Operational Amplifiers with Power Supply


The single feed mode has become very important in systems powered from a single power supply, because it reduces the cost of circuits with operational amplifiers and also makes it easier to use on mobile devices. In this series of articles will look at various ways that can power amplifiers circuits using operational amplifiers with single power supply (single rail).

The diagram of a sound amplifier, which operates with a single power supply shown in Figure 16. In this circuit one end of the signal source connected to the right input terminal of the operational amplifier, and the other end connected to the bias voltage VBias.

The load resistance (RL) and R1 also associated with the bias voltage. The voltage at the right input terminal, VP, is equal to:

Vp = VBias + Vin

The voltage at the reverse input terminal VN, is equal to:

VN = VD + VP but because,

Vd = 0 we have:

VN = VP = Vin + VBias

without an input signal

Vin = 0, so therefore

Vn = Vp = VBias



Figure 16. Amplifier with single power supply.

Figure 16. Amplifier with single power supply.


Figure 17. Input and output waveforms of the amplifier of Figure 16.

Figure 17. Input and output waveforms of the amplifier of Figure 16.


The bias voltage (VBias) is typically equal to half the voltage of the power supply. The output voltage of this amplifier can be calculated as follows:

The current I1, which flows through the resistor R1, is approximately equal to the current I2, which flows through the resistor R2, and thus I1 = I2. The current flowing through R1 is equal to the difference voltage across R1 divided by the value of R1, ie


In this circuit VB is the voltage applied to the right end of R1, and VA is the voltage applied to the left end of R1.

From the figure 16 we see that:
VB = VP but VP = VBias + VIN, thus:
VB = VBias + VIN
We also have VA = VBias so:



Similarly, the current I2 flowing through the resistor R2, is equal to the difference voltage across the R2.
i.e. VR2 = VO - VB divided by the value of R2
We also have VB = VBias + VIN, thus:
 


Figure 18a. AC amplifier with simple power supply.

Figure 18a. AC amplifier with simple power supply.



Since I1 = I2, we obtain:

Rearranging we get:
 
 

By multiplying both sides of this equation with R2 we get:


So in the absence of an input signal, ie when VIN = 0, the above equation becomes

 or VO = VBias

So in the absence of an input signal, the input terminals and the output terminal of the operational amplifier have the same voltage (VBias).

If an input voltage 2 volts peak to peak, is applied to the amplifier shown in Figure 16, R1 = R2, and VCC = 10V, then when the input signal is at the positive peak value of +1V, the output voltage will be:

 




When the input signal changes is at its highest negative value -1V, then the output voltage equals:

 So when the input signal changes from -1V to +1V, the output voltage varies from +3V to +7V.

Figure 17 shows waveforms of the input and output. Notice that the first term in equation 1 represents the voltage gain of a conventional amplifier using a double power supply. The voltage gain is given by the following relationship:


In a conventional non-inverting amplifier, which operates with double supply, the bias voltage VBias is equal to 0 volts, and the output voltage is given by the following equation:

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