# Op Amp in buffer configuration is decreasing output voltage

I have a 4.24V DC signal which I have to apply a buffer in the purpose of have a high impedance between the 5V signal and Op Amp output. My circuit is like following:

The problem is that my output voltage presents aproximattely 3.8V

Why the Op Amp in buffer configuration is decreasing the output voltage?

PS: I'm using an Op Amp Single Power Supply (I tried with LM324 and LM358)

• Read the Datasheet. Commented Jun 17, 2016 at 19:55

Because you're trying to have the op amp output a voltage higher than it's capable of. A typical upper limit is V+-1.5V, so the fact that you're getting 3.8V out on a 5V supply is already better than that.

Either pick an op amp with a rail-to-rail output or use a supply with a higher voltage.

• Could also be a problem with the common mode INPUT range!I just got bit by an opamp who's common mode INPUT could only reach +rail - 3.5V!!! Absolutely floored me. Thought I could use it, as I was powering at 5V, and referencing +IN to 2.5V. Who woulda guessed I couldn't work at rail - 2.5 V. This particular amp was rail to rail on the output. Commented Jun 17, 2016 at 21:38
• @ScottSeidman Yes, my answer to this question points out that there's a 1.5V headroom requirement on the input side of these op amps as well.
– Null
Commented Jun 19, 2016 at 3:37

You have to actually read the datasheet before using a part. Here is the relevant snippet from the LM324 datasheet:

The middle column of numbers applies to the LM324. With 30 V power, it can only go to 26 V output, meaning it requires 4 V of headroom. The datasheet isn't very clear what the headroom requirement is with 5 V power, but the 1.2 V you see should be no surprise.

There is a similar limitation on the input common mode voltage range, with is 0-28 V with 30 V power. Your 4.24 V in may be violating the limit for 5 V power.

• The headroom voltage in most op amps is approximately independent of the supply voltage, but of course it strongly depends on the output current. The 4 V headroom voltage is specified with a load resistance of 2 kohm, which at 26 V draws a current of 13 mA. Probably the OP measured the output voltage at no load, and got a much lower headroom voltage. Commented Jun 18, 2016 at 13:22

As others have already mentioned, the output voltage of a real op amp can never reach (be equal to) the voltages on the power supply rails (+VCC or -VEE). Looking at the schematic diagram for a real op amp, one can understand by inspection why this is.

Figure 1 is a schematic diagram I copied from the National Semiconductor LM324 data sheet (dated August 2000):

Figure 1. National Semiconductor LM324 Op Amp Schematic

The op amp's output stage is comprised of the Darlington transistor pair Q5 and Q6, resistor $R_{SC}$, and transistor Q13.

When current flows through resistor $R_{SC}$, Ohm's Law applies—i.e., there will be a voltage drop across $R_{SC}$ that is directly proportional to the amount of current flowing through it. If the input signal tries to drive the OUTPUT voltage toward $V^{+}$, and if the load circuit connected to the OUTPUT pin pulls a lot of current, then Q6 is driven into saturation (or close to it), presumably Q13 is cutoff (or close to it), and the maximum voltage at the op amp's OUTPUT pin (relative to ground) is approximately

$$V_{OUTPUT} = V_{load} = V^{+} - V_{Q6} - V_{R_{SC}}$$

(n.b. I am assuming the load circuit is connected between the op amp's OUTPUT pin and ground.)

For the conditions mentioned above, the voltage drop across transistor Q6 ($V_{Q6}$) is the collector-to-emitter saturation voltage, and the voltage drop across resistor $R_{SC}$ ($V_{R_{SC}}$) is determined by the amount of current flowing through that resistor (Ohm's Law). If the load circuit connected to the op amp's OUTPUT pin pulls lots of current, then by inspection one can see that the maximum voltage level at the OUTPUT pin will be considerably less than $V^{+}$.

If you look at @OlinLathrop's answer to your question, note that on his data sheet example the Output Voltage Swing $V_{OH}$ spec—i.e., the maximum positive ("high") voltage swing at the OUTPUT pin—is defined as a function of $V^{+}$ AND load resistance $R_{L}$ (see Fig. 2).

Figure 2. LM324 "Output Voltage Swing" specification

This is because the load resistance plays a big role in determining the amount of current that flows through resistor $R_{SC}$, and therefore the voltage drop in the op amp's output stage.

• That's a really good observation of the transistors in the schematics. Commented Jun 17, 2016 at 23:14

Both of the op amps you used -- LM324 and LM358 -- have an input common mode range requirement that the input can't be less than $V_{\text{CC}} - 1.5\text{V}$ away from the $V_{\text{CC}}$ rail. Additionally, these op amps don't have a rail-to-rail output -- they can't drive the output to closer than $1.5\text{V}$ from $V_{\text{CC}}$. You can see both of these specifications on the LM324 datasheet, for example:

With $V_{\text{CC}} = 5\text{V}$ both your input and output exceed these ranges.

The easiest way to fix this may be to increase your supply voltage. If that's not possible you need to find an op amp which can deal with these ranges.