# HELP!!! Solve for diode current this circuit

#### Hang Chong

Joined Nov 3, 2022
8

unable to solve this circuit......new to electronic

#### MrChips

Joined Oct 2, 2009
27,695
Hello "Help!!!"

Welcome to AAC!
Certainly you can come up with a better title. AAC receives dozens of requests for help daily. Imagine how useful a title of "Help!!!" is if all threads were titled "Help!!!".

Besides that, your post has been moved to "Homework Help".
For "Homework Help" you need to show us your best effort before we can guide you with your homework.

#### Hang Chong

Joined Nov 3, 2022
8
base on my calculation,ID2 equal to 0.004A but not sure whether its correct...

#### Hang Chong

Joined Nov 3, 2022
8
Hello "Help!!!"

Welcome to AAC!
Certainly you can come up with a better title. AAC receives dozens of requests for help daily. Imaging how useful a title of "Help!!!" is if all threads were titled "Help!!!".

Besides that, your post has been moved to "Homework Help".
For "Homework Help" you need to show us your best effort before we can guide you with your homework.
thank you, will be extra careful next time

#### MrChips

Joined Oct 2, 2009
27,695
base on my calculation,ID2 equal to 0.004A but not sure whether its correct...

#### Hang Chong

Joined Nov 3, 2022
8
Hi,
9v-D1(0.7V)-D2(0.3V)=8V/2000ohm=0.004A

#### dl324

Joined Mar 30, 2015
15,462
Hi,
9v-D1(0.7V)-D2(0.3V)=8V/2000ohm=0.004A
This is the current in D1. The current in D2 is less than the current in D1.

Your equation doesn't make sense. You have a voltage equated to a current.

This is what you meant:
$$\large I_{D1}=\frac{V_{SUP}-V_{D1}-V_{D2}}{R_1+R_3}=\frac{9V-0.7V-0.3V}{2k\Omega}=4mA$$

#### ericgibbs

Joined Jan 29, 2010
16,803
hi HC,
If there is 0.3V across R2, what is the current through R2.?
E

#### Hang Chong

Joined Nov 3, 2022
8
thank you guys, my latest answer is 0.004A.... after deduction of sum 4kohm

#### MrChips

Joined Oct 2, 2009
27,695
thank you guys, my latest answer is 0.004A.... after deduction of sum 4kohm
The question asks for the current in D2, not D1.

#### ericgibbs

Joined Jan 29, 2010
16,803
hi HC,
Are you sure it is precisely 4mA.?
E
Ref: post #8

#### Hang Chong

Joined Nov 3, 2022
8
9v-D1(0.7V)-D2(0.3V)=8V/4000ohm=0.002A

#### MrChips

Joined Oct 2, 2009
27,695
9v-D1(0.7V)-D2(0.3V)=8V/4000ohm=0.002A
How did you arrive at 4000ohm?

#### Hang Chong

Joined Nov 3, 2022
8

#### MrChips

Joined Oct 2, 2009
27,695
r1+r2+r3=4000ohm isn't it
That is incorrect.
You are just guessing and not using correct circuit theory.

#### Hang Chong

Joined Nov 3, 2022
8
let me try again

#### MrChips

Joined Oct 2, 2009
27,695
Post #6 and #7 are correct.

#### Jerry-Hat-Trick

Joined Aug 31, 2022
208
9v-D1(0.7V)-D2(0.3V)=8V/2000ohm=0.004A
Are you studying physics or engineering? If you are studying engineering you were nearly right, if you'd added that the current through R2 is small enough (an order of magnitude smaller) to ignore, especially as your voltage drops across the diodes are only roughly right.

If you are studying physics you'd recognise that the voltage drop across a germanium diode is typically between 0.25V and 0.30V as it has a softer more gradual knee than the silicon diode which has a typical voltage drop of between 0.6 to 0.7V but it can be higher, and both depend on temperature. For a full answer you'd need the full voltage current characteristic curves of the diodes being used and (I think) an iterative approach to the final accurate answer.

Usually, 0.004A would be written as 4mA