6.3 Open Loop Voltage Comparator

A very common application of an op amp that makes deliberate use of saturation is the voltage comparator. This circuit compares two input voltages and produces a binary output voltage (that is, a voltage that can have only one of two possible values) to indicate which voltage is higher. Consider the circuit shown, with unipolar supply voltage V_CC.  Assume the op amp is ideal, with infinite open-loop gain, A. The output voltage hits the positive voltage supply rail, V_CC volts, whenever the differential input voltage is positive, i.e., when V₊>V_– . Likewise, the output voltage sits at the ground rail, 0 volts, whenever the differential input voltage is negative, i.e., when  V₊<V_– .  If we consider a real op amp, with finite open loop gain, such as A=10⁶, then the output would sit at the positive supply rail whenever V+ exceeds V- by 1 μV, a very small value.

Examples

Example:  Design an op amp comparator circuit that determines whether a test voltage, V_test, exceeds a reference voltage, V_ref.  The circuit should cause an LED to glow whenever V_test>V_ref.

Solution: The op amp is set up as a voltage comparator with inputs V_test and V_ref.  (We are not concerned with where these voltages come from in this circuit; in the next chapter, we will examine this using a more detailed circuit.) When powered with a unipolar supply voltage of +5V, the op amp output will be high (+5V) when V_test>V_ref and  low (0V or gnd potential) otherwise. An LED at the output of the op amp serves as an indicator, glowing when V_test>V_ref. The 100Ω resistor between the LED anode and the op amp output was chosen to provide a current of ~ 30 mA in the LED circuit, which would result in a bright LED.

 

Question: in the example just discussed, what value resistor should be used in the output circuit, in place of the 100Ω resistor, if the voltage supply rail of 12 V is used instead of 5V?  Assume, for this case, that the LED drops 2 V and should be operated at a current of 10 mA.

Answer: The high-level output from the op amp would be 12V. The resistor would drop 10 V, and the appropriate resistance would be R=V/I=10/.01 = 1000Ω=1kΩ.

Note that the output voltage was assumed to saturate at the voltage supply rail, 5 and 12 volts, in these two examples. We are ignoring the ~ 1 V voltage drop between the supply rail and the saturated output in these examples. 

The figure below shows how the previous circuit might be developed using  a pair of voltage dividers at the op amp inputs.  Op amp voltage rails V_CC and ground are connected to the + and – terminals of a 5 volt battery, respectively.  Two voltage dividers, each comprised of a pair of resistors in series, are connected in parallel with the battery.  In this circuit, V_ref=2.5 V and V_test is determined by the setting of variable resistor R. When R>5kΩ, V_test>V_ref and the output LED will glow.

Below we have another view of the same circuit with the ground symbol used in place of the lower node.

Next, we show a wiring drawing of the circuit using an LM358 Dual Op Amp IC. Only one of the two op amps on this IC is used in the circuit.