Hi:

I simluated the following circuit and did not obtain the results that I expected.



The current is 633 uA.

This is what I expected.

RT = R1 + R2 = 4K + 3K = 7K
VR1 + VR2 = V1 - V2 - VD1 = 10V - (-5V ) - 0.7V = 14.3V
IT = VR1 + VR2 / RT = 14.3V / 7K = 2.04mA

VR1 = R1 * IT = 4K * 2.04mA = 8.16V
VR2 = R2 * IT = 3K * 2.04mA = 6.12V

Then I got lost.

Could someone please lend a hand. Thanks.

 

Hi maker,
It helps us if you post your asc file.
E

 

The placement of R1 & D1 is not important. It does not matter which one comes first.
D1 adds complicity. Remove D1 and reduce V1 by 0.56V. V1=9.44.
V2 confuses me. -5V and the source is upside down which makes it a +5V. (double negative)
Can you find the answer with +9.44, +5V and R1 & R2?

 

Hi maker,
It helps us if you post your asc file.
E

HI E:

Here is the .asc.

M

 

The placement of R1 & D1 is not important. It does not matter which one comes first.
D1 adds complicity. Remove D1 and reduce V1 by 0.56V. V1=9.44.
V2 confuses me. -5V and the source is upside down which makes it a +5V. (double negative)
Can you find the answer with +9.44, +5V and R1 & R2?
View attachment 324193

The current in sim is 633 uA but calc says 2.04 mA. Which is correct?

 

To get 633uA = 4.5V/7k.
To get 2mA = 14.4V/7k.

The problem is in your understanding of V2.
---edited---
Eric wants the SPICE file, not a picture.

 

Hi maker,
It helps us if you post your asc file.
E

Sorry Eric I clicked on the wrong file and didn't realize it. Just got back home now.
Here is the file.
M

 

To get 633uA = 4.5V/7k.
To get 2mA = 14.4V/7k.

The problem is in your understanding of V2.
---edited---
Eric wants the SPICE file, not a picture.

Looks to me that there is +15V if you probe V1 to V2. See post #7.
Please explain V2.

 

You drew V2 upside down. It is like putting a battery in upside down.

What is the voltage where the red arrow is.

 

You drew V2 upside down. It is like putting a battery in upside down.

That is the way the circuit was in a professor's notes on his website.

 

What is the voltage at the bottom of R2? You think it is -5V but it is +5V.