Voltage of ideal diodes in parallel?

[Gary Glaenzer]

Voltage of ideal diodes in parallel?

I have this question in my study material and I can't figure out what it is asking. It wants the voltage V or total voltage with diodes in parallel, and I can't figure it out. Here is a diagram of the question.

If you have voltages applied as shown (assuming those are referenced to ground), the voltage at the junction of the diodes' anodes will 2.3 V, assuming ideal silicon diodes with a 0.7V drop.

Therefore, the two bottom diodes will be reversed biased and can be ignored.

The current would be 2.7 V / 1000 ohms or 2.7 mA.

At least, that's what I think the question is meant to ask.

Your mileage may vary...............

 

[GoldDigger]

If you have voltages applied as shown (assuming those are referenced to ground), the voltage at the junction of the diodes' anodes will 2.3 V, assuming ideal silicon diodes with a 0.7V drop.

Therefore, the two bottom diodes will be reversed biased and can be ignored.

The current would be 2.7 V / 1000 ohms or 2.7 mA.

At least, that's what I think the question is meant to ask.

Your mileage may vary...............

You have your diode function backwards from what is in the picture. :happyno:

 

[gar]

161012-1700 EDT

We need an exact statement from cdynasty001 of the wording of the question in the study material.

Gary Glaenzer:

A true ideal diode has a zero forward impedance, and an infinite reverse impedance, and conventionally the transition occurs at zero volts. However, I can expect some might want to define an ideal diode with some other transition voltage than zero. I would rather this be done by inserting a voltage source in series with a true ideal diode.

In the diagram shown current flows from right to left when the right side is more positive than the left side. The resistor shown pulls the right side of the three diodes positive. If the left sides of the diodes are connected to solid ideal voltage sources of the voltages indicated, then the lower diode conducts when the point labeled V is +1 V. At this condition the upper two diodes are reverse biased and do not conduct. Thus, V never goes above +1 V.

.

 

[Besoeker]

I have this question in my study material and I can't figure out what it is asking. It wants the voltage V or total voltage with diodes in parallel, and I can't figure it out. Here is a diagram of the question.

They are not in parallel if they are feeding different loads.

 

[Carultch]

They are not in parallel if they are feeding different loads.

I don't see any way to construct this circuit in practice, without them ultimately being in parallel. Granted, they would be in series first, with whatever it is that is generating the 1V, 2V, 3V points first. But to make a complete circuit, the three paths would eventually rejoin.

 

[Besoeker]

I don't see any way to construct this circuit in practice, without them ultimately being in parallel. Granted, they would be in series first, with whatever it is that is generating the 1V, 2V, 3V points first. But to make a complete circuit, the three paths would eventually rejoin.

The circuits may be in parallel from a common supply. The diodes are not in parallel.

 

[Sahib]

I have this question in my study material and I can't figure out what it is asking. It wants the voltage V or total voltage with diodes in parallel, and I can't figure it out. Here is a diagram of the question.

Assume diodes are ideal. The current through 1k resistance to be found first: it is equal to ((5-3)+(5-2)+(5-1))/1000=(9/1000)A. So voltage drop across 1K resistance is (9/1000)*1000=9V more than source voltage of 5V. An impossibility. So the problem is with wrong data.

 

[Carultch]

Assume diodes are ideal. The current through 1k resistance to be found first: it is equal to ((5-3)+(5-2)+(5-1))/1000=(9/1000)A. So voltage drop across 1K resistance is (9/1000)*1000=9V more than source voltage of 5V. An impossibility. So the problem is with wrong data.

How do you figure the "(5-3)+(5-2)+(5-1)" part of that equation?

 

[steve66]

Assume diodes are ideal. The current through 1k resistance to be found first: it is equal to ((5-3)+(5-2)+(5-1))/1000=(9/1000)A. So voltage drop across 1K resistance is (9/1000)*1000=9V more than source voltage of 5V. An impossibility. So the problem is with wrong data.

Incorrect. You are assuming all the diodes conduct.

 

[GoldDigger]

Incorrect. You are assuming all the diodes conduct.

Worse than that.
If we make all the left side voltages, equal his calculation would tell us that with three diodes to (for example) 1V you would get three times the current through the resistor that you would get with only one diode to 1V.
Clearly wrong.

 

[Sahib]

Caultac: Answer to your question: I figured it out using network analysis. Steve66: All diodes arrows point in same direction and a voltage drop in each diode arrow direction and so each diode must conduct. GD: There is really no contradiction in that.

 

[gar]

161013-1412 EDT


Carultch:

For DC analysis the diodes are not in parallel unless you connect all three of the cathodes directly together. Besoeker gave you a very simple clear response.

For AC analysis all ideal DC voltage sources are replaced by a zero resistance resistor. Then the diodes would be in parallel assuming those voltage values next to the cathodes represent ideal DC voltage sources. . For all ideal DC current sources in an an AC circuit analysis you replace the current source with an open circuit.

You need to do more study on the theory of circuit analysis. From what school did you get your engineering degree? This type of circuit analysis should be taught in early EE classes, and even before that in physics.

.

 

[gar]

161013-1435 EDT

Sahib:

If all of the + voltage values shown on the diagram are ideal voltage sources and referenced to the same common point, then only one ideal diode can conduct with the values shown, and it is the diode whose cathode is connected to + 1 V. The other two diodes are reverse biased under these conditions.

.

 

[GeorgeB]

Assume diodes are ideal. The current through 1k resistance to be found first: it is equal to ((5-3)+(5-2)+(5-1))/1000=(9/1000)A.

Only the diode connecting to the 1V source will conduct; the others will be reverse biased and open. So there will be 4V across the 1k resistor for 4mA

 

[steve66]

Caultac: Answer to your question: I figured it out using network analysis. Steve66: All diodes arrows point in same direction and a voltage drop in each diode arrow direction and so each diode must conduct. GD: There is really no contradiction in that.

Network analysis only works when all the components are linear. It will not work for most diode circuits.

Your equation for finding the current: (5-3)+(5-2)+(5-1) /1000 implies that the voltage on the left side of the resistor is 1 volt, 2 volts, and 3 volts all at the same time. Clearly not possible.

 

[junkhound]

Network analysis only works when all the components are linear. It will not work for most diode circuits.
QUOTE]

You are correct of course that classical linear circuit analysis only works with linear components - I even have a couple of 1950s textbooks that say the analysis can be run with non-linear components by inserting the proper response equations but that the required computing power is not available.

That said, have never run into a circuit where network analysis does NOT work. Diodes, FETs, IGBT, any transistor, LED, variable resistors, saturating magnetic cores etc. etc. can all be described by an equation.
Finite element programs like SPice, PSpice, LTspice, , etc. all use equations to describe the component responses to voltages, current, and time.
One guy I know is such a 'math hound' he distains the finite element programs and writes humungous equations for entire non-linear circuits and runs them on Matlab.

example with 1N4002 actual diodes at 25 C, bias voltages and currents in the schematic,
If the 5 Vdc is replaced with 5V 60 Hz, the voltage and currents are as seen in the waveforms.

Anyone interested the Cadence ORCAD web site has a limited versin of PSpice free for download, and the Linear Technologies website has a more complete analysis program (LTspice) free for download.
Either of these is sufficient for most electrician work, only the free LTspice is capable of more complex circuits such as PWM power supplies, etc.
Why guess at answers when you can be precise in the real world ? I've use PSpice since 1987 and never found an error other than my erroneous inputs.
Note the nanoamp bias currents on D1 and D2, those are reverse leakage currents. The math models for the components take temperature into account, if the temp is changed the leakage currents change and the 1.635 Volts changes also. heh, heh, up or down is left for the student to complete!

 

[junkhound]

PS: It took about 3 minutes to draw the circuit and just milliseconds for the computer to calculate the dc biases and the waveforms.
OTOH, some detailed models that run VFD circuits can take many minutes to run for a pulse by pulse circuit response for a 100kHz PWM circuit.

 

[GoldDigger]

PS: It took about 3 minutes to draw the circuit and just milliseconds for the computer to calculate the dc biases and the waveforms.
OTOH, some detailed models that run VFD circuits can take many minutes to run for a pulse by pulse circuit response for a 100kHz PWM circuit.

From the waveforms it appears that the program applied AC with a DC bias, not pure DC.
What, on the diagram, indicates the sign (direction) of each diode current?

 

[junkhound]

From the waveforms it appears that the program applied AC with a DC bias, not pure DC.
What, on the diagram, indicates the sign (direction) of each diode current?

My apologies for not being more precise in making clear the waveform and the diagram are not the same analysis.
There were 2 DIFFERENT circuits: on the bias voltages and currents (circuit as shown); and, for the waveforms the 5Vdc was replaced with pure 60Hz ac, 5Vpeak.

The waveforms are pure ac, no dc bias, the numbers on the circuit are pure dc, no ac applied.
Current numbers are positive in conventional sense (current flow OUT of positive and denoted as positive current), would have a negative sign otherwise, same with voltages.
The ORCAD PSpice does have weird conventions sometimes, too much to go into here.

 

[steve66]

Network analysis only works when all the components are linear. It will not work for most diode circuits.
QUOTE]


That said, have never run into a circuit where network analysis does NOT work.

Sure, you can make it work on some circuits, with mixed results. But not if one incorrectly assumes which diodes are conducting, as we had in this example.

A simple example of where it won't work is using the superposition theorem on a circuit where 2 different supplies both feed a single diode through separate resistors.

Because the diode curve is non-linear, the actual operating point will not be the same as the sum of the currents by figuring each source independently.

So I guess my main point is that with non-linear elements, we can't guarantee network analysis will give us a correct answer.