What parameters of a real op amp determine the lowest voltage it can amplify?

I am trying to amplify a 0.5mV AC signal (from around 4kHz-8kHz, generated by an electret microphone) to about 50mV. I am thinking of using an inverting amplifier using LM358 op-amp.

The datasheet says that it has an input offset voltage of 2mV. Does it mean that it is not possible to amplify signals on the order of 2mV?

More generally what are the parameters that I should look for when trying to design amplifier for this kind of application?

• Generally they can amplify all input voltages, however the questions is how big their effect is vs. those of other noises that are present at various places. May 10, 2016 at 8:52
• What about gain? Can it differ too much from ideal? May 10, 2016 at 8:54
• the ideal opamp has infinite gain, so I would expect every physical implementation of that to differ from that ideal. May 10, 2016 at 8:56
• I meant the gain of the inverting amplifier calculated assuming the op amp is ideal. May 10, 2016 at 8:59

What parameters of a real op amp determine the lowest voltage it can amplify?

It's all about signal (desired) to noise (not desired) ratio (SNR). The LM358 has an equivalent input voltage noise of 55 nV per sqrt(Hz). Now that probably sounds confusing but it isn't. Let's say your bandwidth is 10kHz$^1$ - the total noise will be 55 nV x sqrt(10,000) = 5.5 uV RMS.

If your signal level is 500 uV then your SNR is 20 log (500/5.5) = 39 dB.

Is this acceptable? I don't know but it would be fine for a telephone conversation.

Does it mean that it is not possible to amplify signals on the order of 2mV?

No the 2mV figure tells you that with a gain of 100 you will see an output offset voltage (an error) of 0.2V - this shouldn't normally be a problem in an AC amplifier.

$^1$ The onus is on the designer to incorporate filtering that sufficiently removes noise above 10 kHz - for instance a 1st order filter (a simple capacitor across the gain setting feedback resistor) is usually enough but, for this type of filter the "noise bandwidth" will be a bit bigger than that determined by the CR components ($\pi/2$ bigger). In other words a 10kHz filter will have a noise bandwidth of 15.7 kHz and this would raise the noise from 5.5 uV RMS to 6.9 uV RMS.

• 5.5uV will be more then OK. I am going to measure the voltage using ATMega32 ADC with a step of 5mV. I think I can handle the offset in code. May 10, 2016 at 9:14
• Remember the 5.5 uV is amplified by 100 and it doesn't produce an offset - it's noise superimposed on top of your desired signal. You need to grasp that. The 200 mV is an offset (a constant value) but the 5.5 uV (x 100) is noise having frequencies from below 20 Hz to over 10 kHz. May 10, 2016 at 9:29

No, the LM358 will not work for your application as described. But not due to the input offset voltage.

The input offset voltage is the voltage that must be applied across the input terminals for the output to be forced to 0V (or whatever the output is referred to and what you're choosing as ground). That might not be the most helpful description though, so I like to actually see the input offset voltage in the circuit.

The input offset voltage can be thought of as a voltage in series with one (but not both) of the inputs. It doesn't matter which one, and the offset voltage can be of either polarity, and fall anywhere in the range specified in the data sheet. It is a source of DC error, as whatever you amplify will be the voltage you're interested in ± that offset voltage which is in series with one of the inputs. So even if you are amplifying 500µV, and your offset is worst case around 2mV, and you want a gain of 100, then 0V input will result in 0.2V on the output. 500µV on the input will result in 0.25V on the output. That is certainly amplifying the signal, but there is just that error superimposed on it.

So the input offset voltage is simply that, a DC voltage offset that is always present, and will be amplified along with whatever else is at the input. If you care about measuring a specific DV voltage level and need that amplified, then the offset does indeed present a very real problem. However, when amplifying AC signals, it is not that much of an issue. It will reduce your dynamic range and maximum gain, as 2mV of DC voltage will be 2V on the output with a gain of 1000, so that is 2V less range your output can swing, so clipping can occur.

The bad news is this offset voltage is going to be fairly random even between devices that were made on the same wafer. One might be 2mV, another 1mV, who knows. Worse, the offset drifts with time and temperature.

The good news is this contributes to DC error. You're amplifying an AC signal. It will be AC coupled. Remember, the DC offset is just that - a DC voltage that the AC signal will be superimposed on. Blocking the DC component with a capacitor will leave you with the AC signal.

So, with only that in mind, the LM358 would work perfectly. Only, it won't. It's too slow. Sorry. It has a gain bandwidth product of 0.7MHz. That means it's gain falls to 1 at that frequency. Since your gain is 100, it will only be able to amplify signals that are 7000Hz (700,000Hz / 100) or less. Signals above that will see substantially less than 100 times the gain, which means you'll have significant frequency-dependent distortion on the output.

Now, if you can settle for a bit less gain, like 80 instead of 100, then you can retain full amplification over your desired frequency range. In that case, the LM358 will work fine. Well, "fine". It has a lot of cross over distortion, but I doubt that will be a significant degrader of audio fidelity when your input is an electret microphone. That is to say, the LM358 is a fairly crappy audio amplifier, but it works great if your audio is even crappier, which it definitely is going to be coming from a narrow range electric microphone.

Reduce your gain to 80 and don't expect CD quality audio or anything, and then the LM358 should work ok.

Besides gain bandwidth product (which is probably the most important parameter for amplifying AC signals, in that if it is too low, then you simply cannot use that op amp and will need to find a faster one. But as long as you are not asking it to surpass this, it stops being particularly important. It's an all or nothing parameter), things like CMRR (which is how well the op amp will reject signals that are common to both inputs - in other words, all the crap that ISN'T the signal), input voltage noise, and voltage gain into a load are probably pretty important.

Input voltage noise is amplified by your gain just like your signal and offset, so if your input noise is 50µV, that is much more like the hard limit you are imagining offset voltage to be. Any signal that is 50µV or less will just be drowned out in the noise, and there will be random variations of 50µV and everywhere below constantly superimposed on the input voltage, so keep that in mind. I like to imagine noise, offset, all that stuff as voltage sources in series with an input, and indeed, they can be modeled that way. But I find it helps me from an intuitive understanding position too.

Voltage gain, usually measured in seemingly silly units like V/mV, is how much gain you can expect with a certain amount of loading on the output. Many op amps will begin to lose a lot of their amplification if you start demanding they drive too low an impedance load, so just check that to be sure.

Anyway, op amps are thankfully a bit easier to use and there are less critical parameters if you're using them in an AC amplifier application. As long as its not too slow, not too noisy, and the input can drive the input impedance and the opamp's output can drive whatever impedance is loading the output, you generally won't run into any surprises. Things get a lot nastier if you're using them for DC stuff.

• The Gain-bandwidth-product is typically 1.1 MHz. The 700 kHz (worst case) figure applies with an output capacitance of 100 pF and a load of 2 kohm which might be a little excessive for just a signal amplifier. I would definitely consider using a better amp if better performance was needed however. May 10, 2016 at 9:50

Transforming the signal from 0.5mV to 50mV requires a gain of 100. The 2mV of input offset on the opamp will shift the output of the amp by that amount times the gain so in this case could contribute up to 200mV of DC offset to the signal. This offset may be a -200mV or +200mV depending on the polarity of the input offset and the gain configuration of either inverting or non-inverting.

If you can AC couple the signal to the next stage then this offset may not be of much concern as long as the amplified signal stays in the middle of the amplifier output range.

If you require a DC coupled signal then you could select a better grade of amplifier with small maximum offset voltage or you could setup an offset nulling adjustment with a trimpot to cancel out the amplifier offset.

From your question, you apparently a gain of 100.

The $V_{OS}$ term will also multiplied by that gain, so the amplifier output will have a DC offset of +/- 200mV (typical at 25C) but it could be as high as +/- 700mV. Note that the output DC offset could be either positive or negative.

If this offset is of no concern because the 50mV ac signal will not approach the power rails when riding on this offset provided you supply the device with a suitable supply that ensures the output does not approach the positive rail closer than 1.5V and the negative rail by 2V, then you should be fine if you only care about the ac signal in the output.

Note that because the maximum feedback resistance usually used is about 100k, the input resistor would be 1k, and this is the load your input signal would see; you will need to ensure your input signal can drive this load.