Help me with nulling the 741 offset

hi there, please help me

i try to nulling the 741 offset by connect a 10K potentiometer with outside lead of the potmeter connected to the pin 1 and 5 of 741, and the wiper of the potmeter connected to the negative supply as described at :

**broken link removed**

i try to simulate it on multisim, and it can't work, and then i try it on circuit, it also can not work, please help me out.

thank you

741 offset null

If you connect 741's input as shown on the attached picture - the offset null adjustment has to work ..

Regards,
IanP

offset null 741

thanks for the quick reply

i tried it, but i didn't use the negative supply at Vee, i use 0 and +15V supply at Vee and Vcc, it can't work with this configuration, i am not yet tried using the negative supply, i'll try it tomorrow, but, is it possible to set the offset if i using the +15 V and 0 V for the DC input? i did simulate the circuit shown on the attached file, but it can't work on the simulation, i will try it again tomorrow.

i have another question about the logarithmic amplifier, what the logarithmic amplifier would be if we reverse the feedbeck diode? i mean if we connect the cathode of the diode to the inverting input, and the anode to the output pin. is it would let the circuit turn into open loop?

thank you

offset nulling circuit

The offset circuit will work with 0V at -Vcc. If you look at the internal schematic, you will see that the offset adjustment pins are simply connected to the emitters of two transistors, which are then grounded through two resistors. All you are doing is connect the two sides of the pot in parallel with each of the resistors and by adjusting the pot you adjust the effective resistance each transistor sees to -Vcc (or GND in your case).

However, when you adjust the circuit you have to make sure both inputs are connected not to GND, but to a voltage (the same for both), which is within the common mode range of the opamp which needs to be at least 1.5V higher than -Vcc (I think, but check the datasheet to be sure.)

741 offset

thanks,

i just tried it using +15 for Vcc and -15 for Vee, and i connect the circuit as shown on the attached file, but still it can't work, i can't set the offset to 0, it just can be changed between +15 to -14, i tried to connect the input + and - to the Ground, and to the non ground, but the result is still the same, the offset can't be set to 0. i didn't find any offset nulling circuit inside the datasheets of 741. are there any other methods to nulling the offset?

thank you all

offset nulling

i am sorry, mistyped on this sentence :

it just can be changed between +15 to -14

must be :

it just can be changed between +15 and -14

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the circuit shown in the attached schematic was open loop, i tried to design the circuit in close loop by adding a negative feedback from output to inverting input, and i connect the V+ and V- to the ground, when i didn't connect the pin 1 and 5 to the 10K pot, it gave me Vout ≈ 4.5 Volt on Vcc ≈ +16V and Vee ≈-16 V.

when i connected the pin 1 and 5 to the pot, and from pot to the Vee, i can set the offset (4.5 V) of the circuit until 0 V, but it can't work on open loop circuit, is it because the open loop gain is very high, and since the pot isn't very precision, only a small differential voltage appeared on the both V+ and V- , the output goes to Vcc or Vee?

since the offset nulling circuit works on close loop circuit, but doesnt work on open loop circuit, does the offset nulling circuit from the link still considered to be correct? because if we connect a load to the output of that circuit, the load would act as feedback to the input, and the open loop circuit on that picture would become close loop. but, without the load, the offset still can't be set.

please help me answer the question about log amp as shown in the messages above..

thank you

offset 741

Try using a high resistance [R2] between 6 and 2, and a low resistance [R1] between 2 and gnd. The gain of the system is then ~R2/R1, and you can easily adjust the voltage at 6 to be zero! But be sure that the GAIN is not too high [not more than say thousand].

About the log amp: the unit will work for negative voltage in put, but will behave as a comparator [OPEN LOOP gain being high] for positive in put voltages.