# Diode limiter circuits

I am given the following circuit and told to assume that the resistance of the potentiometer is zero. I am asked to plot the characteristic vo=f(vi), indicating the conduction state of the diodes.

Shouldn't D1 block vi due to its direction? I assumed so, and therefore I found vo from the series of the two resistors and D2.

simulate this circuit – Schematic created using CircuitLab

I assumed that the voltage across D2 was 0.7V, and therefore to find vo, I did the following:

• (10 - 0,7)/10k = 0,93mA
• vo = 0,93mA * 10k = 9,3V

• Set Vi to, for example, 1 V, or 2 V. What's the voltage at the junction of D1 and D2 now? Plotting V0 as a function(Vi) means setting Vi to various values, and seeing what V0 is as a result of that. Apr 22 at 15:51
• This kind of circuit will be piecewise linear, so not very much of a curve. Think about the points of inflection (where the current through a given diode just drops to zero. For example, if Vf of the diodes is 0, then there will be an inflection point at Vi = 5.0V. For Vi < 5.0V, D1 will conduct. You can ask the same question about D2. If your hw requires non-zero Vf then you will have to include that detail in the calculations. Apr 22 at 16:28
• Click "simulate this circuit" and hover the mouse over the schematic to check your calculations. Draw another schematic for the other situation - when the input voltage is low. Apr 22 at 21:12
• Just looking at the complete circuit, assume what happens when Vi is at 0V (or ground). Your equivalent circuit (assumption) would only be valid when D1 is reverse biased, (but it could also be forward biased). Even then your voltage and current calculations are missing the fact that there are "two" 10k resistors (and a diode) in series. For the whole circuit you might assume that you want to plot Vo as Vi ranges from -5V to +5V.
– Nedd
Apr 23 at 1:32

You got off to great start, ignoring D1 for the time being, and focussing on that chain of R1, D2 and R2.

Before I address where your calculation failed, let me do a more formal analysis, to establish the truth:

KVL says that all the voltages across R1, D2 and R2 must add up to the total voltage between the two ends of the chain, 10V:

$$V_{R1} + V_{D2} + V_{R2} = 10$$

KCL tells us that the current down the entire chain (as long as we ignore current branching away via D1) is the same in every element, which I can write as:

$$I_{R1} = I_{D2} = I_{R2} = I$$

The last thing we do is apply Ohm's law to glue everything together. We can relate $$\I\$$ to $$\V_{R1}\$$ and $$\V_{R2}\$$ like this:

$$V_{R1} = I \times R_1$$

$$V_{R2} = I \times R_2$$

Substitute these last two into the first equation, and also plug in $$\V_{D2}=0.7V\$$:

$$I \times R_1 + 0.7 + I \times R_2 = 10$$

That's enough to find $$\I\$$, which I will leave for you to do. Once you know $$\I\$$, it's trivial to find $$\V_{R1}\$$ and $$\V_{R2}\$$ from the Ohm's law equations above.

The value of $$\v_o\$$ is also found with an application of KVL, although you might not have thought of it in that way. The potential at the top of R2 must be higher than the potential at the bottom of R2, by an amount $$\V_{R2}\$$:

\begin{aligned} v_o &= 0V + V_{R2} \\ \\ &= V_{R2} \end{aligned}

Maybe you can already see where you went wrong. Your initial calculation for current through R1 was incorrect, because you failed to account for all the elements in the chain.

Intuitively, the voltage remaining after the diode has "removed" or "taken up" its share of 0.7V is $$\10V - 0.7V = 9.3V\$$. However this "remainder" is shared by both R1 and R2.

You seem to have assumed that this voltage would appear across only R1, but actually it is shared by all remaining elements.

From that perspective, current flowing must be due to 9.3V across R1 and R2:

$$I = \frac{9.3V}{10k\Omega + 10k\Omega}$$

When you know $$\v_0\$$ with D1 absent, you can start thinking about the influence of $$\v_i\$$ and D1.

I'll redraw the circuit with a named node x at the junction of the two diodes:

simulate this circuit – Schematic created using CircuitLab

The potential $$\v_x\$$ is 0.7V higher than $$\v_o\$$:

$$v_x = v_o + 0.7$$

Let me define a value $$\v_{xu}\$$ to be the potential at x with D1 absent. Since $$\v_x\$$ will be varying, I prefer to have a different variable referring to its "unloaded" value, a constant with value $$\v_o + 0.7\$$ when $$\v_i\$$ and D1 are disregarded.

It is only possible for $$\v_i\$$ to influence $$\v_x\$$ (and $$\v_o\$$) when D1 is conducting, forward biased. It's pretty clear that for $$\v_i > v_{xu}\$$ it can't possibly have any influence, since D1 is reverse biased. That portion of the graph must be flat, since $$\v_o\$$ is unchanging.

It's also pretty easy to see that for D1 to be conducting (with 0.7V across it), $$\v_i\$$ must be lower than 0.7V below $$\v_{xu}\$$. In other words, when $$\v_i\$$ drops to $$\v_{xu} - 0.7V\$$ the diode becomes fully conductive.

As $$\v_i\$$ falls further, it drags $$\v_x\$$ with it, such that $$\v_x = v_i + 0.7V\$$. This is the relationship that determines the section of graph $$\v_i < (v_{xu}-0.7)\$$.

Don't forget, you're plotting $$\v_o\$$ vs. $$\v_i\$$, not $$\v_x\$$.

• Isn't that better? Or maybe you have some reason not to (for example, to remain a mystery to the OP)... If so, you can rollback it. Apr 23 at 10:50
• @Circuitfantasist I didn't want to answer his homework completely! LOL I've been scolded for doing that before Apr 23 at 11:23
• Simon, Done! Sorry for the attempt... Apr 23 at 11:49
• So for D1 to be conducting, this has to happen: vi = vxu - 0,7. So i just need to find vxu ( through my calculations vxu = 10 - 0,7 - 4,65-> vxu = 4,65v.) According to those calcules, i can find the maximum value that vi can have in order to allow D1 to conduct, which is vxu - 0,7 = 3,95V. Knowing this, i plotted some values for Vi (from 0 to 4V) in order to find Vo. The result i got was a line simillar to y=x until vi=4v, and after that a horizontal line for vo ( vo= 4,65V, because vo = Vr2), since the value for vo will not change if D1 is not conducting. Is that right? Apr 23 at 14:24
• @Aleat that sounds right, yes Apr 23 at 22:55