I'm working on an interface that will be connected instead of a speaker, input of my interface is differential, and output is single ended.

I'm using the following schematics : enter image description here

R3 is the load to simulate the speaker, the R1/R7/R4 group is here to decrease the level.

I would like to easily adjust the gain by acting on R2 (knowing that I should act the same on R5) but I need a major value change to see a minor change on the output, I'm using Proteus to do the simulation.

Do you see an alternative to my scheme? or can I use a transformer to move from differential to single ended then use an AOP to adjust the gain?

  • \$\begingroup\$ but I need a major value change to see a minor change on the output I do not understand, the gain and therefore the output voltage is directly proportional to the values of R2 and R5 (actually their value relative to R8 and R6). So if R2 and R5 are doubled in value the output signal should also double. Why not change the value of R7 and leave the gain fixed? \$\endgroup\$ – Bimpelrekkie Oct 18 '18 at 9:19
  • \$\begingroup\$ the differential signal can reach 15Vpp as the AOP is powered by 6V I'm reducing the input with those resistors, i can increase them to reach 3Vpp at the AOP entrance and use the AOP gain to reduce but I'm facing the same issue, i need major value changes to reduce a bit the levels. \$\endgroup\$ – PtiBruno Oct 18 '18 at 9:43
  • \$\begingroup\$ What you're doing now doesn't make much sense, R1,R4 and R7 reduce the signal by a factor of about 100, then you amplify it again by a factor of 10. Why not reduce the 15 Vpp by a factor of 5 (make R7 about 20 k) and then make gain = 1 a round the opamp by making R2,R5, R6, R8 all 100 kohm. \$\endgroup\$ – Bimpelrekkie Oct 18 '18 at 11:06
  • \$\begingroup\$ about the factors I'm ok with you, the final idea is to replace R2 with a digipot and adjust the output levels by changing his value the largest way possible ( from 0 to AOP VDD), but I'm not able to put 2 digipot ( in order to replace R2 and R5), So that is why I'm wondering if my design is correct and how can I improve it to archieve my needs. \$\endgroup\$ – PtiBruno Oct 18 '18 at 11:13
  • \$\begingroup\$ I made some comments and suggestions, however from your comments I do not get the feeling that these comments are understood by you. \$\endgroup\$ – Bimpelrekkie Oct 18 '18 at 11:16

You can use variable gain amplifier like this one, that you can connect either to a potentiometer, or to a DAC, it will come much cheaper than a digipot and solve lots of the issues related to it to the design.

There are plenty of those, controlable from an analog source, or digital like I2C, SPI...


I'd suggest varying R7 instead of R2. R7 is a single value in your circuit that will directly affect the gain. Because R7 is so low compared to R1 and R4, the effect will be close to linear from about +20 to -infinity dB.

  • \$\begingroup\$ Hi, yes it is the easist solution, but I would like to use a digital potentiometer and I'm limited to the digipot power supply (5.5V) referenced to ground and at this stage i'm still in differential. \$\endgroup\$ – PtiBruno Oct 18 '18 at 13:22
  • \$\begingroup\$ Is the voltage across R7 outside the 5.5V range? If not, you can use it as long as it supports differential mode. \$\endgroup\$ – Cristobol Polychronopolis Oct 18 '18 at 15:10

Are you doing a linear or transient Spice analysis? In your final application will the source and your circuit have a common ground or will your circuit float relative to the source? These factors affect how you need to simulate your input signal. Have you arranged the the two sources to be 180 out of phase? Also, be careful with the power supply. TL074 is not a particularly low voltage part. Datasheet says "(1) VCC+ and VCC– are not required to be of equal magnitude, provided that the total VCC (VCC+ – VCC–) is between 10 V and 30 V." It also recommends that you should operate inputs at least 4 V above negative rail. I hope this helps.

  • 1
    \$\begingroup\$ Hi,the source is floating, for the simulation I just put a sinus, one is 8V amplitude, the other is -8, so same as floating 16Vpp.thanks for the remark regarding the 4V above negative rail.I'm doing frequency response and analog reponse. \$\endgroup\$ – PtiBruno Oct 18 '18 at 13:20
  • \$\begingroup\$ I think your pair of voltage sources give to a differential signal with a ground reference. This works with the "analogue analysis" graph, which is time domain. I don't know how to generate a floating source in Proteus or if it's possible. If it were truly floating Spice would flag an error because the left hand part of the circuit would be floating and therefore at an undefined potential. You'd need at least a very high value res to ground. As far as the frequency domain is concerned, it calculates response in dB relative to a source and it doesn't matter what you set its amplitude to. \$\endgroup\$ – Steve Hubbard Oct 19 '18 at 5:19
  • \$\begingroup\$ Continuing on from last comment, the frequency response is probably not very meaningful, although you could one referenced to each source and combine the results by hand (one has 180 degree phase). You're biggest issue may be power. The power symbols you are using may only be schematic labels, not providing bias at all. Use DC from the generators menu. Then add some voltage probes and click the play button in the lower left corner to check bias point is sensible. Finally check that you are using the TL074 device that has a Spice model. I have only touched upon simulation, not your design. \$\endgroup\$ – Steve Hubbard Oct 19 '18 at 5:27
  • \$\begingroup\$ HI Steve, in fact i'm not able to generate a floating input with proteus, I will try that and see what happen, simulation process is probably my problem something badly configured. \$\endgroup\$ – PtiBruno Oct 19 '18 at 11:04

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