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Topic: LM741 (Read 6108 times) previous topic - next topic


I have a problem using LM741 for an non-inverting amplifier.
The schematic is attached.

A little explanaition :

The main problem is that i need to power the lm741s with a single 5 V power from the arduino. The second lm741 is the one who will amplificate the signal received from the sensor (here represented as a power supply).
The first lm741 instead it's configured as buffer, his job is putting the input pins of the lm741 at a 2,5 volt reference value, that is the reference of the second lm741.

Now the issues i have :

Depending on the gain I choose for the amplifier the lm741 will saturate at different voltages, for example

R_2 = R_1 -> gain = 2 -> high saturation voltage = 4,7 ; low saturation voltage = 0,9

R_2 = 10 * R_1 -> gain = 11 -> high saturation voltage = 4,2 ; low saturation voltage = 1,8

R_2 = 100 * R_1 -> gain = 101 -> high saturation voltage = 4 ; low saturation voltage = 3

So the saturation values change depending on the gain, note that in some case the reference value of the second lm741 (that should be 2,5 V) it's not included in the saturating range. This lead to the second issues, if instead of the sensor i put an external voltage supply, that provide 0,1 V i expect the lm741 to saturates high, and instead saturates low. I think it could be because the reference value of the second lm change as the saturation voltages do.

What could be the problem? Maybe the supply power i provide it's not sufficient?

here's the datasheet of my lm741: http://www.ti.com/product/ua741

(Pls note that the name of the op amps in the schematic is not matching the one im actually using)

Tell me if you need some more infos.

Thank you in advance.


I am sure there is a better op-amp for this. The LM741 was not designed to be used on a 5v single supply. The output will not swing rail-to-rail, and the common-mode input voltage range limitation may cause unpredictable result if exceeded.

What is that .op device on the left? What is its function here?


R1 and R2 set the reference voltage. R3 and R4 set the feedback voltage. Because you've connected R3 to the reference voltage, a change in Vref will also change the amplification (as this is controlled by the feedback voltage). If you want to use this as an amplifier with a 2.5 V offset, connect R3 to GND, not Vref. Then use R3 and R4 to control the gain. An LM741 won't go rail to rail (0-5V) in any case.




@ Surfer tim
the .op time isn't anything, don't care about it


I'm quite sure R_3 must connected to V_ref = 2,5 V:

V- and V+ are the voltages at the - and + pins of the second op amp, then

V- = V+ and I- = I+ = 0 (because of the properties of the op amp)
So if I apply Ohm's law at the node between R3 and R4
{V_ref - (V_ref + V_sensor) } / R3 = (V_ref+V_sensor - V_out) / R4

If you solve for v_out :

V_out = V_ref + (V_sensor ) (1+ R4/R3)

That is exactly what I need to read from arduino analog pin, no?


Sep 29, 2012, 04:02 pm Last Edit: Sep 29, 2012, 04:37 pm by SurferTim Reason: 1
So I have this correct, U2 is just providing the "virtual ground" on a +-2.5v split power supply?

edit: The output will oscillate around 2.5 volts ("virtual ground") with that gain of 2. I don't know about the sensor output or design.

Due to the design of the 741, the output will not get lower than 1 volt, nor higher than about 4 volts.


You don't need U2 to provide an offset: there are other simple ways of doing this, for example, using a divider, or a voltage reference, a diode, or an led, etc.

If you have to stay with a single rail supply, you have to ac-couple the signal. That can create an issue from a frequency response point of view - which may be a non-issue for your application.

I can post a schematic if that helps - hint: configure the opamp into a non-inverting amplifier.


yes it works as you described. The output of U3 it's directly going to analog pin 0

What you mean for "ac-couple" the signal?

I must say that, a part for some noises problem, and the fact that depending on the gain the second op amp changes its saturating level, the circuit works : if i put a led in place of "sensor" the output change if i switch on the lights.


What is that sensor? An led is not a good choice for that either.

If the sensor is polarized (provides only a positive voltage at the '+' pin), you should change the two resistors on the '+' input to U2 so the quiescent output (sensor at zero output) is at about 1 volt. That would almost double the useful range on the analog input.


What you mean for "ac-couple" the signal?

You will need a way to introduce a DC offset to your ac signal and amplifying just the ac signal.


No ac coupling if you want to determine the light level. With ac coupling, you would only be able to determine if the light level was changing, not the current level.



I'm quite sure R_3 must connected to V_ref = 2,5 V:

it should be, my mistake.


I think maybe is better that I read some more about electronics for measurement and things like that :)


A Voltage follower based on an LM662 would be perfect end easy to do. The second amp could provide modest gain and be capable of a 0 - 5V swing... From a 5V supply  because of rail to tail output and the inputs work well from ground to Vcc or nearly so. There are many similar parts most newer ones will be probably better and many for different causes, I just like the 10 mA output current... Got bored one day and made a relay operated timer where the relay was the switched source current for turning off the op-amp... Put an LED on it and let people try to ... go figure..

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