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How do I get to 139 Ohms? A worked solution would be appreciated.

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    \$\begingroup\$ Well a worked on question would be equally appreciated. We won't solve homework or related questions without any work done on it. \$\endgroup\$ – Arsenal Mar 23 '17 at 13:31
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    \$\begingroup\$ Hint: You know the output voltage (2.4 = 3 x 0.8V) and you know the answer is 139 ohms so you can work out how much current would flow through the diodes. You know the diodes have a voltage drop of 0.7V at 1mA. So the question is - what is the forward current for a diode with a voltage drop of 0.8V? \$\endgroup\$ – JIm Dearden Mar 23 '17 at 13:31
  • \$\begingroup\$ $100 would be appreciated. Put effort in to your question if you want us to put effort in to an answer. \$\endgroup\$ – Bort Mar 23 '17 at 13:34
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    \$\begingroup\$ I'm voting to close this question as off-topic because its HW with no attempt to solve \$\endgroup\$ – Voltage Spike Mar 23 '17 at 15:33
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My take is that there are unstated assumptions that have to be made. Going back through your course notes may help reveal those assumptions.

To get 2.4V at the output we need 0.8V accross each diode and 7.6V accross the resistors. The question then becomes what current do we need?

We need to get 0.8 volts across the diodes. To do that we need a current flow of more than 1 ma, but how much more?

The shokley diode equation is.

$$I=I_\mathrm{S} \left( e^\frac{V_\text{D}}{n V_\text{T}} - 1 \right)$$

At these voltages we can ignore the -1 term as negligable.

$$I=I_\mathrm{S} e^\frac{V_\text{D}}{n V_\text{T}} $$

Lets plug in the values we have for 1 ma.

$$10^{-3}=I_\mathrm{S} e^\frac{0.7}{n V_\text{T}} $$

And now the value we are trying to calculate

$$I_{0.8\mathrm{V}}=I_\mathrm{S} e^\frac{0.8}{n V_\text{T}} $$

Now to make use of some power-maniupulation rules and substitution.

$$I_{0.8\mathrm{V}}=I_\mathrm{S} e^\frac{0.8}{n V_\text{T}}=I_\mathrm{S} e^\frac{0.7}{n V_\text{T}}e^\frac{0.1}{n V_\text{T}} =10^{-3}e^\frac{0.1}{n V_\text{T}} $$

We need some values for the idealaty factor \$n\$ and the "thermal voltage" \$V_\text{T}\$. Lets try taking the ideality factor as 1 and the thermal voltage as 0.02585 V (wikipedias value for "room temperature")

$$I_{0.8\mathrm{V}}=47.87*10^{-3}=0.04787$$

Now we also need to assume there is no load on the circuit and therefore that the diode current equals the resistor current.

$$ R = V/I = 7.6 / (0.04787) = 158.7 $$

Which is in the same ballpark as your teacher's answer, I suspect they made slightly different assumptions about the thermal voltage or ideality factor.

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