Well, I am new here, and new to this hobby and I do not know if this the right sub-forum to post my question to.
I have constructed this circuit on a prototyping board:
https://www.circuitlab.com/circuit/875t83/receiver_1_4/
Each amplifier works fine as long as nothing else is connected to the chip. Once I connect a single resistance to the chip, the output of the amplifier goes high. The output on one of the amplifiers rises to 0.5 V while the second falls to -0.5 V. Both work properly but with the mentioned offsets. Now my question(s):
1-Why does this happen? 2-Can I avoid these effects or.. 3-That is normal and I have to eliminate these offsets, or... 4-Probably use two single OP-AMP chips instead
Thanks in advance
PS. I tried using TL084 instead of the LM324: same behavior.

 

Your described behavior is abnormal. I see no reason they should not all work properly. Are you sure all grounds, including the power ground are connected together?

Edit: What is the purpose of the 1 megohm resistors across the power supply?

 

Edit: What is the purpose of the 1 megohm resistors across the power supply?

Voltage divider to create ground.

 

Voltage divider to create ground.

Your supplies are already connected to ground. The resistors serve no purpose.

 

Well, I am new here, and new to this hobby and I do not know if this the right sub-forum to post my question to.
I have constructed this circuit on a prototyping board:
https://www.circuitlab.com/circuit/875t83/receiver_1_4/
Each amplifier works fine as long as nothing else is connected to the chip. Once I connect a single resistance to the chip, the output of the amplifier goes high. The output on one of the amplifiers rises to 0.5 V while the second falls to -0.5 V. Both work properly but with the mentioned offsets. Now my question(s):
1-Why does this happen? 2-Can I avoid these effects or.. 3-That is normal and I have to eliminate these offsets, or... 4-Probably use two single OP-AMP chips instead
Thanks in advance
PS. I tried using TL084 instead of the LM324: same behavior.

What value is the resistance and how are you connecting it to the chip. ?

 

Have you calculated the gain of the two stages? What output voltage would you expect if not constrained by power supply voltage and operational limits of the amplifiers for a 0.5v input?

 

Have you calculated the gain of the two stages?

Then multiply the gain by the input offset voltage of the LM324 which is typically 2mV, maximum 9mV @ 25°C.

Timescope

 

Voltage divider to create ground.

As crutschow mentioned, you don't need these provided you have things wired as shown. But the fact that you describe their purpose this way makes me want to verify that you really do have the "ground" you are using for your signals tied directly to the middle junction of the battery stack.

 

Your supplies are already connected to ground. The resistors serve no purpose.

Yes you are right. It is a mistake in the sketch. Actually I use a power supply (transformer) in reality and use the resistances to create the ground.

 

As crutschow mentioned, you don't need these provided you have things wired as shown. But the fact that you describe their purpose this way makes me want to verify that you really do have the "ground" you are using for your signals tied directly to the middle junction of the battery stack.

That was a mistake in the sketch, not in reality. Thanks

 

Then multiply the gain by the input offset voltage of the LM324 which is typically 2mV, maximum 9mV @ 25°C.

Timescope

The gain is about 100. I am using 100 K Ohm feedback resistance and 1 K Ohm connected to the inverting terminal as shown in the schematic. 2-9 mV offset leads to 0.2-0.9 V, which agrees with my case: 0.5 V. The strange point though is this offset vanishes completely when I use one amplifier only and nothing else is connected to the chip, and appears once a component (resistance) is connected. The other strange point is that I get 0.5 V (amplified) offset on one amplifier and -0.5 on the other. Assuming these offset voltages are unavoidable, how can eliminate them in the circuit? I mean what to do to compensate these values and reach zero level again? Should I put a diode with 0.5 V drop? or is it better to use a voltage divider?

 

Yes you are right. It is a mistake in the sketch. Actually I use a power supply (transformer) in reality and use the resistances to create the ground.

That was a mistake in the sketch, not in reality. Thanks

What was a mistake in the sketch? Having the resistors there or having the junction between them tied to the midpoint of the supplies?

Are you using some kind of single power supply and using the resistors to create your ground or are you using some kind of bipolar supplies with the ground tied directly to the midpoint of the supplies, making the resistors extraneous? These are two very different things -- which is one of the reasons why it is very helpful to see actual schematics for the circuit you are actually working with.

If you are using the resistors to create your ground, then consider that an imbalance of just 1μA will move your "ground" by 1V.

You need a low impedance ground. One thing you might consider is using one of the other opamps to buffer your ground reference by configuring it as a voltage follower and using the ground reference created by your resistors as in the input and using the output as your ground.

 

The gain is about 100. I am using 100 K Ohm feedback resistance and 1 K Ohm connected to the inverting terminal as shown in the schematic. 2-9 mV offset leads to 0.2-0.9 V, which agrees with my case: 0.5 V. The strange point though is this offset vanishes completely when I use one amplifier only and nothing else is connected to the chip, and appears once a component (resistance) is connected. The other strange point is that I get 0.5 V (amplified) offset on one amplifier and -0.5 on the other. Assuming these offset voltages are unavoidable, how can eliminate them in the circuit? I mean what to do to compensate these values and reach zero level again? Should I put a diode with 0.5 V drop? or is it better to use a voltage divider?

The input offset voltage can be either possitive or negative. Isn't your gain about -100? It's an inverting amplifier, correct (I don't have the circuit up in front of me)? Try putting a 1kΩ resistor between the non-inverting input and ground.

 

Are you using some kind of single power supply and using the resistors to create your ground

Exactly

which is one of the reasons why it is very helpful to see actual schematics for the circuit you are actually working with.

The sketch has been corrected in the link above.

You need a low impedance ground. One thing you might consider is using one of the other opamps to buffer your ground reference by configuring it as a voltage follower and using the ground reference created by your resistors as in the input and using the output as your ground.

Thanks foe the hint. I will try it today.

 

The input offset voltage can be either possitive or negative. Isn't your gain about -100? It's an inverting amplifier, correct (I don't have the circuit up in front of me)? Try putting a 1kΩ resistor between the non-inverting input and ground.

The circuit can be seen in:

https://www.circuitlab.com/editor/#?id=875t83

Actually it is a non inverting arrangement with a gain of about 100 (100 k ohm and 1 k ohm resistances). When I use one op-amp only of the four and nothing else connected to the chip (except power), I get almost zero offset and the amplifier is working fine. Whenever I connect any additional component to the chip, that amplifer goes mad. I ignored it and went on to build the circuit as shown - two amplifiers used and two terminated as shown. What I finally got is offset of 0.5 V on one amplifier, and -0.5 on the other.

 

Besides make the virtual ground a much lower impedance you also need decoupling capacitors from each power rail to your virtual ground point.

 

The circuit can be seen in:

https://www.circuitlab.com/editor/#?id=875t83

Actually it is a non inverting arrangement with a gain of about 100 (100 k ohm and 1 k ohm resistances).

Try inserting a 1kΩ resistor in series with the input signal.

But I suspect your biggest problem is the ground impedance.

 

Besides make the virtual ground a much lower impedance you also need decoupling capacitors from each power rail to your virtual ground point.

But go easy on them. If you use too large a value you might make the opamp break into oscillation. How easy you need to go depends on the opamp. There are ways to counter this tendency if it turns out to be an issue.

 

But go easy on them. If you use too large a value you might make the opamp break into oscillation. How easy you need to go depends on the opamp. There are ways to counter this tendency if it turns out to be an issue.

It is working much better now after using one amplifier as a buffer. I added .1 uF decoupling capacitors as recommended, but have not noticed much difference. Can I use the same circuit with 4 AAA cells instead of the power supply? Problem number 2: I noticed 0.3 V offset at the output of one of the amplifiers. After inspecting the circuit I discovered that the two inputs are not identical. One of them is 1 mV while the other is 9-10 mV, which is amplified to the mentioned .3 V. The final objective is to compare the two amplified signals. Is there a way to eliminate this offset?

Thanks in advance

 

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Problem number 2: I noticed 0.3 V offset at the output of one of the amplifiers. After inspecting the circuit I discovered that the two inputs are not identical. One of them is 1 mV while the other is 9-10 mV, which is amplified to the mentioned .3 V. The final objective is to compare the two amplified signals. Is there a way to eliminate this offset?

You can cancel it by summing in an opposite voltage at the input. This can be adjusted by using a 10kΩ pot between the two supplies and connecting the wiper through a resistor to the op amp input. That requires stable power supplies to maintain a stable offset.

The alternative is to use a better op amp with lower offset.