This small reverse leakage current is the same to or less than Izmin?
If Iz1min (1mA) is different from Iz2min (2ma). What is the current series (Iz serie) in the circuit. The diode 1 or 2 ?. Which predominates?
Thanks again !

You mentioned microamp (uA) currents in your original post, not milliamp currents.

With a regular diode, the reverse leakage current is often referred to as Is, or the saturation current. To be honest, I'm not sure what it's referred to with Zener diodes. My books are at work -- I'll check tomorrow AM when I get into work, unless someone else (or you) beats me to it.
 
Oh and on your question about which to use -- look at the I-V curves for diodes and Zener diodes. Near the 0V crossing, you will see that there is a reverse saturation current associated with the different values of bias voltage. If 2 diodes have different values of Is versus reverse voltage bias, how will the series combination play out? The current will be the same Is through both of them, but the reverse voltages will have to balance out to match the respective diodes' Is-V curves, no?
 
Thank you very much for your time. I had a mistake typing the value of the current in my second question.
I understand perfectly what you are trying to indicate me in your answers. I share with you your reasoning.
To cut to the chase, I will put a document (.pdf) in which I will show you the results of this problem that I found on the Internet
In the answer a), it is observed that the ecuation of (Idiode) equals the current of Diode 1 and the current of Diode 2. They also make a calculation, obtaining the current of diode 2, I don’t understand this equalization.
And on answer b), Diode 1 is ON and Diode 2 is OFF and the current of the circuit is the current 2. Why?
The analysis that is showed in the resolution of the problem, I looked for it on the internet and I’ve been looking for a logical analysis of the Zener Diode and it does not match.
Thankful for your time.
 

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Antonio G said:
Thank you very much for your time. I had a mistake typing the value of the current in my second question.
I understand perfectly what you are trying to indicate me in your answers. I share with you your reasoning.
To cut to the chase, I will put a document (.pdf) in which I will show you the results of this problem that I found on the Internet
In the answer a), it is observed that the ecuation of (Idiode) equals the current of Diode 1 and the current of Diode 2. They also make a calculation, obtaining the current of diode 2, I don’t understand this equalization.
And on answer b), Diode 1 is ON and Diode 2 is OFF and the current of the circuit is the current 2. Why?
The analysis that is showed in the resolution of the problem, I looked for it on the internet and I’ve been looking for a logical analysis of the Zener Diode and it does not match.
Thankful for your time.

I only looked at the attachment briefly, but it looks like they are making an error. You need to use the Is values for each diode (1uA and 2uA), and solve the equation for the two diodes in reverse bias before breakdown. Also, the diode equation they are using does not seem to include the Zener breakdown region.

I could be wrong, and will look more a bit later. But the Zener knee is nowhere sharp enough to say that there is 100V across one diode and 20V across the other. You need full current to make the 100V reverse Zener voltage, which is not consistent with saying the other diode is at 20V reverse. If it is, you are still near Is, nowhere near a full reverse Zener breakdown current...
 
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