1. Medical electronics II

2. Voltage follower op-amp circuit
Voltage Follower. Buffer.
Voltage follower op-amp circuit

3. The voltage gain of this configuration is 1
The voltage gain of this configuration is 1. The output voltage follows the input.
So what is the usefulness of this op-amp circuit?
Let’s look at the input and output resistance characteristics of the circuit. As we have discussed, the resistance at the input terminals of the op-amp is very large. Indeed, for our ideal model we have taken the value of that resistance to be infinite. Therefore the signal Vin sees a very large resistance which eliminates any loading of the signal source. Similarly, since the output resistance of the op-amp is very small (zero ideally), the loading is also eliminated at the output of the device. In effect this is a resistance transformer.

4. In order to see the importance of this buffer circuit let’s consider the case where the input signal is a source with an output resistance Rs and is connected to a load with resistance RL. In Figure 16(a) the signal source is connected directly to the load RL.
Figure 16. (a) Source and load connected directly. (b) Source and load connected via a voltage follower.

5. From Figure 16(a), the voltage divider formed by Rs and RL gives a value for VL which is a fraction of Vin given by
For example, if RL = 1kΩ and Rs = 10 kΩ, thenVL ≈ 0.1 Vin which represents a considerable attenuation (loading) of the signal source.
If we now connect the signal source to the load with a buffer amplifier as shown on Figure 16(b). Since the input resistance of the amplifier is very large (no current flows into the terminal), the voltage at the non-inverting terminal, Vp, is equal to Vin. In addition, since the output resistance of the op-amp is zero, the voltage across the load resistor VL = Vo = Vin. The load now sees the input voltage signal but it places no demands on the signal source since it is “buffered” by the operational amplifier circuit

6. Common Mode Voltage
Most sensors and medical transducers convert the measured quantity (temperature, pressure, sound etc…) to an electrical signal which need to be amplified to a useful level. However, this electric signal always contains Dc offset, drift and noise
The Dc offset, drift and noise are common
in the two sensor terminal voltages, and
this is called common mode voltage Vcm

7. The sensor output signal can be represented by a common mode voltage Vcm and a differential voltage
Vd = (Vs+) – (Vs-)
Thus we need an amplifier which responds only to the differential voltage ((Vs+) – (Vs-)) and amplify it and blocks the common mode voltage Vcm.

8. Common Mode Rejection Ratio (CMRR)
As we have seen, the usefulness of the op amp is derived from its ability to amplify differential signals. In fact the ideal op amp has an infinite gain for differential voltage signals. It is desired that signals that are common to both inputs “Common Mode (CM) Signals” be rejected by the amplifier. An ideal op amp has the ability to completely reject those CM signals; thus having infinite Common Mode Rejection (CMR) ability.
For an amplifier subjected to a differential voltage VD and a common mode voltage VCM, the output voltage is

9. Where GD is the differential gain and GCM is the common mode gain of the amplifier. In practice we would like to minimize GCM. The ability of an amplifier to reject the CM signal is expressed by a parameter called the Common Mode Rejection Ratio (CMRR) which is defined as the ratio of the differential gain to common mode gain as follows.
In the open loop configuration, our standard 741 op amp is characterized by a CMRR of 90dB for signal frequencies less than 100 Hz. At higher frequencies CMRR degrades considerably falling to 40dB at 100 kHz.
In general CMRR is not of concern in the inverting amplifier configuration.

10. Ideal Op-amp in an open loop configuration
An ideal op-amp is characterized with infinite open–loop gain
A→∞
The other relevant conditions for an ideal op-amp are:

11. Ideal op-amp in a negative feedback configuration
When an op-amp is arranged with a negative feedback the ideal rules are:
These rules are related to the requirement/assumption for large open-loop gain A → ∞, and they form the basis for op-amp circuit analysis.
The voltage Vn tracks the voltage Vp and the “control” of Vn is accomplished via the feedback network.

12. Operational Amplifier Circuits as Computational Devices
mathematical operations ranging from addition and subtraction to integration, differentiation and exponentiation.1 We will next explore these fundamental “operational” circuits.

13. Figure 1. Summing amplifier
A basic summing amplifier circuit with three input signals is shown on Figure 1.
Figure 1. Summing amplifier
Current conservation at node N1 gives

14. By relating the currents I1, I2 and I3 to their corresponding voltage and resistance by Ohm’s law and noting that the voltage at node N1 is zero (ideal op-amp rule) Equation (1.1) becomes
And so Vout is
The output voltage Vout is a sum of the input voltages with weighting factors given by the values of the resistors. If the input resistors are equal R1=R2=R3=R, Equation (1.3) becomes

15. The input resistance seen by each source connected to the summing amplifier is the corresponding series resistance connected to the source. Therefore, the sources do not interact with each other.

16. Figure 2. Differential Amplifier
This fundamental op amp circuit, shown on Figure 2, amplifies the difference between the input signals. The subtracting feature is evident from the circuit configuration which shows that one input signal is applied to the inverting terminal and the other to the non-inverting terminal.
Figure 2. Differential Amplifier

17. Before we proceed with the analysis of the difference amplifier let’s think about the overall behavior of the circuit. Our goal is to obtain the difference of the two input signals ( Vin2 – Vin1). Our system is linear and so we may apply superposition in order to find the resulting output. We are almost there once we notice that the contribution of the signal Vin2 to the output is
and the contribution of signal Vin1 is
And the output voltage is

18. Note that in order to have a subtracting circuit which gives Vout=0 for equal inputs, the weight of each signal must be the same. Therefore
which holds only if
The output voltage is now
which is a difference amplifier with a differential gain of R2/R1 and with zero gain for the common mode signal. It is often practical to select resistors such as R4=R2 and R3=R1.

19. Figure 3. Differential amplifier
The fundamental problem of this circuit is that the input resistance seen by the two sources is not balanced. The input resistance between the input terminals A and B, the
Figure 3. Differential amplifier
Since V+ = V- , Vin = R1 I + R3 I and thus Rid = 2R. The desire to have large input resistance for the differential amplifier is the main drawback for this circuit. This problem is addressed by the instrumentation amplifier

20. Thank You