Power Amplifiers Tutorial

  
THEORY DESIGN


This tutorial explains the basic theory of operational amplifiers and voltage general purpose various design techniques that will enable you to design low frequency voltage amplifiers, and how to choose the right type of operational amplifier that meets the electrical specifications of the design.


INPUT AND OUTPUT RESISTANCE OF A REVERSAL OPERATIONAL AMPLIFIER

It can be proved that a circuit of a reversal amplifier shown in Figure 1 can be represented by an equivalent circuit given in Figure 2. In Figure 2 the resistance Rm is given by the relation:



where Aol is the gain of the open-circuit of the operational amplifier. The value of RM is very small because the Aol is too big. The resistor RM, which is also called Miller resistance appears in parallel with the internal resistance (Ri) of the operational amplifier.

Example. We calculate the value of RM, where R2 = 20Kohm and AOL = 200,000.



The resistance Rm is in parallel connection with Ri of the operational amplifier and the impedance is very small, almost negligible. So all the input resistance of the amplifier is R1.


Figure 1 and figure 2.


At the amplifier in Figure 1, the resistance R1 = 5000ohms, and the internal resistance of the operational amplifier is Ri = 200000ohm, So to calculate the input impedance of the amplifier Rin we do the following calculation.

From the equivalent circuit of the amplifier we see that the input impedance is equal to R1 in series with the parallel combination of Rm and Ri, that is:



But:



And because (Rm / Ri) = Rm = 0.25ohm so:

Rin = 5000ohm + 0,25ohm ~ 5000ohm and for that Rin = R1

The total output impedance, Ro(T) of the reversal amplifier is very small and approximately equal to a output resistance Ro of the operational amplifier, which is given by the manufacturer.
 


INPUT AND OUTPUT RESISTANCE OF A RECTAL OPERATIONAL AMPLIFIER


The overall input resistance of a rectal amplifier shown in Figure 3 is very high and is given by the following equation:


where Ri is the differential input impedance of the operational amplifier, Aol is open loop voltage gain, and Acl is the closed loop voltage gain of the operational amplifier. Since the closed loop voltage gain (Acl), is much smaller than the voltage gain of open loop (Aol), the above equation shows that the input impedance of the rectal amplifier is much greater than the internal input impedance of the operational amplifier.


Figure 3.


If we want amplifiers with high input impedance we use rectal amplifiers. Notice that the voltage gain open loop (Aol) is not constant and depends on the frequency (decreases when the frequency increases), for this reason, the input impedance because it depends on Aol, also depends on the frequency, and decreases as the voltage gain open loop (Aol) decreases with increasing frequency. The output impedance of rectal amplifier is given by:



Where Ro is the internal resistance of the operational amplifier. The resistance Rout of the amplifier is much lower than the output impedance of the operational amplifier.

Example. Determine the input resistance (Rin) and the output resistor (Rout) in a rectal amplifier shown in Figure 3. Ri = 2Mohm, Ro = 50ohm, Aol = 200,000, R2 = 99Kohm, R1 = 1Kohm.

Solution: The voltage gain of the amplifier is:



The input resistance is:



And the output impedance is:


 


VOLTAGE FOLLOWER



Figure 4.


Figure 4 shows the basic circuit of a voltage follower. The input voltage, V1, is applied to the correct (+) input of the amplifier. Because the voltage between the (+) and (-) terminals of the operational amplifier can be considered to be zero, we can assume that the input voltage V1, is equal to the output voltage, and have the same polarity, ie the output voltage follows the input voltage, Vo = V1.

In this case the gain of theamplifier is given by the relation:



This amplifier has a negative feedback because the output is connected to the (-) input. This amplifier though does not amplifies (Acl = 1) is very useful because the input impedance of the amplifier (not the resistance of the operational amplifier) is very high, while the output impedance is very low and allows the input voltage to be transferred to the output without the load resistance to lower the source voltage applied to the input.

The voltage follower is a special case of rectal amplifier with R2 = 0, and so the same relationships applied to the rectal amplifier are valid and for the following voltage setting Acl = 1 and R2 = 0.
 


INVERSAL AMPLIFIER


Figure 5.


Figure 5 shows the circuit of an inverse amplifier that can be used to sum the two signals applied to the input (any number of inputs can be used). The input signals V1, and V2, ARE applied through resistors R1 and R2 in the inversal input of the operational amplifier. There is also a feedback resistor Rf. The inverse input terminal of the operational amplifier behaves as a virtual ground (i.e., is at 0 volts), thus:



The input currents I1 and I2 are joined to the summing point, and because the current drawn by the reverse input of the operational amplifier is almost zero, the current flowing through the resistor Rf is equal to I1 + I2, and creates a voltage drop across the RF given by:



So the output voltage is the sum of two input voltages.
 


POWER CONVERTER TO VOLTAGE (TRANSRESISTANCE AMPLIFIER)


Figure 6.


Figure 6 shows a power converter to a voltage, which also called transresistance amplifier. In an ideal amplifier we assumed that all of the current from the source signal passes through the feedback resistor Rf. Furthermore the voltage between the (+) and (-) terminals is assumed to be zero, so the output voltage appears across the resistor Rf, and is equal to Vo = - Is x Rf. So the output voltage is directly proportional to the input current. Notice that the output voltage is independent of the load Rl.
 


VOLTAGE TO CURRENT CONVERTER


Figure 7.


The circuit shown in Figure 7 is used as a voltage to current converter amplifier. If we assume that an ideal operational amplifier, the voltage between the (+) and (-) terminals is zero volts. Therefore, the voltage across R1 is equal to Vin. This voltage generates a current through the load resistor Rl and is given by the relation:



The above relationship shows that the output current is proportional to the input voltage and is independent of the load impedance. One disadvantage of the circuit is that the load is suspended, means that neither of the terminals are connected with the earth. The circuit shown in Figure 7 can also be used as power source. If a constant current source is required, then the input voltage may be provided by a battery or by a regulated DC voltage power supply.
 


CURRENT BOOSTER

The output terminal of one operational amplifier of general perpuse can provide about 5 to 25mA. To increase the load current, we use a circuit at the output of the amplifier that composed of two transistors as shown in Figure 8a. In this circuit when the output of the operational amplifier is positive and greater than 0.6 to 0.7 volts, the transistor T1 is biased (ie conducting), and the transistor T2 is reverse biased (ie does not conduct). The output terminal of the operational amplifier gives only the base current to the transistor T1, which is much smaller than the load current Il. The relationship between the current base of T1 and ll is given by the following relationship:



wherein B is the current gain of the transistor in common emitter connection, ie B = Ic / Ib.

If we assume that the terminal of the operational amplifier can provide a maximum output current of 20mA in the base of the transistor, the circuit of the current amplifier can provide a maximum load current equal to: Il(max) = 20mA x b and If b = 100, then the maximum load current is:



Naturally, the transistor must be able to provide safely this current, otherwise destroyed.

When the output of the operational amplifier is negative and greater than -0.6 to 0.7 volts, then the transistor T1 is reverse biased (not conducting) and the transistor T2 is properly biased and conducting, as shown in Figure 8b. In this case the transistor T2 provides current to the load.