Problem with a diode-circuit design

Question

I have a problem with a diode circuit design problem:

[Source: Microelectronic Circuit Analysis and Design.]

Basically a circuit where the current through the Si diode is linear if:

\begin{equation} 2V < V_{in}<10V \end{equation}

And:
\$i_D = 0mA\$ if \$V_{in}=2V\$,
\$i_D=10mA\$ if \$V_{in}=10V\$

My attempt:

If V1 is 2V, then the voltage drop across R1 should be 1.3V and all the current flows through R2, which are in series, so values for R1 and R2 can be taken simultaneously. But if Vin = 10V, I don't know what to do.

Answer 1

Kind of hoped you'd write up what you worked out in comments. (Very good!) But as I already have the image prepared, I will write up your thoughts below.

Yes, as you worked out you need three diodes (each dropping \$700\:\text{mV}\$) in order to prevent current until the voltage reaches \$2.1\:\text{V}\$. After that, you have a slope of about \$800\:\Omega\$. Just as you worked out.

So the circuit would look about as you see on the left, below, with the resulting transfer curve shown on the right of it:

Now, your curve actually shows an exact \$10\:\text{mA}\$ at exactly \$10\:\text{V}\$. So, technically, the resistor should be \$\frac{10\:\text{V}\,-\,3\,\cdot\,700\:\text{mV}}{10\:\text{mA}\,-\,0\:\text{mA}}=790\:\Omega\$.

So you may want to use that value, instead.

But the result is that you demonstrated a proficiency in fully understanding how to approach your problem and then solve it. Excellent!

Answer 2

Solution could be something like this ... Ideal conditions.

where R1 = ~ 800 Ohm and R2 = ~ 431 Ohm ...

Simulation could say something like this ... Diode generic.