0
\$\begingroup\$

I'm building a simple non-inverting amplifier using UA741,The gain should be of about 11,and the input is 500m v dc so the output should be in the range of 5.5 volts however i'm getting about 0.45 volts.I'v tried changing the bais voltage many times from 15 to 32 volts but the result is the same. The Circut

PLease Help....

\$\endgroup\$
2
  • \$\begingroup\$ Seems like you forgot to connect battery's negative to ground. \$\endgroup\$ Nov 3, 2016 at 17:57
  • \$\begingroup\$ @RohatKılıç connecting the neg of the battery to ground won't help - it's not a rail to rail device (it's a dinosaur actually). \$\endgroup\$
    – Andy aka
    Nov 3, 2016 at 18:01

2 Answers 2

1
\$\begingroup\$

You need to ensure that the inputs of the 741 (any op-amp, in fact) are within the input common mode range. On the 741, that range is within 3V of each supply rail, according to the datasheet. The output also cannot swing all the way to either supply rail.

If you use supplies of +/-15V (the common must be grounded) then the circuit should work, however you show only a single power supply of +30V with no ground reference. If it is, in fact, floating as you show the circuit will not work, and it will not work if either side of the 30V supply is grounded. You also cannot use a voltage divider because current flows to the ground through R1. You could use another op-amp connected as a unity-gain buffer and a pair of equal value resistors (eg. 10K each) as a 'rail splitter'.

schematic

simulate this circuit – Schematic created using CircuitLab

Or swap the ancient LM741 out for half an LM358 and ground the (-) of the 30V supply and things will be okay (for positive inputs only).

\$\endgroup\$
4
  • \$\begingroup\$ Thanks a lot that really worked....But the gain should be 11 and in the output i'm getting 3.4 volts.How is that...?? \$\endgroup\$ Nov 3, 2016 at 19:37
  • \$\begingroup\$ What really worked? Something is obviously very wrong. \$\endgroup\$ Nov 3, 2016 at 19:41
  • \$\begingroup\$ Do you have any idea of what i should do now..? \$\endgroup\$ Nov 3, 2016 at 19:54
  • \$\begingroup\$ Maybe document the actual circuit you built and what voltages you measure at each node \$\endgroup\$ Nov 3, 2016 at 20:07
1
\$\begingroup\$

Earth (aka ground, aka 0 volts) is normally associated with the mid-point of the power rails so, either split the 30V in two halves and call the middle node "earth" or make a mid rail by using two resistors forming a potential divider (R1 can also become 2 resistors; a 2 kohm resistor to battery negative and a 2 kohm resistor to battery positive).

Alternatively (for R1) connect it to either rail via a high value capacitor (assuming that the application is for AC signal amplification such as audio): -

enter image description here

In the circuit above the input can be referenced to the most negative end of the battery - R3 and R4 form a rail splitter centreing the non-inverting input. C2's impedance needs to be significantly smaller than R1's resistance at all signal frequencies.

Here's the "inverting" version of the same circuit: -

enter image description here

Pictures taken from here and that site will probably be worth studying to learn other op-amp basics.

See also Choosing resistor values for Op-amp and inverting/non-inverting configuration for a similar configuration

\$\endgroup\$
8
  • \$\begingroup\$ I'm not sure I follow your question @Spehro \$\endgroup\$
    – Andy aka
    Nov 3, 2016 at 18:16
  • \$\begingroup\$ @Spehro 2x 2k resistors splitting the rails is the same as 1x 1k resistor to the midpoint voltage. If DC isn't an issue a capacitor to either rail and keeping the original 1k resistor will also work. The OP still needs to generate a midpoint for the the input. \$\endgroup\$
    – Andy aka
    Nov 3, 2016 at 18:22
  • \$\begingroup\$ I'm not sure i understand you perfectly...so can you attach a schematic design of what i should do. \$\endgroup\$ Nov 3, 2016 at 18:32
  • \$\begingroup\$ @MohamedRushdy done. \$\endgroup\$
    – Andy aka
    Nov 3, 2016 at 18:41
  • \$\begingroup\$ This won't work for low frequency inputs, obviously. \$\endgroup\$ Nov 3, 2016 at 18:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.