# Op amp is not amplifying a 2 mV signal

I trying to amplify an electrical signal created by electrodes on skin, which create a potential of around 2 mV. Therefore, I am using a LM324N operational amplifier (powered by 9 V battery) and creating the same circuit as the following one:

This circuit was made for a LM358N op amp but some search on google shows that the LM324N can also be used, but can it?

The problem is that I get this signal on output:

I tried reading the raw signal with my oscilloscope, and got that ECG crystal clear, but with a potential of 2 mV.

I tried to substitute the 100k ohms resistor with 10k ohms to have a gain of ten as I can read 10 mV with my oscilloscope but I still have this square signal and not the ECG one amplified.

• where did you get the idea that it should work like that? Commented Aug 27, 2021 at 20:45
• @MarkoBuršič Well, shouldn't the op amp work like that? I'm very new to amplificator, I found about the formula that says V_s = V_e*(1+100k ohms/1k ohms) so it should amplify the signal to one that has an amplitude of 101*1mV = 101mV Commented Aug 27, 2021 at 20:51
• That circuit is wrong. I don't know why they have the right "lead" labelled where it is. That completely bypasses the rest of the circuitry and makes the op amp behave as an amplifier with infinite gain that cannot actually output infinite voltage so saturates at the power supply voltage whenever it does (i.e. that's what a comparator is). That right "lead" label should be at the same location as the ground. Commented Aug 27, 2021 at 20:53
• If measuring an AC signal, it doesn't make sense to short it out on the terminals. Commented Aug 27, 2021 at 21:05
• The opamp is not powered then it does nothing. Depending on how it is powered then its + input is not biased at the correct DC voltage then it does nothing. It is missing a series input coupling capacitor then its input might be at its negative DC supply voltage so it rectifies the input signal. Of course the second "lead" must be connected to the signal ground and 0V of the power supply, not to the - input. Since the input level is low, shielded audio cable must be used at the input. Commented Aug 27, 2021 at 21:15

Welcome Tom -

I think the author intended a connection on the negative terminal. This is usually not inferred when two lines cross and it could be better represented like this:

I couldn't put a picture in the comments, but give it a try.

Good luck!

• Hey John, I made the initial circuit with the connection of the cross and it didn't work either Commented Aug 27, 2021 at 21:22
• I would try removing the capacitor - then maybe put a large resistor to ground from the op amp positive terminal. Commented Aug 30, 2021 at 13:15

This isn't how you go about amplifying a biopotential.

First, since your "answer" can be positive or negative, you need to pay close attention to your supply rails, and you either need a positive and negative supply, or you need to create an offset to amplify around mid-rail. You've done neither.

Second, you need a differential amplifier. You kind of have a differential amplifier, but not a very good one. You can build one with op amps, but the far better option is to use an instrumentation amplifier.

Next, biopotential electrodes have many inherent offsets you need to deal with, and the offsets on different electrodes can be different. In fact, the difference between electrodes (that has nothing to do with ECG) can be big enough to saturate an amplifier with a gain of 100, and that's most likely what you're observing.

A more typical arrangement is a differential amp stage with modest gain, maybe about 5, usually implemented with an instrumentation amplifier, followed by high pass filtering to remove offsets and low pass filtering to remove noise. After that you'll have a higher gain op amp stage, usually about 20, maybe with some additional filtering.

The input common mode voltage range extends below ground, but the output voltage range does not. This means that small output signals will not appear at the output, and large ones will be pseudo-half-wave rectified.

First, you don't show any power or decoupling connections. Use two 9 V batteries, one for Vcc and one for Vee. Connect your circuit grounds (GND) to the midpoint between the batteries.

Second, add a 1 M resistor from the non-inverting input to GND. This establishes a DC operating point midway between the two power rails, centering the output signal.

The circuit still will be nowhere near a true differential amplifier usually used for this type of interface, but it might do something.

I am not an experienced electrical engineer, so don't take anything I say as necessarily true, but instead check everything. If any reader is an experienced electrical engineer, please check what I wrote. I'll be happy to accept corrections.

First of all, your amplifier circuit looks really weird. The inputs to the op amp are simply labeled "lead", but it's not clear if they are the same signal on two wires; a differential signal, where one is the negative of the other; or perhaps one is a signal and the other is cable (signal) ground. In any case, this circuit will not work, as several others have already pointed out:

• In a (non-differential) amplifier, there is only one input signal, not two. In your case, it looks like that ought to be the left lead. Connecting both leads (or one lead twice) to the inputs of an op amp is weird (outside of differential amplifiers).
• I'm going on about differential amplifiers because it seems to me that an ECG lead might well pick up common-mode noise, i.e., noise that affects both the signal lead and the signal ground equally and in the same direction, and differential amplifiers are very good at getting rid of such noise. But the picture you show is not a differential amplifier.
• The feedback voltage seems to be added to the right lead without a coupling capacitor. That means that if the feedback voltage contains a nontrivial DC component (as it seems it should in your case, see below), current could flow into your lead. If there aren't any protections in place in the ECG probe to prevent that from happening, the result could be unpleasant.

Then you say that your signal has a "potential" of 2 mV, but the picture on your oscilloscope shows a hefty DC bias of nearly 25 mV (if I'm reading this correctly) with a 2 mV signal riding on top of it. Do you want to amplify that bias as well, or just the raw AC signal?

What you also don't show is how you have connected the power. Have you connected the positive rail to the positive terminal of the battery (+9 V) and the negative to the negative terminal (0 V)? Or have you created a balanced supply of +/-4.5 V with a divider?

If you want to amplify the DC bias along with the signal in a non-differential amplifier, you can use the circuit depicted here. Don't worry if the op amp shown is not what you have. It's a bog-standard noninverting amplifier with gain of 101.

However, if I'm right about that +25 mV DC bias, that will be amplified as well. With a gain of 101, your output signal will have a bias of about +2.5 V, with a 0.2 V signal riding on top of that. If you want to remove that, put a small capacitor on the output. You can use an electrolytic cap, just make sure that the polarity is right.

I've put a bypass capacitor on the power connection in the circuit, which might or might not be needed, depending on what else that battery is powering.

• You’ve accidentally drawn shorts across your (probably-unnecessary) decoupling caps. The + opamp input needs a path to a voltage reference or ground so that its bias current has a place to flow. The entire voltage swing at the input of the opamp needs to end up between the rails of the opamp so either have to bias up the input above ground, or add a negative supply for the opamp. Commented Sep 6, 2021 at 19:37
• The leads originally drawn in are most likely two different leads. Commented Sep 6, 2021 at 20:28
• Thanks @td127 for pointing out the shorts. I have also removed the superfluous cap to ground. The signal seems well biased above 0V (it's range seems about 24-28 mV), so a negative supply seems not to be needed. Commented Sep 6, 2021 at 21:06
• @ScottSeidman: I thought so too, but then the circuit makes even less sense to me than it does when assuming that only one signal is to be amplified. Commented Sep 6, 2021 at 21:15

To build up an ECG machine with just an operational amplifier is utterly absurd. You need several instrument amplifiers and several electrodes in order to correctly deduce the signals from heart. In case you want to build one by yourself maybe you find some good ideas here: https://www.mikrocontroller.net/articles/EKG_mit_XMC_%C2%B5C

In case you wanted just to play with an opamp I think the closest schematic to your proposal is the non invertig opamp see below.

simulate this circuit – Schematic created using CircuitLab

Adding to Scott's response, the amplification of biopotentials requires converting ionic currents occurring at the skin interface to an electric current. This is typically done through Ag/AgCl (Silver/Silver Chloride) electrodes (those patches that the cardiologist puts on your chest when you get an EKG/ECG). In terms of the electronics conversion circuit, an instrumentation amplifier is the most 'affordable' and 'reliable' way to accomplish this. For instance, the INA128/29 or similar can be configured as the following image shows:

Additionally, care should be taken when powering such a circuit. A battery pack or an insulated power supply is mandatory when testing it in yourself or somebody.