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Here is an example of an op amp used for amplification:

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What I understand:

  • C1 is used as AC coupling to remove any DC bias.
  • Vref is used to apply a DC voltage to be able to amplify negative voltage values.
  • R2 and R1 these are used for gain.
  • C2 is used such that the gain formed by R2 and R1 only apply to AC signals as C2 at low frequencies looks like open circuit for DC

What I think I understand:

  • Since there is no input current, ideal op amp, R3 has no voltage across it. Hence input voltage is Vs+2.5 .
  • v+ = v- = Vs+2.5
  • Since the gain is 1+ R2/R1 then this means that Vout = Vs+2.5 ( 1+ R2/R1)

What I Know I got wrong:

  • The analysis above is only true for AC signals? But how could I do I arrive at the "general" answer for all types of signals. I am not sure how to approach this.
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    \$\begingroup\$ What you're forgetting is the last point you understand : C2 looks like an open cct at DC. Thus, at DC,"R1" (+ C2) becomes infinite thus gain = 1 and Vout = Vin = 2.5V.The general formula for gain will include the impedance of C2, therefore replace R1 with (R1 + 1/jwC2) where w stands for omega = 2*pi*f \$\endgroup\$ Oct 22, 2020 at 13:03

2 Answers 2

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In order to simplify things a bit, you need to consider A.C. and D.C. gain separately for this A.C. coupled circuit.

At D.C. capacitor C2 is effectively 'open' i.e. high impedance. The value of R1 is no longer important and the feedback is 100%, therefore there is no gain. This means that the output 'sits' at Vref (+2.5V in this example) in the 'quiescent' state. This amplifier configuration is termed a voltage follower and the D.C. that's added to the A.C. signal (Vref) is usually referred to as an 'offset' which does indeed allow A.C. signals to be amplified.

For A.C. signals the gain is indeed 1+(R2/R1) and the output requires A.C. coupling in the same way as the input. As there's no gain at D.C. there's no requirement to consider the offset as part of the input signal as it 'drops out of' the in band A.C. gain equation.

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  • \$\begingroup\$ I understand from this the gain for the Ac+Dc combination. But what would the overall input voltage be when we consider both DC and AC voltages? . In essence, If I input an Ac signal as Vs, what would Vout be? I am struggling to understand what exactly the input voltage would be. Is it just Vs+2.5? \$\endgroup\$ Oct 22, 2020 at 16:04
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    \$\begingroup\$ You need to concentrate on the wanted AC voltage gain. Vout = (1+(R2/R1)).Vs \$\endgroup\$ Oct 22, 2020 at 16:08
  • \$\begingroup\$ Yes. I understand thank you. So the output will be fixed at 2.5V and whatver amplified Ac signal there is will just be around 2.5V instead of 0V. I understand the 'intuition' of it. What I dont understand is the maths part of it. If I wanted to write down Vout as a combination of say Vs R3, R2, R1, C2 how exactly would I formulate this? I am trying to approach it from a gain x input voltage but this from what I am getting is just gain x Vs+2.5? how does R3 come into play with the input voltage and gain? \$\endgroup\$ Oct 22, 2020 at 16:13
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    \$\begingroup\$ You need to concentrate on the wanted AC voltage gain. Vout = (1+(R2/R1)).Vs . Vref plays no important part in the wanted in band signal or gain. It's merely an offset that allows for negative input swings with a single supply amplifier. Transistor amplifiers were usually of this configuration before split supplies became commonplace. I suggest that you consider the signal path separately for DC and in band AC. If you don't, then the gain equation runs riot with complex numbers representing C1 and C2's reactance. \$\endgroup\$ Oct 22, 2020 at 16:16
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To analyze this circuit at frequencies where Xc1 ~= R3 or Xc2 ~= R1 where Xc = \$\frac{1}{2\pi f C}\$ you have to use a more sophisticated approach, but if you consider capacitors as shorts at high frequency and as open at 'low' frequency then you have AC gain = (R2/R1+R2) and DC gain = 1 (since the capacitor is very high reactance 1+R2/(XC2+R1) ~= 1).

For the general steady-state sinusoidal case you have to use complex numbers, and you will end up with a magnitude and a phase so rather straightforward. It's really just the same calculation as with resistors (voltage divider at the input and Vo/V+ = 1+Za/Zb) but with complex numbers. You can plot magnitude and phase on a Bode plot.

(You can't just add Xc and R like they were real numbers and get the correct answer because Xc is not a real number- they add in quadrature)

For signals in general, it's not so easy. Laplace transforms or numerical simulation can be used, and you'll need to know the initial conditions (eg. the voltage on each capacitor).

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