# Solving this op-amp circuit

I would appreciate some help solving the following circuit. I've been trying for a very long time and can't get the same answer the book has. I'm trying to get $V_o$ in terms of $I_{os}$.

The answer is $V_o = I_{os}R_2$ when $R_3 = R_1 || R_2$.

My attempt is simply:

$$V_+ = 0.5 I_{os} R_3 \\ V_- = -0.5 I_{os} R_1 + V_o \frac{R_1}{R_1 + R_2}$$

since $V_- = V_+$, $$0.5 I_{os} R_3 = -0.5 I_{os} R_1 + V_o \frac{R_1}{R_1 + R_2} \\ V_o = (R_1 + R_2) R_3 I_{os}$$

But this is wrong. Even in the simulation that I conduct, I see that $R_1$ has no baring on $V_o$. What am I doing wrong? simulate this circuit – Schematic created using CircuitLab

• I don't think the circuit matches the final formula and also you have "lost" R3 in your equations at the point you equate -V and +V. EDIT - OK you've fixed that now – Andy aka Mar 30 '13 at 14:30
• To get such a simplified result, there must be relationships between the various R's that you've not given us. I suggest that $R_1=R_3=2R_2$ – placeholder Mar 30 '13 at 18:05
• and if I do use those values I get $i_{os}R_3$ or $i_{os}2R_2$ by inspection. – placeholder Mar 30 '13 at 18:13

The answer is actually pretty simple :

Voltage difference between $V_+$ and $V_-$ is zero. No current flows into the opamps input. Output voltage is the sum of opamp input voltage and voltage across resistor R2. Also, output voltage is dependent on R1.

*NOTE: Ix is the current source between the two input pins. If your source is 0.5Ios, simply substitute the Ix with 0 5Ios. Thanks goes to Andy_aka who caught my nomenclature mistake.*

Thus:

$V_+ = R_3 * I_{X}$

$V_- = V_+$

$I_1 = V_+\ / R_1$

$I_2 = I_1 + I_{X}$

$V_{R2} = R_2 * I_2$

$V_O = V_{R2} + V_{R3}$

or

$V_O = I_{X} * (R_3 + R_2 + R_3 * R_2 / R_1)$

• I dont understand this V+ = R2 * Ios. Could please elaberate n explain this? – Durgaprasad Mar 31 '13 at 7:22
• That's easy to explain - when I copied down the circuit I mixed up R2 and R3. Fixed. – SunnyBoyNY Mar 31 '13 at 12:39
• It's not Ios - it's 0.5*Ios that is the constant current – Andy aka Mar 31 '13 at 14:12
• That's just semantics, really. I take the current source as Ios. The output voltage is directly proportional to the current anyway. The OP wants to understand how to derive the equation for output voltage and this is exactly what he needs. – SunnyBoyNY Mar 31 '13 at 14:43
• Second thought -- you're right, my answer was confusing. I have changed Ios to Ix to make it clear. Thanks. – SunnyBoyNY Mar 31 '13 at 15:11

Visualize the op-amp as an inverting stage. The - input is a virtual earth (almost: it is off 0V because of the small voltage at the + terminal). Displacement notwithstanding, it is a mixing point. From the point of view of the - terminal, the current source is just an extra input which feeds the mixing point. Because the mixing point is a voltage source, R1 and the current source isolated from affecting each other.

Let's draw a schematic for this concept. The current that drawn out of the mixing point at the - terminal is pumped into R3, but that is irrelevant and we can separate that by splitting the current source into two.

If the current sources are removed, the behavior of the circuit is very clear. The + terminal is grounded, no current flows across R1 because both ends are at 0V, and the output is 0V.

Next, we can think about happens if the two current sources are added one at a time.

When the second current source, the one feeding R3, is introduced, the effect is that the mixing point is lifted from 0V. Because that happens, there is now a potential difference across R1, since its other end is grounded, and a current flows. The same current flows across R2, creating a potential difference there, which gives us $V_o$.

Then, when the current source in parallel with R1 is introduced, there is an additional current flow at the mixing point. This additional current does not disturb the voltage of the mixing point. It simply adds to the current that flows across R2, which adds to the R2 potential drop. We can simply adjust the previous $V_o$ with that drop.

R1 is relevant with regard to the displacement of the + terminal voltage. R1's top is pinned to ground, and so there is a current flow from its bottom which contributes to the flow across R2. If we remove R1, we remove that current and the output changes. R1 is not relevant to the effect of the top current source, because it is in parallel with it. That source adds a certain amount of current to R2, and therefore adds to the voltage drop, regardless of how much R2 current already comes from R1.

Here's how I ended up doing it: \begin{align*} V_o &= I_{R_1} R_1 + I_{R_2} R_2 \\ &= R_1 \left( \frac{1}{2} I_{os} \frac{R_3}{R_1} \right) + R_2 \left(\frac{1}{2} I_{os} + \frac{1}{2} I_{os} \frac{R_3}{R_1} \right) \\ &= \frac{1}{2} I_{os} \left( R_3 + R_2 + \frac{R_3 R_2}{R_1} \right) \end{align*} now if $R_3 = R_1 || R_2 = \frac{R_1 R_2}{R_1 + R_2}$, \begin{align*} V_o &= \frac{1}{2} I_{os} \left( \frac{R_1 R_2}{R_1 + R_2} + R_2 + \frac{R_1 R_2}{R_1 + R_2} \frac{R_2}{R_1} \right) \\ &= \frac{1}{2} I_{os} \left( \frac{R_1 R_2}{R_1 + R_2} \left( 1 + \frac{R_2}{R_1} \right) + R_2 \right) \\ &= \frac{1}{2} I_{os} \left( \frac{R_1 R_2}{R_1 + R_2} \left( \frac{R_1 + R_2}{R_1} \right) + R_2 \right) \\ &= \frac{1}{2} I_{os} ( 2 R_2 ) \\ &= I_{os} R_2 \end{align*}

I did make a mistake in my question. I did not realise that $R_3 = R_1 || R_2 = \frac{R_1 R_2}{R_1 + R_2}$ so that is why I'm not choosing my answer is the correct one @SunnyBoyNY had it right given the information that was presented in the question.

Just for people's interest, in order to minimise the bias current of an op-amp (not the offset current), $R_3 = R_1 || R_2$. Making $R_3 = R_1 || R_2$ will make the output, $V_o$, independent of the bias current; however, it will not remove the effect of the offset current. Nonetheless, it is better to zero the affect of the bias current (than the offset current) on the output because it generally has a greater affect on the on the output than the offset current.