Timeline for What's the need to set the amplifier input common mode voltage to mid–supply voltage in this non-inverting microphone pre-amplifier circuit?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Nov 26, 2022 at 20:27 | vote | accept | Scipio | ||
| Nov 26, 2022 at 16:49 | comment | added | td127 | A rail-to-rail opamp is still not capable of going negative, just closer to zero than a standard opamp. You need to use a bipolar supply or put the capacitors back in. C3 is just part of a 130kHz low pass filter and is not critical. But C4 is a DC blocker that is mandatory for this single supply circuit: without it there will be a 2.5V DC component across R4 which then gets amplified by R5 resulting in a large DC offset on the output. | |
| Nov 26, 2022 at 16:28 | comment | added | Scipio | @MathKeepsMeBusy ok, understood! So i tried to simulate part of that circuit in LTSpice and the output signal isnt centered at the input common mode quiscent voltage... You can see it in my new post edit. Any idea why? | |
| Nov 26, 2022 at 15:00 | comment | added | Math Keeps Me Busy | @G0tBlackOps the gain is frequency dependent in this case. The combination of the 2 resistors and 2 capacitors form a band-pass filter. This is useful in audio circuits because noise at frequencies above or below the audible range can adversely affect overall signal quality. Whether or not an input causes amplifier saturation depends not only on the instantaneous signal amplitude, but also on the instantaneous noise amplitude. So it is useful to attenuate at frequencies outside of the audible band. You could omit those capacitors, but at the risk of possible distortion. | |
| Nov 26, 2022 at 14:49 | comment | added | Scipio | @td127 why use capacitors C2 and C3? The gain doesnt depend on them | |
| Nov 26, 2022 at 3:01 | comment | added | td127 | The output of the opamp comes out centered around 2.5V (or whatever VCC/2 is). That means there's a DC component of 2.5V and a small AC voltage (the audio) superimposed on it. The AC component then passes through the final output capacitor C6, but the DC component is blocked by it. The DC level of the output side of C6 is set by whatever R6 is connected to, ground in this case. So, the final output Vout is your amplified audio, centered around ground. | |
| Nov 26, 2022 at 2:41 | comment | added | Math Keeps Me Busy | @G0tBlackOps Yes. (Assuming you keep the power supply at 5V). | |
| Nov 26, 2022 at 2:40 | comment | added | Scipio | Ok, so if the mic outputs a negative signal and it passes through that opamp the negative part of the signal will be clipped, that's why we use those 2 resistors forming a voltage divider. So the output signal should be come out centered in 2.5V? Also, i understand why the first capacitor in the inpu part is there for. But that last RC circuit in the ouput what is it for? The gain doesnt depend on anybof the output capacitors | |
| Nov 26, 2022 at 2:08 | history | edited | Math Keeps Me Busy | CC BY-SA 4.0 |
added further answers
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| Nov 26, 2022 at 1:58 | history | answered | Math Keeps Me Busy | CC BY-SA 4.0 |