Parameters of an OPAMP

Example - 1

A 100 PF capacitor has a maximum charging current of 150 µA. What is the slew rate?

Solution:

C = 100 PF=100 x 10^-12 F
I = 150 µA = 150 x 10^-6 A

Slew rate is 1.5 V / µs.

Example - 2

An operational amplifier has a slew rate of 2 V / µs. If the peak output is 12 V, what is the power bandwidth?

Solution:

The slew rate of an operational amplifier is

As for output free of distribution, the slews determines the maximum frequency of operation fmax for a desired output swing.

so    
So bandwidth = 26.5 kHz.

Example - 3

For the given circuit in fig. 1. I_in(off) = 20 nA. If V_in(off) = 0, what is the differential input voltage?. If A = 10⁵, what does the output offset voltage equal?

Fig. 1

Solutin:

I_in(off) = 20 nA
V_in(off) = 0

(i) The differential input voltage = I_in(off) x 1k = 20 nA x 1 k = 20µ V

(ii) If A = 10⁵ then the output offset voltage V_in(off) = 20 µ V x 10⁵ = 2 volt

Output offset voltage = 2 volts.

 Example - 4

R1 = 100Ω, R_f = 8.2 k, R_C = 10 k. Assume that the amplifier is nulled at 25°C. If V_in is 20 mV peak sine wave at 100 Hz. Calculate E_r, and V_o values at 45°C for the circuit shown in fig. 2.

Fig. 2

Solution:

The change in temperature ΔT = 45 - 25 = 20°C.

Error voltage = 51.44 mV

Output voltage is 1640 mV peak ac signal which rides either on a +51.44 mV or -51.44 mV dc level.

Example - 5

Design an input offset voltage compensating network for the operational amplifier µA 715 for the circuit shown in fig. 3. Draw the complete circuit diagram.

Fig. 3

Solution:

From data sheet we get v_in = 5 mV for the operational amplifier µA 715.

V = | V_CC | = | - V_EE | = 15 V

Now,

If we select R_C = 10Ω, the value of R_b should be
                  R_b = (3000) R_C = 30000Ω = 304Ω

Since R > R_max, let R_S = 10 R_max where R_max = R_a / 4. Therefore,

If a 124Ω potentiometer is not available, we may prefer to use to the next lower value avilable, such as 104Ω, so that the value of R_a will be larger than R_b by a factor of 10. If we select a 10 kΩ potentiometer a s the R_a value, R_b is 12 times larger than R_a, Thus

R_a = 10 kΩ potentiometer
 R_b = 30 kΩ
R_c = 10Ω.

The final circuit, which also includes the pin connections for the µA 715, shown in fig. 4.

Fig. 4

The ideal OPAMP :

An ideal OPAMP would exhibit the following electrical characteristic.

1. Infinite voltage gain A_d
2. Infinite input resistance R_i, so that almost any signal source can drive it and there is no loading of the input source.
3. Zero output resistance R_O, so that output can drive an infinite number of other devices.
4. Zero output voltage when input voltage is zero.
5. Infinite bandwidth so that any frequency signal from 0 to infinite Hz can be amplified without attenuation.
6. Infinite common mode rejection ratio so that the output common mode noise voltage is zero.
7. Infinite slew rate, so that output voltage changes occur simultaneously with input voltage changes.

There are practical OPAMPs that can be made to approximate some of these characters using a negative feedback arrangement.

Equivalent Circuit of an OPAMP:

Fig. 5, shows an equivalent circuit of an OPAMP. v₁ and v₂are the two input voltage voltages. R_i is the input impedance of OPAMP. A_d V_d is an equivalent Thevenin voltage source and R_O is the Thevenin equivalent impedance looking back into the terminal of an OPAMP.

Fig. 5

This equivalent circuit is useful in analyzing the basic operating principles of OPAMP and in observing the effects of standard feedback arrangements

v_O = A_d (v₁ – v₂) = A_d v_d.

This equation indicates that the output voltage v_O is directly proportional to the algebraic difference between the two input voltages. In other words the OPAMP amplifies the difference between the two input voltages. It does not amplify the input voltages themselves. The polarity of the output voltage depends on the polarity of the difference voltage v_d.

Ideal Voltage Transfer Curve:

The graphic representation of the output equation is shown in fig. 6 in which the output voltage v_O is plotted against differential input voltage v_d, keeping gain A_d constant.

Fig. 6

The output voltage cannot exceed the positive and negative saturation voltages. These saturation voltages are specified for given values of supply voltages. This means that the output voltage is directly proportional to the input difference voltage only until it reaches the saturation voltages and thereafter the output voltage remains constant.

Thus curve is called an ideal voltage transfer curve, ideal because output offset voltage is assumed to be zero. If the curve is drawn to scale, the curve would be almost vertical because of very large values of A_d.