# Ideal diode questions final soon help

• frixis
In summary: D1 is cut off from the circuit. Please remember this information for later use!D1 is cut off from the circuit.
frixis
Ideal diode questions final soon urgent help

its actually an example in my book (sedra and smith, 5th ed) which i fail to understand completely!
Question:
the question is :
Assuming the diodes to be ideal find V and I in the circuits.

I obviously went through the example but it totally escapes me.
Relevant info :
They've first assumed that D1 is conducting and proceeded from there. I'm lost.
i tried to assume that d1 is not conducting and proceeded from there but i didnt know how to check my answer at all.

please help. I'm soooooooooooooo lost and its like a bug in my system. i think i understand basic diodes but this assumption stuff is really confusing.

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I'd assist you, but your attachment is still "pending approval", so I cannot view it at the moment. Sorry

Gnosis said:
I'd assist you, but your attachment is still "pending approval", so I cannot view it at the moment. Sorry
here's another link hope this works.

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frixis said:
its actually an example in my book (sedra and smith, 5th ed) which i fail to understand completely!
Question:
the question is :
Assuming the diodes to be ideal find V and I in the circuits.

I obviously went through the example but it totally escapes me.
Relevant info :
They've first assumed that D1 is conducting and proceeded from there. I'm lost.
i tried to assume that d1 is not conducting and proceeded from there but i didnt know how to check my answer at all.

please help. I'm soooooooooooooo lost and its like a bug in my system. i think i understand basic diodes but this assumption stuff is really confusing.

What is the forward voltage drop across an ideal diode. Using that number and assuming D1 is conducting, and recognizing that D1's anode is grounded, what is the voltage at D1's cathode?

zero?

frixis said:
zero?

berkeman said:

both.

frixis said:
both.

Good. That defines ("pins") that node's voltage at ground. Now you can solve for the rest of the currents and voltages.

k but suppose i said that d1 was not conducting. and d2 was. then my circuit wud be different. how wud i check which assumption is right?

frixis said:
k but suppose i said that d1 was not conducting. and d2 was. then my circuit wud be different. how wud i check which assumption is right?

(Please spell out the word "would" -- txt speak abbreviations are not allowed on the PF. Thanks.)

You ask a fair question. There are at least two general ways that you can figure out which diodes are in forward condution, and which are cut off in reverse bias. The first is probably the quickest, at least in this problem. Power up your supplies one at a time, and figure out what happens in the circuit for each step. Like, leave your -10V supply off (at GND), and power up your +10V supply in the circuit. What are the voltages and currents that result? Now power up your -10V supply -- what changes now that both supplies are on? You can power up either supply first in this circuit, it won't make a difference in the final answers.

The other way you can do it is to make a table, and solve for the voltages and currents for each entry:

D1 D2
Off Off
Off On
On Off
On On

If you assume that a diode is off for a particular choice in the table, but when you calculate the circuit voltages that result, and you get that the cathode voltage is lower than the anode voltage, that would change your assumption, right?

There’s no need to power up the circuit to derive the voltages and currents. The required values can be derived strictly via calculations.

I’ll assume silicon diodes are being used with an “ideal” forward voltage drop of .7 volts.

I’ll discuss schematic ‘a’, which has the 10k ohm resistor connected to the +10 source voltage and the 5k ohm resistor connected to the -10 source voltage.

With the resistor values provided, the indicated diode directions, and the source voltages and polarities shown, D1 and D2 are both forward biased and developing .7 volts drops. D2 will conduct as configured in schematics ‘a’ as well as schematic ‘b’, but how can you verify the conduction state of D1? Here’s how to make that determination:

For the moment, assume D1 ISN’T part of the circuit. In this case, the total source voltage is the difference between the +10 volts and the -10 volts hence, a voltage potential of 20 volts. D2 is installed in a direction that allows it to be forward biased and there’s more than enough voltage for it to develop a .7 voltage drop across its cathode/anode, and so it does. This means the remaining 19.3 volts (20 volts - .7 volts = 19.3 volts) is the voltage drop that’s applied across the total series resistance of 15k ohms (10k ohms + 5k ohms = 15k ohms). Therefore, we can derive the series current as follows:

19.3 volts / 15k ohms = .001286667 amps

NOTE: Remember, this is the current with D1 REMOVED, so it’s not the current sought after with D1 installed.

Now that we have the current (.001286667 amps), we can derive the voltage drop developed across the 10k ohm resistor:

(.001286667 amps) (10k ohm) = 12.86666667 volts (more than enough voltage drop)

A voltage drop of 10 volts or greater is required in order to place the forward biased cathode of D2 at a potential of at least -.7 volts (as measured from ground). If the voltage drop across the 10k ohm resistor were any less than 10 volts, the cathode of D2 would be less negative than -.7 volts, which would prevent D1 from being biased into forward conduction when placed in the circuit.

So, we calculated a voltage drop of 12.86666667, which is more than enough to ensure that D1 will forward conduct, so now we continue with D1 installed in the circuit.

As soon as D1 begins to conduct, it clamps the cathodes of D1 & D2 at a potential that is .7 volts below ground therefore, -.7 volts. The difference between the -10 source voltage and the -.7 cathode is -9.3 volts and this voltage is dropped across the 5k ohm resistor therefore, current through the 5k ohm resistor is calculated:

-9.3 volts / 5k ohms = -.00186 amps

Meanwhile, the anode of D2 is .7 volts higher than its cathode, which places the anode of D2 at a potential of precisely 0 volts (virtual ground). This leaves the entire +10 source voltage across the 10k ohm resistor therefore, current through the 10k ohm resistor is calculated:

+10 volts / 10k ohms = .001 amps

And there you have it. I could take you through the entire workings of schematic ‘b’, but you’ve been given all you need to know to figure out which diodes are forward biased and what's happening in the circuit. I hope you found this helpful.

thats exactly what my teacher said. thanks! so so so much.
he also said that u pick the diode that has the highest potential difference. :D
Gnosis said:
There’s no need to power up the circuit to derive the voltages and currents. The required values can be derived strictly via calculations.

I’ll assume silicon diodes are being used with an “ideal” forward voltage drop of .7 volts.

I’ll discuss schematic ‘a’, which has the 10k ohm resistor connected to the +10 source voltage and the 5k ohm resistor connected to the -10 source voltage.

With the resistor values provided, the indicated diode directions, and the source voltages and polarities shown, D1 and D2 are both forward biased and developing .7 volts drops. D2 will conduct as configured in schematics ‘a’ as well as schematic ‘b’, but how can you verify the conduction state of D1? Here’s how to make that determination:

For the moment, assume D1 ISN’T part of the circuit. In this case, the total source voltage is the difference between the +10 volts and the -10 volts hence, a voltage potential of 20 volts. D2 is installed in a direction that allows it to be forward biased and there’s more than enough voltage for it to develop a .7 voltage drop across its cathode/anode, and so it does. This means the remaining 19.3 volts (20 volts - .7 volts = 19.3 volts) is the voltage drop that’s applied across the total series resistance of 15k ohms (10k ohms + 5k ohms = 15k ohms). Therefore, we can derive the series current as follows:

19.3 volts / 15k ohms = .001286667 amps

NOTE: Remember, this is the current with D1 REMOVED, so it’s not the current sought after with D1 installed.

Now that we have the current (.001286667 amps), we can derive the voltage drop developed across the 10k ohm resistor:

(.001286667 amps) (10k ohm) = 12.86666667 volts (more than enough voltage drop)

A voltage drop of 10 volts or greater is required in order to place the forward biased cathode of D2 at a potential of at least -.7 volts (as measured from ground). If the voltage drop across the 10k ohm resistor were any less than 10 volts, the cathode of D2 would be less negative than -.7 volts, which would prevent D1 from being biased into forward conduction when placed in the circuit.

So, we calculated a voltage drop of 12.86666667, which is more than enough to ensure that D1 will forward conduct, so now we continue with D1 installed in the circuit.

As soon as D1 begins to conduct, it clamps the cathodes of D1 & D2 at a potential that is .7 volts below ground therefore, -.7 volts. The difference between the -10 source voltage and the -.7 cathode is -9.3 volts and this voltage is dropped across the 5k ohm resistor therefore, current through the 5k ohm resistor is calculated:

-9.3 volts / 5k ohms = -.00186 amps

Meanwhile, the anode of D2 is .7 volts higher than its cathode, which places the anode of D2 at a potential of precisely 0 volts (virtual ground). This leaves the entire +10 source voltage across the 10k ohm resistor therefore, current through the 10k ohm resistor is calculated:

+10 volts / 10k ohms = .001 amps

And there you have it. I could take you through the entire workings of schematic ‘b’, but you’ve been given all you need to know to figure out which diodes are forward biased and what's happening in the circuit. I hope you found this helpful.

## 1. What is an ideal diode?

An ideal diode is a hypothetical electronic component that only allows current to flow in one direction, from the anode to the cathode. It has zero voltage drop when conducting and infinite resistance when reverse-biased.

## 2. How does an ideal diode differ from a real diode?

An ideal diode has no voltage drop and infinite resistance, while a real diode has a small voltage drop and finite resistance when conducting. This means that an ideal diode can act as a perfect conductor in one direction, while a real diode has some limitations.

## 3. What are the applications of ideal diodes?

Ideal diodes are commonly used in power supply circuits to prevent reverse current flow, in rectifier circuits to convert AC to DC, and in logic gates to control the flow of current.

## 4. What are the limitations of ideal diodes?

One of the main limitations of ideal diodes is that they do not exist in reality. Real diodes have some imperfections, such as voltage drop and reverse leakage current. Additionally, ideal diodes cannot handle high levels of current or voltage, so they are not suitable for high-power applications.

## 5. How can I simulate an ideal diode in a circuit?

In circuit simulation software, you can use a "diode ideal" component to simulate an ideal diode. This will allow you to observe the behavior of an ideal diode in a circuit and compare it to a real diode.

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