# How to bias non-inverting op-amp amplifiers

Often when we use Op-amp or operational amplifiers in electronics circuits we want to bias the operational amplifier so that the output signal swings around a desired DC voltage so that maximum voltage swing is obtained without clipping the signal. This is often the case when single supply is used with the op-amp wherein the desired biased voltage at the input is between the supply voltage and ground.

Here we consider how to bias non-inverting op-amp amplifier when single supply is used. Amplifiers can be designed based on whether we want to allow both DC and AC signal to pass through the amplifier or if we just want to pass the AC signal and block the DC signal. Amplifier designed to allow to pass both the DC and AC signal are called DC coupled amplifiers and amplifiers designed to block DC signal and allow only AC signal to pass are called AC coupled amplifiers. Here we show how to bias non-inverting op-amp for both DC coupled and AC coupled operational amplifiers.

### DC coupled Non-inverting amplifiers

A generic circuit diagram of DC coupled non-inverting op-amp amplifier with dual power supply is shown below.

The gain of the non-inverting amplifier is given by,

$A = 1 + \frac{R1}{R2}$

The resistor R3 at the non-inverting terminal which is equal to parallel combination of R1 and R2 is used to make the voltage at the terminals.

$R3 = \frac{R1 R2}{R1+R2}$

Consider the following example circuit with +12V and -12V dual power supply. Let R1 = 50KOhm and R2 = 10KOhm. This gives gain of A = 6 and the value for R3 is 8.3KOhm.

Let us apply input sine signal of amplitude 100mV and frequency of 1KHz. Let us consider that the input signal has voltage offset of 1V. That is the input sine wave is riding on 1V DC. Then the output and input signal waveform is as follows.

### Biasing of DC coupled Non-inverting amplifiers with Single Supply

If we want to use single power supply with DC coupled non-inverting amplifier we have put bias circuit at the input. The bias circuit is connected at the output because the output of the op-amp is very strong.

The following shows circuit diagram of non-inverting amplifier with single supply voltage.

Let the single supply voltage be +5V, gain A = 11 with R1=100KOhm and R2=10KOhm and R3 =R1||R2=9kOhm we have the following circuit.

If the input signal is as before a sine wave of amplitude of 100mV, frequency of 1KHz and offset at 1V, then we get the following input and output waveform.

The output signal is not available because the signal is outside the range of the op-amp and power supply limitation.

Split resistor biasing

The circuit diagram of dc coupled non-inverting op-amp amplifier with split resistor is shown below.

The gain of this split resistor biased dc coupled non-inverting amplifier is,

$A = 1 + \frac{R1}{R_{||}}$

where, $$R_{||}= R2||R4 = \frac{R2 R4 }{R2+R4}$$

The bias voltage at the junction of R2 and R3 is given by,

$V_{b} = \frac{R2}{R2+R4} V_{cc}$

With $$V_{b} = 2.5V$$  and $$V_{cc} = 5V$$, we get R2=R4

If we want gain of A =11 as before, by selecting R1=100KOhm then $$R_{||}= 10KOhm$$ and thus solving for R2 and R4 we get,

R2=R4= 20KOhm

Let the input signal be sine wave with amplitude and frequency of 100mV and 1KHz. Let the input signal be at offset voltage of 1V. Then the input and output waveform is shown below for this circuit,

Voltage divider biasing

The following is circuit diagram of DC coupled non-inverting amplifier with voltage divider biasing.

The gain of this biased non-inverting amplifier is still,

$A = 1 + \frac{R1}{R2}$

The resistors R4,R5 forms the voltage divider biasing circuit and the capacitor C1 is the bypass capacitor. The value of R4 and R5 are chosen so that the bias voltage at their junction is the mid-voltage between 5V and ground which is 2.5V.

$V_{b} = \frac{R5}{R4+R5} V_{cc}$

Let $$V_{b}=2.5V$$, then solving above equation we get R4=R5.  Let choose R4=R5=10KOhm.

The cutoff frequency for C1 is given by,

$f_{c} = \frac{1}{2 \pi (R4||R5) C1}$

with C1=10uF, R4=R5=10KOhm and therefore, R4||R5 = 5KOhm we get the cutoff frequency of $$f_{c} = 3Hz$$

Let the input signal be sine wave with amplitude and frequency of 100mV and 1KHz. Let the input signal be at offset voltage of 1V. Then the input and output waveform is shown below for this circuit,

Thus using split resistor biasing or voltage divider biasing in dc coupled op-amp amplifier we cannot get output waveform arbitrarily.

### AC coupled Non-inverting amplifiers

An ac coupled non-inverting amplifiers have a capacitor at its input to block the dc signals and allow only ac signal with desired frequencies to pass through it. The following shows the generic ac coupled non-inverting op-amp amplifier with dual power supply.

The gain of this ac coupled non-inverting op-amp amplifier is given by,

$A = 1 + \frac{R1}{R2}$

For ac coupled amplifier there must be a DC path to ground for the input which is provided by the resistor R3. The coupling capacitor C1 along with resistor R3 forms a high pass filter whose cutoff frequency is given by the following equation.

$f_{c} = \frac{1}{2 \pi R3 C1}$

The value of resistors R3 and capacitor C1 are selected to allow input signal frequencies of interest to pass through it.

Consider that dual supply voltage is 12V. Let the gain be A = 11, this will set R1=100KOhm and R2=10KOhm. Let the cutoff frequency fc be 1.6Hz and let select C1=10uF, this will set R3 as 10KOhm. With these values we get the following ac coupled op-amp amplifying circuit.

As before let the input signal be a sine wave of amplitude 100mV, frequency of 1KHz and with dc offset of 1V. Then we get the following input and output signal waveform.

### Biasing of AC coupled Non-inverting amplifiers with Single Supply

When we want to use single supply, the above circuit must to be biased with the bias voltage in the mid range between the positive supply voltage and ground. If the positive supply voltage is +5V then the circuit is biased at the 2.5V to get maximum swing in the output.

Split resistor biasing

One way to bias the ac coupled non-inverting op-amp amplifier is to use the split resistor biasing technique. The following shows split resistor biasing technique with single supply voltage.

Here, the resistors R4 and R5 forms a resistor split biasing. The values of these resistors are usually selected to bias the input signal at voltage between the supply voltage and ground which in this case is 2.5V. So the dc bias voltage at the junction of the resistors R4 and R5 is around 2.5V. Thus the dc bias voltage is given by,

$V_{b} = \frac{R3}{R3+R4} V_{cc}$

With $$V_{b} = 2.5V$$  and $$V_{cc} = 5V$$, we get R3=R4. Let R3=R4=10KOhm.

The parallel combination of this resistors R3 and R4 along with the capacitor C1 sets the cut-off frequency which is given by,

$f_{c} = \frac{1}{2 \pi (R3||R4) C1}$

Let C1=10uF then with R3=R4=10KOhm we get cutoff frequency of fc=3Hz

In biasing the ac coupled amplifier a decouple capacitor C2 is also required. The cutoff frequency is given by,

$f_{c} = \frac{1}{2 \pi R2 C2}$

with C2=10uF and R2=10KOhm the cutoff frequency is 1.6Hz.

If the input signal is again a sine wave of amplitude 100mV, frequency of 1KHz and with dc offset of 1V, then we get the following input and output signal waveform.

As seen in the above waveform graph, the output is swings around the biased voltage 2.5 approximately.

Voltage divider biasing

Another biasing technique for ac coupled amplifier is to use the voltage divider biasing for single supply amplifiers. The following circuit diagram illustrates voltage divider biasing with single supply.

Here the resistors R4 and R5 sets up the bias voltage at the junction of R3 and C3. The capacitor C3 is a decoupling capacitor. The value of the resistors R4 and R5 are usually chosen so that the bias voltage is at the mid-value of the positive power supply and ground for single supply voltage.

$V_{b} = \frac{R3}{R3+R4} V_{cc}$

With $$V_{b} = 2.5V$$  and $$V_{cc} = 5V$$, we get R3=R4. Let R3=R4=10KOhm.

The decoupling capacitor value is chosen for cutoff frequency of interest which is given by,

$f_{c} = \frac{1}{2 \pi (R3||R4) C2}$

with C2=10uF, R3=R4=10KOhm, the cutoff frequency is 1.6Hz.

Similarly, the coupling capacitor C1 is chosen according to the cutoff frequency required and is given by,

$f_{c} = \frac{1}{2 \pi R3 C1}$

with C1=10uF, R3=10KOhm we get cutoff frequency of 3Hz.

Again a decouple capacitor C2 is also required for baising ac coupled amplifier. The cutoff frequency is given by,

$f_{c} = \frac{1}{2 \pi R2 C2}$

with C2=10uF and R2=10KOhm the cutoff frequency is 1.6Hz.

Let again apply an input sine wave of amplitude 100mV, frequency of 1KHz and with dc offset of 1V, then we get the following input and output signal waveform.

We can see that the output waveform swings around the bias voltage of around 2.5V.

One can also use the non-inverting op-amp biasing calculator to calculate the component values.