1
\$\begingroup\$

I'm trying to use a non-inverting operational amplifier to amplify a voltage by a gain of 2 (using 2 10 ohm resistors). I'm using a 12 volt battery (made by combining 8 1.5v batteries) to power the op amp, and I'm using the 3.3v pin on a Raspberry Pi to connect to the non-inverting input of the op amp.

Here is the circuit on PartSim: https://www.partsim.com/simulator/#99446 My circuit

The output of the op amp is reading 0 volts when I connect the multimeter to the output of the op amp and the ground terminal used on the Raspberry Pi.

To be specific, the op amp I'm using is the UTC3580. (Datasheet)

Why is the op amp outputting 0 volts when it should be outputting around 6.6 volts?

\$\endgroup\$
  • 2
    \$\begingroup\$ What is Ioh max rating? Also 10 Ohms is not appropriate for OA feedback. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Dec 30 '17 at 21:57
  • 1
    \$\begingroup\$ Why are you using 10 Ohms for OA feedback? I doubt that your op-amp can drive that much current into the feedback network to begin with... \$\endgroup\$ – ThreePhaseEel Dec 30 '17 at 21:59
  • 2
    \$\begingroup\$ With 10 ohms resistors there's a significant chance that you destroyed the opamp as well. Next time use 10 kohm resistors (that's a 1000x higher value) unless it is clear you need lower or higher values. \$\endgroup\$ – Bimpelrekkie Dec 30 '17 at 22:27
  • 3
    \$\begingroup\$ VTC unclear until there's a schematic. A link to some proprietary javascript-only page is not a substitute and will most likely go away in the near future, making this question pretty pointless. \$\endgroup\$ – pipe Dec 31 '17 at 12:42
  • 1
    \$\begingroup\$ @pipe I just edited and included a screenshot of the circuit, didn't think of that, thanks! \$\endgroup\$ – Tavon Jan 1 '18 at 18:58
5
\$\begingroup\$

As well as having a grossly too low value in your feedback circuit, you must connect the V- to ground. As others have said, 10K is a reasonable starting value for an LM358/LM324 type of op-amp. There are reasons to go much higher and much lower, but for starters this will do.

The green arrows show the (main) current path through the feedback divider. The red lines show the connection which already exists inside the RPi and the connection to V-.

enter image description here

Without the supply connection, the output cannot push current through the divider and you'll get about 0V across the resistors.

\$\endgroup\$
9
\$\begingroup\$
using 2 10 ohm resistors

Seriously!? That's essentially a dead short for this opamp. Note the datasheet specifies the output voltage swing with no more than 2 kΩ load. You really can't expect it to function properly with two orders of magnitude lower load impedance.

You can lift your arm with a 20 pound dumbell in your hand. How well can you do if I have a 2000 pound elephant sit on it instead?

\$\endgroup\$
6
\$\begingroup\$

Most likely because your feedback resistors are far too low.

Their 20 R load on the op-amp output is trying to draw 330 mA from it at the 6.6 V you want. From your datasheet, its maximum output current is something like 45 mA.

Make both feedback resistors 10 K and try it again.

\$\endgroup\$
3
\$\begingroup\$

With a sensor output of 1 to 20 milliVolts, and the need for lowest possible random noise floor, using 10 ohms Rg and Rfeedback would be wise. Such use would require 2 milliamp from the opamp, and even low-power opamps provide that much Iout.

Normal opamps have short-circuit values of 10 to 50 milliAmps. To supply 300 milliAmps would require a large (discrete?) unity-gain buffer after the opamp.

I'd use 1Kohm or 10Kohm, as the other answers suggested. If the bandwidth is 1,000KHz, knowing the noise density of 10Kohm is 12 nanoVolts/rtHz, the output noise contributed by the resistors will be Av=2 * 12nV/sqrt(2) * sqrt(1e6) or Vout noise = 2 * 9nV * 1000 = 1.8 microVolts RMs, across the bandwidth from DC to 1MHz (assuming your opamp performs well up to 1MHz).

Here is a noise simulation, done in Signal Chain Explorer

enter image description here

I've edited the answer to display the Thermal Distortion when 10 ohms and 10 ohms (Av = +2) are used. The opamp has default 5volt VDD, and 2mA Iout with (5volts -0.04volts) across the opamp's output transistors produces 1uV of low frequency distortion, as the output transistors send their heat across the die to upset biasing conditions of the input differential-pair transistors.

At 1uV distortion with 40,000uV output, the Thermal Distortion is -92dB. Problem with Thermal Distortion is it becomes Amplitude Modulation of all the tonal content; any higher tones become tiny AM_radio signals with sidebands carrying the low-frequency signals.

To view the conditions for defining Thermal Distortion, click on the "thermal" tab on the opamp stage; you'll see the cause of opamp Thermal Distortion is change in Input Offset Voltage at 1uV/degreeC; you'll understand the reason for only low-frequency signals causing Thermal Distortion as the Thermal Resistance and the Thermal Capacitance (using specific-heat of silicon) define the die's thermal timeconstant (bandwidth of the die's thermal response), assuming an non-distributed-RC model.

enter image description here

================================================

By the way, the tool has "Gargoyles" mode. In top center, click Gargoyles on, then click UPDATE on right side. You've just activated magnetic field, electric field, power supply and ground_plane INTERFERERS. To view the active interferers of each type, click on the label: HFI/EFI/PSI/GPI.

Notice the SNR drops, perhaps significantly, when the trash-injectors, the Gargoyles, are enabled. The HFI interferer, by default a switch regulator with dI/dT of 10MegaAmp/second current slewrate, couples into a 14mm by 1.5mm loop, which is the PCB trace between stages.

To view the interconnects, find and click "Show Interconnects" at top left. You can see the details of the PCB trace (the interconnect) between any two stages by clicking on the BLUE region between stages. Again, the default trace length is 14mm, on a 0.06" (1.5mm) double-sided PCB with GND.

To better understand what is happening, click "Analysis Details" on middle right.

\$\endgroup\$
  • 2
    \$\begingroup\$ +1 Upvoted for interesting 'signal chain explorer reference', but y'know the LM358 has about 40-50nV/rt-Hz noise so 12nV of uncorrelated noise is negligible (a few percent). \$\endgroup\$ – Spehro Pefhany Dec 31 '17 at 7:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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