Solving for currents in diode circuit

Question

I’m trying to solve a diode circuit problem, and I'm asked to get the value of I1, I2, I3 and V3 and V1. I’m applying the equations considering an ideal diode with 0.75 V drop. The equations I'm getting are for both the mesh I put in the image:

(1) 5 - i1*1000 - 0.75 - i3*5000 = 0 (2) 10 - i2*1000 - i3*5000 = 0 (3) (this one I'm not sure it is right) i2 + i1 = i3

Solving, I obtain: i1 = -2.2 mA, I2 = 3.5 mA, I3 = 1.3 mA

The problem is: I’m checking the teacher solutions on the exam and they are: i1 = -1.4 mA i2 = 2.83 mA i3 = 1.43 mA

which is the equation im doing wrong?

Answer 1

Well, the very first thing I would do to provide a quick sanity check is to treat the two resistors on the left as a voltage divider and compute the Thevenin equivalent, which is \$V_{_\text{TH}}=\frac{25}{3}\:\text{V}\$ and \$R_{_\text{TH}}=\frac{25}{30}\:\text{k}\Omega\$.

This does mean that the diode is forward biased, so that's good.

This also means that the diode current should be \$I_{_\text{D}}=\frac{\frac{25}{3}\:\text{V}\,-\,5\:\text{V}\,-\,750\:\text{mV}}{\frac{25}{30}\:\text{k}\Omega\,+\,1\:\text{k}\Omega}=1.4\overline{09}\:\text{mA}\$.

This magnitude matches the teacher's solution quite well, suggesting the fuller solution details may also be correct.

So let's look at how I'd set this up (before I bother reading what you wrote) and see where I go with it. Then I'll compare, afterwards.

Here's what I'd do:

kvl1 = Eq( 10 - i2*1e3 - i3*5e3, 0 )          # KVL, left side
kvl2 = Eq( 5 - i1*1e3 + 0.75 - i3*5e3, 0 )    # KVL, right side
kcl3 = Eq( i3, i1 + i2 )                      # KCL, at joining vertex
solve( [ kvl1, kvl2, kcl3 ], [ i1, i2, i3 ] ) # solve it
{i1: -0.00140909090909091, i2: 0.00284090909090909, i3: 0.00143181818181818}

That's pretty close to the teacher's solution, though there is slight difference in the third significant place for one of the values.

Now, I'll look at your work. Hmm....

(1) 5 - i11000 - 0.75 - i35000 = 0

I've highlighted at least one problem. Compare with my own KVL2 above. You subtracted when you should have added. Note the polarity of the diode? The anode is more positive than the cathode, when it is on. (And it is.) So you need to add that value, not subtract it.

Answer 2

Since the voltage 10V is higher than 5V, the diode D1 would be forward biased and current will flow from \$V_3\$ to \$V_2\$.

The direction of current \$I_1\$ would be in the opposite direction. Use -\$I_1\$ for current in the opposite direction. Assume that forward voltage drop across the diode is 0.7V.

You have two unknowns \$V_3\$ and \$V_1\$.

\$V_3\$ - \$V_1\$ = 0.7 Use the currents as \$I_2\$ going down and -\$I_1\$ going up, \$I_3\$ going down as in the diagram. Use KVL equations which include the 10V, 5V voltage sources and \$V_3\$ and \$V_1\$. With these, you can get currents in the branches.

Answer 3

I’d say that the diode voltage drop is in the opposite direction. Since the voltage drop is given in the forward direction and you are doing the voltage sum in the opposite way, you actually have to sum it.