Calculate the current values in this diagram (ASSUMING IDEAL DIODES)

Thread Starter

Mojo Pin__

Joined Apr 13, 2019
19
Hi,

I have attached a picture of a problem my lecturer posed to calculate the current in the circuit.

I have tried using KCL and KVL and I get values:


node Voltage => 30V x (5/15) = 10V


I1 = (30V-10V)/10k

I1 = 2mA



I2 = (10V-0V)/5k

I2 = 2mA



ID2 = ???

so with no resistance at D2 as it is ideal does this imply infinite or zero current? I assume current flows given that the voltage at the anode is 15V and the cathode side is 10V.



Any help would be greatly appreciated.

Kind regards,



Keith






Captureanalog.PNG
 

Thread Starter

Mojo Pin__

Joined Apr 13, 2019
19
hi,
Assume ideal diode has zero forward voltage drop and infinite reverse resistance, recalc your Node voltage it is not 10v.

E
Thanks for your help! The writing on the image is actually the professor's working so I just assumed it was the correct way to calculate the voltage. I must admit I'm even more confused now. Can you advise how I would calculate the node voltage please?
 

Thread Starter

Mojo Pin__

Joined Apr 13, 2019
19
Hi! Thanks for your hints! Does that just mean that the voltage at the node is 15V? I think the issue I have is understanding how 2 voltage sources combine at a node.

I know there is a voltage drop across the 10k resistor but will the remaining voltage add to the 15V coming from the left branch or is it just 15V?

I also still don't really understand how knowing the node voltage will help me figure out the current ID2, sorry I am so confused.

Thanks!

Thanks again for your advice
 

mvas

Joined Jun 19, 2017
538
I think, what the prof is saying is ...
Without D2 connected, the Node Voltage would be ~10 volts = 30 x ( 5 / 15 )
Therefore, when D2 is connected, it will be Forward Biased and it will affect the Node Voltage.

So, I ask you, "How does D2, an ideal diode, affect the Node Voltage?"
 

Thread Starter

Mojo Pin__

Joined Apr 13, 2019
19
I think, what the prof is saying is ...
Without D2 connected, the Node Voltage would be ~10 volts = 30 x ( 5 / 15 )
Therefore, when D2 is connected, it will be Forward Biased and it will affect the Node Voltage.

So, I ask you, "How does D2, an ideal diode, affect the Node Voltage?"
Ah ok, so the ideal diode is forward biased and behaves like a closed switch with no resistance and would not drop any of the 15V supply.

So the node would be 15V plus the voltage remaining from the top branch? Or is it just 15V?
 

ericgibbs

Joined Jan 29, 2010
8,753
hi,
Yes I agree.
Without D2 connected, the Node Voltage would be ~10 volts = 30 x ( 5 / 15 )

So what current has to flow in D2 and the 5k when it is connected in order to raise the Node voltage to 15V.??

Hint: consider how the current thru the 10k will be reduced when the Node V=15v

E
 

Thread Starter

Mojo Pin__

Joined Apr 13, 2019
19
hi,
Yes I agree.
Without D2 connected, the Node Voltage would be ~10 volts = 30 x ( 5 / 15 )

So what current has to flow in D2 and the 5k when it is connected in order to raise the Node voltage to 15V.??

Hint: consider how the current thru the 10k will be reduced when the Node V=15v

E

Ok I see, that is a good way of looking at it, thank you.

I just tried to calculate the currents with that in mind and I get:

I2 = 3mA
I1 = 1.5mA
ID2 = 1.5mA
 

Thread Starter

Mojo Pin__

Joined Apr 13, 2019
19
hi,
You should now confirm your results by applying them to that circuit.
E
I'm sorry I'm not sure what you mean. If I draw the node with the 3 branches and the two entering are 1.5mA and the one leaving is 3mA is that a sufficient confirmation?

Thanks again for all your help
 

mvas

Joined Jun 19, 2017
538
Ah ok, so the ideal diode is forward biased and behaves like a closed switch with no resistance and would not drop any of the 15V supply.

So the node would be 15V plus the voltage remaining from the top branch ? Or is it just 15V?
How could the Node Voltage ever be higher than 15 volts?
The voltages at the Node do not "add", the currents do though ...

The SUM of the currents at the NODE are Zero = KCL, right?
I1 + Id2 + I2 = 0 or
I1 + Id2 = -I2

You have current flowing in from D2
You have current flowing in from D1 and the 10K
So, where does all of this current go? - out through the 5K

You got it !
 
Last edited:

ericgibbs

Joined Jan 29, 2010
8,753
hi,
It is always a good idea when you have calculated a result to check it out on the original circuit.
eg:
I2 = 3mA, so 3mA * 5K =15V Node
V2-V1= 15V , so I1 = 15v/10k =1.5mA
etc.....

If the checks do not match, it means you have a miscalculation in your first results.

E
 

Thread Starter

Mojo Pin__

Joined Apr 13, 2019
19
How could the Node Voltage ever be higher than 15 volts?
The voltages at the Node do not "add", the currents do though ...

The SUM of the currents at the NODE are Zero = KCL, right?
I1 + Id2 - I2 = 0 or
I1 + Id2 = I2

You have current flowing through D2 ...
You have current flowing through D1 and the 10K
So, where does all of this current go ?
Hi, thank you so much for that response, I know it was a dumb question but I couldn't seem to figure that out.

I do believe I have calculated the 3 currents now:

I2 = 3mA
I1 = 1.5mA
ID2 = 1.5mA

Does that appear to be correct to you as well?

Thanks!
 

Thread Starter

Mojo Pin__

Joined Apr 13, 2019
19
hi,
It is always a good idea when you have calculated a result to check it out on the original circuit.
eg:
I2 = 3mA, so 3mA * 5K =15V
V2-V1= 15V , so I1 = 15v/10k =1.5mA
etc.....

If the checks do not match, it means you have a miscalculation in your first results.

E
I see what you are saying now, thanks for the advice, upon checking as you advised with the circuit I do believe these values are correct but if I have made an error please let me know.

And sincere thanks for taking time to help me today
 

mvas

Joined Jun 19, 2017
538
Yes,
When you multiply the current through each resistor times the resistor's value then you get the proper voltage drop ( 15 Volts each ) and
the SUM of the currents at the NODE is ZERO per KCL ...
I1 + ID2 + I2 = 0
because technically I2 is negative vs I1 + ID2.
 

Thread Starter

Mojo Pin__

Joined Apr 13, 2019
19
Yes,
When you multiply the current through each resistor times the resistor's value then you get the proper voltage drop ( 15 Volts each ) and
the SUM of the currents at the NODE is ZERO per KCL ...
I1 + ID2 + I2 = 0
because technically I2 is negative vs I1 + ID2.
That's great, thank you so much, you've really helped make it clearer for me.

have a great day
 

MrAl

Joined Jun 17, 2014
6,513
Hi,

I have attached a picture of a problem my lecturer posed to calculate the current in the circuit.

I have tried using KCL and KVL and I get values:


node Voltage => 30V x (5/15) = 10V


I1 = (30V-10V)/10k

I1 = 2mA



I2 = (10V-0V)/5k

I2 = 2mA



ID2 = ???

so with no resistance at D2 as it is ideal does this imply infinite or zero current? I assume current flows given that the voltage at the anode is 15V and the cathode side is 10V.



Any help would be greatly appreciated.

Kind regards,



Keith






View attachment 175083

This is actually a problem where the highest resulting voltage wins.
Calculate the voltage with diode D1 forward and D2 reverse biased, then calculate the voltage with D2 forward and D1 reverse biased, whichever one causes the highest resulting voltage wins and the putput is then that highest voltage.

It would be more of a challenge if the other diode also had a resistor in series with it.
 
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