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An amplifier can be used to adapt the range of a signal to a requirement, to make it more robust for transmission, or to satisfy interface requirements (like input/output impedance).
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3
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What's the need to set the amplifier input common mode voltage to mid–supply voltage in this...
I don't understand the need to set the amplifier input common mode voltage to mid–supply voltage in the circuit below. What does it do and why is it needed? … The author of these images is Texas Instruments: Non-inverting microphone pre-amplifier circuit
Edit:
So i tried doing part of that circuit in LTSpice with an opamp that is also rail to rail at the output …
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0
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Why does the output of this differential amplifier always get 0 V?
I used a differential amplifier to try to get rid of the offset that came along with the desired signal.
Vcc = 5 V
In simulation everything looks good. … But when I place everything on a breadboard the output signal of the differential amplifier is always 0 V on the oscilloscope no matter what. …
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1
answer
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How does this non-inverting amplifier remove the offset of the input signal?
I'm trying to find out how the circuit inside the red line works.
It seems to remove the offset of the signal Vrect and amplify it.
I would like to know how exactly that works. Is it possible to lim …
0
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3
answers
110
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Why does the majority of the microphone current have to flow through R1 in this amplifier?
In this circuit:
Why does the part underlined in red matter? I don't understand, why do I want more current flowing through R1?
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2
answers
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Output of RC circuit is just a straight line
It consists of a pre-amplifier, a filter, an RC circuit that acts like an exponential average circuit, a peak detector and perhaps I still need to do another amplifier in order to use the entire range … If I were to connect this circuit to an ADC, how exactly would I build an amplifier that goes to the input range (0V to 3.3V) of the ADC? …
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Finding the gain expression for this CMOS operational transconductance amplifier
So here's the circuit:
I think M2 is a common source amplifier and M4 a common gate amplifier, so they form a cascode amplifier where the gain is given by : Av = -gm2 * r0, r0 is the impedance at the … node where the drain of the common gate amplifier is at (Vop).
r0 = ( [rds8 // (rds2a * ( gm4a / gds4a) )] * ( gm6 / gds6 ) ) // ( rds2 * (gm4 / gds4) )
I think the low frequency gain is this and there …