cdynasty001
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Voltage of ideal diodes in parallel?
- Thread starter cdynasty001
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kwired
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Voltage is measured between two points, something is missing here, kind of depends on what they intended to ask about with the problem.
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If the forward voltage of each diode is .7V, which of the three diodes will be conducting in the circuit shown? Which will be reverse biased and therefore non-conducting?
Most definitely. :happyyes:
To have different voltages on the left side of the diodes could depend on the balance (not shown portion) of the circuit. It is also possible the diodes have different forward bias voltages.
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Actually not. See previous post.
Pretty much all conventional silicon diodes are .7V, Schottky idiodes .2V, and germanium .3V.
Okay, so what is V equal to?Actually not. See previous post.
Pretty much all conventional silicon diodes are .7V, Schottky idiodes .2V, and germanium .3V.
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V = 1.7V. if silicon. V= 1.0V if "ideal":happyyes:
What is I through the bottom diode?
gar
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MY guess.
You assume that the +5 is referenced to a common that has a potential of 0 V.
Since the statement that the diodes are defined as ideal this means that an ideal the diode is either an infinite impedance or 0 impedance, and this impedance change occurs at some voltage drop across the diode. Normally you would assume that ideal means the transition is at zero V, but I don't think that was meant here.
In the open circuit state some output voltages are defined. Obviously a slight load on a diode is required to get the low voltages shown. From the values shown you can determine the diode voltage drop. D1 has a drop of 2 V, D2 of 3 V, and D3 of 4 V.
Connect all cathodes together (parallel the three diodes) and apply a load. Which diode conducts (the other two diodes are reversed biased)?
If the cathodes in parallel are connected to common, then V is what? Easy since you know each of the diode drops.
Reading the previous posts, then assume the diodes are really ideal ( 0 voltage drop ) and that the cathodes are connected in parallel to a voltage sources of V1, then V = V1. But since you can not parallel 3 different ideal voltage sources without infinite current I don't think this analysis would apply.
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MY guess.
You assume that the +5 is referenced to a common that has a potential of 0 V.
Since the statement that the diodes are defined as ideal this means that an ideal the diode is either an infinite impedance or 0 impedance, and this impedance change occurs at some voltage drop across the diode. Normally you would assume that ideal means the transition is at zero V, but I don't think that was meant here.
In the open circuit state some output voltages are defined. Obviously a slight load on a diode is required to get the low voltages shown. From the values shown you can determine the diode voltage drop. D1 has a drop of 2 V, D2 of 3 V, and D3 of 4 V.
Connect all cathodes together (parallel the three diodes) and apply a load. Which diode conducts (the other two diodes are reversed biased)?
If the cathodes in parallel are connected to common, then V is what? Easy since you know each of the diode drops.
Reading the previous posts, then assume the diodes are really ideal ( 0 voltage drop ) and that the cathodes are connected in parallel to a voltage sources of V1, then V = V1. But since you can not parallel 3 different ideal voltage sources without infinite current I don't think this analysis would apply.
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gar
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GoldDigger:
I did indicated that a slight load would be required on the diodes to get those voltage readings. Obviously the question as originally presented is not clear. So we can make whatever assumptions we want.
If at the diode outputs those are solid voltages and we throw away the word parallel, and assume the usual meaning of ideal, then V = 1 V.
A better title might be --- What is V assuming ideal diodes?
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GoldDigger:
I did indicated that a slight load would be required on the diodes to get those voltage readings. Obviously the question as originally presented is not clear. So we can make whatever assumptions we want.
If at the diode outputs those are solid voltages and we throw away the word parallel, and assume the usual meaning of ideal, then V = 1 V.
A better title might be --- What is V assuming ideal diodes?
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gar
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cdynasty001:
Provide the precise wording of the question in your study material.
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cdynasty001:
Provide the precise wording of the question in your study material.
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I see what you are saying...
But in order to see it this way, I believe we would have to accept the voltage references as all being "textbook" ideal. When the voltages and diodes are not "textbook" ideal (which is where I started), we get all types of "real life" scenarios which say there's not enough information.
So I believe we must conclude the entire depiction references ideal components and values and go with the 1.0V answer.
junkhound
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What gar said, what is the exact wording of the question? Real life is never so simple.
V = 1.7 V if nominal Si diode. Otherwise, you need each diode type and temperature for a precise answer. Really hot and the leakage current of D1 and D2 changes the answer by a couple of decimal digits down. Really really cold and it could be 5 volts.
I assume you are studying general diode circuits, and you should assume each diode has a 0.7 volt drop.
As mentioned, you have to figure out which diodes will conduct, and which won't. Replace the conducting diodes with a 0.7 volt battery. Replace the ones that don't conduct with an open circuit.
Start with any diode - assume it conducts. Figure out if the others will conduct or not, replace the diodes with opens or batteries, and see if the circuit makes sense. (Kirchoffs voltage and current laws have to be met.)
If that doesn't work, start with a different diode.
Eventually you should get it.
As mentioned, you have to figure out which diodes will conduct, and which won't. Replace the conducting diodes with a 0.7 volt battery. Replace the ones that don't conduct with an open circuit.
Start with any diode - assume it conducts. Figure out if the others will conduct or not, replace the diodes with opens or batteries, and see if the circuit makes sense. (Kirchoffs voltage and current laws have to be met.)
If that doesn't work, start with a different diode.
Eventually you should get it.
kwired
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Maybe OP is leaving something out, maybe there is more in the study guide that narrows down what they were after, but as is presented to us, we need to know what other point to measure to, and possibly know a little more about the rest of the circuit, there is no source drawn in there so as is there is no current, no current no voltage drop across anything.
Having + voltages on each end kind of indicates this is just a segment of something larger though, but we still don't know what other point we are measuring to.
I started out looking at it this way. Perhaps your first post influenced me, IDK. Nevertheless, I believe you will realize what I did given a bit more time. HINT: One of the first things I did not do was remember that the thread title says "ideal" diodes.
All the voltage would be measured to the same reference point. Call it ground if you want.
No additional info. needed.
If the diodes are ideal, then any conducting diodes would be replaced with a short circuit instead of a 0.7 volt battery.
The answer is pretty obvious.
Carultch
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An ideal diode by definition doesn't have a voltage drop during forward bias, and has zero current during reverse bias.
Therefore, the answer is V = +1 Volts.
Replace the diode to +1 Volts with a short circuit. Now you see how the voltage is caried directly to point V.
When you do this, the remaining diodes end up in reverse bias, and thus would carry zero current. This is necessary such that there can be a voltage difference across these ideal diodes. All current through the resistor flows through the diode to the +1V point.
If you try to make the diode to +3 Volts carry current, then you now have a voltage drop of 1 Volt to the +2V point, and a voltage drop of 2 Volts to the +1V point, both in the forward bias direction. When this happens, you short-circuit the current until as much current flows as possible, which will bring the V point down to +1V.
In the real world, you'll have a system of three nonlinear equations that model the I-V curve to determine the actual balance of current flowing through each diode, and you'll find a voltage of slightly greater than +1V at point V.
With ideal voltage supplies, the upper two diodes are reverse biased because the bottom diode pulls down the voltage to 1V because of the resistor.
Quite simply (IMO), a 0.7 bias voltage for a diode is saying it is not an ideal diode. :happyyes:
Correct.
I would have said the answer was 1.7 volts, but once you pointed out the title stated "ideal diodes", the answer becomes 1 volt.
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