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[Question] Circuit with diodes and zener diodes for various voltages

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krd777:
Hi everyone!

I'm having some doubts here and i can't find the proper answer anywhere.

It is I can't remember the models for diode circuit analysis.

I've started to try to get the answers but I need to make sure they're correct.

So:

if I have a silicone diode D1 with a junction voltage Vj of 600mV and a resistance Rr of 10 ohm.
When I connect it in series with a voltage source Vdc and a resistor R1 of 1k ohm.
To analyze this circuit, when Vdc is 5V and the diode is directly polarized (ON).

-Can I substitute the diode for a voltage source equal do Vj in series with a resistor equal to Rr to represent the voltage drop along D1?

Also, if Vdc is equal to Vj, can I assume the diode is ON but the current through it is zero? Or should it be behaving like open circuit?

Also, if Vdc is smaller than Vj but bigger than zero, Should I assume the diode as being only a resistor? Or should it be behaving like open circuit?

Also, if Vdc is maller than zero should I assume the diode is OFF and behaving like an open circuit? (I say this because the reverse current is so small in the junction)

Sorry for all these questions but I didn't pay attention in class and I can't find it anywhere on my books. So I'm trying to design these models myself.
My biggest doubts is what to consider when Vdc is between zero and Vj. Should it be a resistor like Rr? Doesn't seem to make much sense.


Thanks in advance and if I explained it badly, please tell me, I can try better.

DJPhil:
Fair warning, this is off the top of my head with a little googling.


--- Quote from: krd777 on December 01, 2010, 04:12:42 pm ---if I have a silicone diode D1 with a junction voltage Vj of 600mV and a resistance Rr of 10 ohm.
When I connect it in series with a voltage source Vdc and a resistor R1 of 1k ohm.
To analyze this circuit, when Vdc is 5V and the diode is directly polarized (ON).
Can I substitute the diode for a voltage source equal do Vj in series with a resistor equal to Rr to represent the voltage drop along D1?
--- End quote ---
Approximately, but not accurately for analysis (or probably schoolwork). The circuit has no Thevanin equivalent, for example, due to the VI curve of the PN junction.


--- Quote from: krd777 on December 01, 2010, 04:12:42 pm ---Also, if Vdc is equal to Vj, can I assume the diode is ON but the current through it is zero? Or should it be behaving like open circuit?
--- End quote ---
This I don't quite understand. Practically speaking there will be some leakage current, but I can't reconcile the diode being ON with zero current as a sane state of affairs. Semantics aside, this might make a lot more sense if you study a VI curve. The more ideal the diode model you use, the steeper the curve. An ideal diode curve is zero to Vj, then instantly vertical. A diode with Vj applied is beginning it's linear region, and should be passing current equal to Vdc/Rr+R1. I doubt this would satisfy a professor though.


--- Quote from: krd777 on December 01, 2010, 04:12:42 pm ---Also, if Vdc is smaller than Vj but bigger than zero, Should I assume the diode as being only a resistor? Or should it be behaving like open circuit?
--- End quote ---
This is a bit easier. Now you'd be sitting on the bottom of the VI curve and the only factor is leakage, which is much smaller than I = Vdc/Rr. This is almost certainly regarded by academia to be an open circuit.


--- Quote from: krd777 on December 01, 2010, 04:12:42 pm ---Also, if Vdc is maller than zero should I assume the diode is OFF and behaving like an open circuit? (I say this because the reverse current is so small in the junction)
--- End quote ---
Yes, right up until you hit zener (avalanche) breakdown, at which point the diode starts conducting heavily.


--- Quote from: krd777 on December 01, 2010, 04:12:42 pm ---Sorry for all these questions but I didn't pay attention in class and I can't find it anywhere on my books. So I'm trying to design these models myself.
--- End quote ---
I'd dig in to wikipedia, or google, with a handful of keywords. VI curve, zener, linear region, and the like should get you started. You can also lean on the open textbook at All About Circuits for formulas and the like.

Hope that helps. :)

scrat:
Diode circuits are quite confusing, if you don't find a "method" to procede solving them. Particularly if you try solving circuits with diodes and opamps...

The simplest way that I have found is to suppose the diode as open, as a first thing, which is usually easy to calculate (in the OFF state current is known, while in the ON you only know it's positive). Then, if the resulting voltage is higher than the threshold, the OFF state is not consistent, and you are sure the diode is ON. In this last case, you substitute the diode with the voltage+resistor.
This is right also for Zener zone, you just have to consider the proper voltage value and direction.
For some circuits with multiple diodes, you should consider all of the cases, preferably starting with the OFF state.

I also found useful to consider current direction when inductances are connected to diodes (as usual in power switching circuits).

Besides all of this, one always have to use good sense, and discard solutions that are not consistent, or intuition to rapidly solve.
A rule that always applies to physical systems is that if the model is right and you find one solution, that is the RIGHT solution. So, if you analyse one of the diode states and find it is consistent, then you have found the solution.

BTW, null current and voltage equal to the threshold is at the limit between OFF and ON state, but a component with null current is simply equivalent to an OPEN circuit.

alm:
My $0.02 (not fact checked either):


--- Quote from: krd777 on December 01, 2010, 04:12:42 pm ----Can I substitute the diode for a voltage source equal do Vj in series with a resistor equal to Rr to represent the voltage drop along D1?

--- End quote ---
The IV curve of a resistor is linear, and an (ideal) diode is exponential, so it won't be a very good approximation. But it's a first order approximation, which may do in some cases. In other cases, it will get you in trouble, like your next question.


--- Quote from: krd777 on December 01, 2010, 04:12:42 pm ---Also, if Vdc is equal to Vj, can I assume the diode is ON but the current through it is zero? Or should it be behaving like open circuit?

--- End quote ---
The ideal pn junction has an exponential IV curve (as long as Vj >> VT*someconstant, which is something like 25mV at room temperature), so at Vdc = Vj(somecurrent), the current will just equal to somecurrent. Eg. the forward voltage may be 300mV at 1uA, 500mV at 1mA and 700mV at 1A (made up numbers, check a real datasheet for actual data). So if you apply 500mV, the current should be 1mA.


--- Quote from: krd777 on December 01, 2010, 04:12:42 pm ---Also, if Vdc is smaller than Vj but bigger than zero, Should I assume the diode as being only a resistor? Or should it be behaving like open circuit?

--- End quote ---
The exponential curve will allow some current to flow unless the forward voltage approaches VT. A resistor sounds like a pretty bad approximation.


--- Quote from: krd777 on December 01, 2010, 04:12:42 pm ---Also, if Vdc is maller than zero should I assume the diode is OFF and behaving like an open circuit? (I say this because the reverse current is so small in the junction)

--- End quote ---
The reverse leakage current will flow, which is a constant (first order approximation, depends on temperature and probably tons of other factors in real life).


--- Quote from: DJPhil on December 01, 2010, 08:52:17 pm ---
--- Quote from: krd777 on December 01, 2010, 04:12:42 pm ---Also, if Vdc is equal to Vj, can I assume the diode is ON but the current through it is zero? Or should it be behaving like open circuit?
--- End quote ---
This I don't quite understand. Practically speaking there will be some leakage current, but I can't reconcile the diode being ON with zero current as a sane state of affairs. Semantics aside, this might make a lot more sense if you study a VI curve. The more ideal the diode model you use, the steeper the curve. An ideal diode curve is zero to Vj, then instantly vertical. A diode with Vj applied is beginning it's linear region, and should be passing current equal to Vdc/Rr+R1. I doubt this would satisfy a professor though.

--- End quote ---
Most diode models I know at least model some kind of exponential function, the step curve is mainly used at the very basic level (and things like digital systems were it usually doesn't matter).

krd777:
Alright everyone, thank you so much for the answers. Based on what you guys said, I'll use a voltage source and resistor in series to replace the diode for the ON mode. For OFF mode I shall replace it with an open circuit. Same way of thinking for Zener region.

I'll confront my teacher with these models and see what she might say. Then I'll come back to let you know if they were accepted. Byt the way, using these models gave me correct results, so following what has been said above, I assume tthey're correct.

Again, thank you very much and I shall be back with answers.

Best regards,

krd777

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