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Design Article

Rail-to-rail input amplifier application solutions

Bonnie Baker

11/8/2012 4:10 PM EST

Bottom line: inverting-amplifier circuit
Bottom line: inverting-amplifier circuit

  • Composite input: performs without distortion with VB outside the PMOS/NMOS transition region
  • Charge-pump input: performs without distortion
The noninverting amplifier’s common-mode voltage (Figure 4) is equal to the input voltage, VIN. The noninverting amplifier circuit gains the input signal with a voltage level-shift implemented with VB.


Figure 4 The noninverting gained amplifier configuration allows the amplifier’s common-mode voltage to change with input signals.

For proper operation, the noninverting amplifier circuit may require a bias voltage, VB, to keep the output range at VO between the power-supply rails. Here VB establishes the common-mode voltage of the amplifier. The transfer function for the circuit is expressed as



The input common-mode voltage (VIN) can vary between ground and VS, as long the voltage at VO remains between the power-supply rails. VB can be assigned so that the input signal is not required to go across the composite input stage’s crossover region. If the input signal always remains less or more than the composite input-stage’s transition voltage, the circuit will not create distortion from the composite’s input stage, crossover phenomena.

Bottom line: noninverting-amplifier circuit

  • Composite input: the input signal (VIN) must not travel into the transition region of the PMOS and CMOS differential input stages
  • Charge-pump input: performs without distortion in reaction to common-mode voltage or VIN changes

With a unity-gain buffer (Figure 5), the input signal (VIN) can traverse from one rail to the other rail. In this circuit, VIN establishes the amplifier’s common-mode voltage. If the circuit designer uses the amplifier in a voltage follower or buffer configuration, the composite amplifier exhibits some limitations in linearity due to the input-stage topology. If the circuit designer wants a distortion-free output from this amplifier circuit, an op amp with a composite input topology may limit the input range of the buffer.


Figure 5 The input signal, VIN, of the buffer circuit changes the amplifier’s common-mode voltage. Composite input op amps can create distortion at the output, VO, as VIN changes.

The transfer function of the circuit in Figure 5 is shown here:



The input common-mode voltage (VIN) can vary between power-supply rails. If VIN travels across this entire range, the composite input amplifier produces distortion at the circuit output, VO. If the input signal always remains less than the composite input stage’s transition voltage, the circuit does not create distortion from the composite’s input stage, crossover phenomena. Alternatively, an amplifier with a charge-pump input stage does not create this unwanted distortion across the entire input range.

Bottom line: buffer circuit

  • Composite input: the op amp’s input range (VIN) must not travel into the transition region between the PMOS and CMOS differential input stages
  • Charge-pump input: performs without distortion in reaction to common-mode voltage changes

Conclusion


Manufacturers use several input topologies in their designs of rail-to-rail input amplifiers. It is true that circuit designers can use rail-to-rail input amplifiers in virtually any op-amp configuration. However, if you choose to use a rail-to-rail input amplifier, it pays to understand the CMRR impact on your circuit. The composite input amplifier produces a crossover distortion, requiring special considerations in some classes of circuits. The charge-pump input amplifier does not produce this same distortion.
References
  1. Edgar Sánchez-Sinencio, “Rail-to-Rail Op Amps,” Ax-09, TAMU, AMSC
  2. Datasheet: www.ti.com/opa365-ca
Also see:




johnboy50

11/27/2012 5:00 PM EST

Am I wrong ? Should not the second term or the equation be Vb (R1+R2)/R1 ????

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Bonnie Baker Texas Instruments

1/3/2013 7:19 PM EST

Hi johnboy50,
I am not sure what your question is. It seems to me that the second term in the equation is Vb(R1+R2)/R1. Please clarify.

Bonnie

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willdo

1/30/2013 10:45 AM EST

Bonnie,

Doesn't the the plot to figure 3 actually show (VO - VB) vs. VIN, meaning the output would be clipping for VIN greater than 1.25V? Also the minus sign at VIN is missing in the operational equation.

For VIN=VB the circuit becomes a voltage follower. At VIN=0V the output voltage would aim to VO= 3VB=7.5V.

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willdo

1/31/2013 11:22 AM EST

Sorry, of course wanted to say "for VIN smaller than 1.25V"...

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