Switchable Signal Inverter/Non-Inverter Circuit with Op-Amp and Switch

This circuit is designed with an operational amplifier (op-amp) and a switch, allowing the signal phase to either be inverted or remain unchanged as it passes through the circuit. In other words, the output signal can be 180 degrees out of phase with the input signal (inverted) or 0 degrees/360 degrees out of phase (non-inverted). Essentially it acts as a phase shifter circuit.

Circuit Operation:

The input signal is connected to both terminals of the op-amp. Specifically, the signal is directly connected to the inverting terminal, while the non-inverting terminal is connected to the signal via a switch.

• When the switch is positioned such that the non-inverting terminal is grounded, the circuit operates as an inverting op-amp amplifier. The gain of the inverting op-amp is given by:

  $$A = -\frac{R_{in}}{R_f}$$

  In this case, $= R$, so the gain is:

  $$A=-1$$

  Therefore, when the non-inverting terminal is grounded (via the switch), the output signal is the inverted version of the input signal. For example, the input signal (a sine wave) will be inverted, and its phase will be shifted by 180 degrees. This results in a cosine wave because of the negative sign in the gain equation.

Mathematical Derivation of Phase Inversion:

Let the input signal $V_{in}$ be represented as:

$$V_{in}=A_c\sin (\omega t)$$

where $A_c$is the amplitude and $\omega = 2\pi f$ is the angular frequency of the input signal.

When the switch is set to ground the non-inverting terminal, the output signal $V_{out}$ of the op-amp is:

$$V_{out} = -A V_{in} = -1 \times V_{in} = -1 \times A_c \sin(\omega t)$$

Since $\sin(\omega t)$ can be written as $\sin(180^\circ - \omega t)$, the output becomes:

$$V_{out} = A_c \sin(180^\circ - \omega t)$$

Using the trigonometric identity $\sin(180^\circ - \theta) = \cos(\theta)$, we get:

$$V_{out} = A_c \cos(\omega t)$$

Thus, the output is a cosine wave, 180 degrees out of phase with the input sine wave. This is shown in the picture below.

The following video illustrates this phase inversion of the input signal.

• When the switch is moved to the upper position, the input signal is applied to both the inverting and non-inverting inputs of the op-amp simultaneously. The voltage gain for the inverting channel remains:

  $$A_v(\text{inv}) = -1$$

  The voltage gain for the non-inverting channel(see non-inverting op-amp amplifier) is:

  $$A_v(\text{non}) = \frac{R}{R + 1} \approx +1$$

  The total voltage gain is the algebraic sum (superposition) of the two gains:

  $$A_v = A_v(\text{inv}) + A_v(\text{non}) = -1 + 1 = 0$$

  Therefore, the circuit is a switchable inverter/non-inverter. It can produce a voltage gain of either +1 or -1, depending on the switch position. This means that the output voltage will have the same magnitude as the input voltage, but the phase can be switched between 0° and 180°. For example, if $V_{in} = 15 \, \text{V}$, the output $V_{out}$ will be either +5 V or -5 V, depending on the switch position. This phase inversion or phase shift of digital signal is illustrated in the pictures below.