![\"Komparator](\"https:\/\/i0.wp.com\/transistorgrab.de\/figures\/komparator_mit_hysterese-1.png\" "\\"Komparator")
<\/a>\n<\/a><\/div>\n
\nOn the left side are two voltage dividers. R4<\/em> and R5<\/em> divide the input voltage V1<\/em> and generate the voltage V3<\/em>. R6<\/em> and R7<\/em> divide the input voltage V2<\/em> and generate V4<\/em>.
\nWith R3<\/em> and R2<\/em> the histeresis will be set, the use of R1<\/em> is only this of a current limiter to following circuits.
\nThe maximum output voltage of this circuit depends on the supply voltage of the OP.
\nFor the following observations all possible losses are neglegted. <\/span><\/div>\n
![<\/a>](\"http:\/\/transistorgrab.de\/figures\/komparator_invertierend.png\"){kind=link}
![\"Komparator](\"http:\/\/transistorgrab.de\/figures\/komparator_mit_hysterese-1.png\" "\\"Komparator")
<\/a>\n<\/a><\/div>\n
For the following observations the positive supply voltage is defined to 5 Volts, the negative supply voltage is defined to 0 Volt(supply voltage not shown in the figure). V3<\/em> is the reference voltage and defines when the comparator will switch. If the voltage V4<\/em> on the inverting input (\u2212) sinks below the voltage V+<\/em> then the output voltage Va<\/em> of the OP will raise to the value of the positive supply voltage. If the voltage V4<\/em> is higher than V+<\/em> then the output voltage will be the value of the negative supply voltage.<\/p>\n R2<\/em> causes a positive feedback.<\/p>\n Lets take a look on the following cases:<\/p>\n 1st case<\/strong>: V4<\/em> is higher than V+<\/em> 2nd case<\/strong>: V4<\/em> is less than V+<\/em> V+<\/em>=2.5\u00a0V+((5\u00a0V\u22122.5\u00a0V)*(10\/20))=3.75\u00a0V. When the value of V4<\/em> is higher than that voltage the comparator switches. 1st possibility<\/strong><\/p>\n If you swap the input terminals of the OP, thus connecting V+<\/em> with the inverting input (\u2212) of the OP and V4<\/em> with the non-inverting input (+), the following behaviour occurs: 1st case<\/strong>: V4<\/em> is greater than V+<\/em><\/p>\n That leads to: The voltage on (+) is higher than the voltage on (\u2212), the output has therefore the value of the positive supply voltage.(Va<\/em>=+5\u00a0V). V+<\/em> is calculated as in the case 2<\/em> above: V+=<\/em>3.75\u00a0V. For the comparator to switch the voltage on (+) must be lower than the voltage on (\u2212), that means V4<\/em> must be below 3.75\u00a0V. <\/strong><\/p>\n 2nd case:<\/strong> V4<\/em> is less than V+<\/em><\/p>\n That leads to: The Voltage on (+) is lower than the voltage on (\u2212), because of that the output is on ground level(Va=<\/em>0V). The voltage\u00a0 V+<\/em> is calculated as in the 1st case<\/em> of the inverting Schmitt-Trigger: V+=<\/em>1.25\u00a0V. For the comparator to switch the voltage on (+) must be higher than V+<\/em>. If V4<\/em> is higher than 1.25\u00a0V the compatator will switch. Let’s pretend V4<\/em> is 2.5\u00a0V. What additional possibilities are there?<\/p>\n 2. possibility We connect the reference voltage with the inverting input (\u2212) of the OP and connect the voltage to be monitored on V3.<\/em> What happens now? With the given values for the devices we can calculate what value has V+ <\/em>depending on V3<\/em>.<\/p>\n Let’s have a look on the both possible cases.<\/p>\n 1st case<\/strong>: V4<\/em> is higher than V+<\/em> The voltage divider R3\/R2 is now between V3<\/em> and 0\u00a0V. The value of R2<\/em> is 10\u00a0k\u03a9, the value of R3<\/em> too. Ohm’s Law dictates: V+\/R2 = V3\/(R2+R3)<\/em> thus V3 = V+*(R2+R3)\/R2 = 2,5\u00a0V*<\/em>(20)\/10 = 5\u00a0V.<\/p>\n If V3<\/em> is more than 5\u00a0V the comparator will switch.<\/p>\n 2nd case<\/strong>: V4<\/em> is less than V+<\/em> The voltage divider R2\/R3<\/em> is now between 5\u00a0V and V3<\/em>. V+ = V3 + (<\/em>5\u00a0V\u2212V3) * R3\/(R2+R3) = V3 + (<\/em>5\u00a0V \u2212V3) * <\/em>0,5.<\/p>\n After conversion of this equation we get V3 = <\/em>2*V+<\/em> \u22125\u00a0V = 2*2,5\u00a0V\u22125\u00a0V = 0\u00a0V.<\/p>\n With the given values for R2<\/em> and R3<\/em> this comparator is not usable, because in case 1 V3<\/em> needs to be higher than 5\u00a0V and in case 2 less than 0\u00a0V.<\/p>\n But we excluded these conditions at the start of our observations.<\/p>\n Conclusion:<\/strong> In the 1st case<\/strong> we want to achieve that V+<\/em> is higher than 2.5\u00a0V when V3<\/em> is less than 5\u00a0V.<\/p>\n We define that V+ <\/em>shall be 2.5\u00a0V when V3 =<\/em> 4\u00a0V.<\/p>\n It makes no sense to change both resistors.<\/p>\n We let R2=<\/em>10\u00a0k\u03a9. V3\/(R2+R3)=V+\/R2<\/em>. Thus R3 = (V3\/V+) * R2\u2212R2 = (<\/em>4\/2,5) * <\/em>10\u00a0k\u03a9\u221210\u00a0k\u03a9 = 6\u00a0k\u03a9.<\/p>\n For the 2nd case<\/strong> that leads to: <\/em><\/p>\n V+ = V3 + (Va\u2212V3) * R3\/(R2+R3)<\/em>.<\/p>\n Converted for V3<\/em>: <\/em><\/p>\n V3 = ((R2\u00a0*\u00a0V+)\u00a0\u2212\u00a0R3*(Va\u2212\u00a0V+))\/R2 =<\/em>((10k\u03a9*2,5\u00a0V)\u22126\u00a0k\u03a9*(2,5\u00a0V))\/10\u00a0k\u03a9 =(2,5\u00a0\u2212\u00a01,5)\u00a0V\u00a0=1\u00a0V.<\/p>\n The calculations shows that this approach brings the desired success.<\/p>\n It is necessary that the resistors R2 and R3 are not <\/strong>of the same value.<\/p>\n R3<\/em> needs to be less than R2<\/em>.<\/p>\n The exact values depend on the desired threshold and on the desired hysteresis.<\/p>\n The components for the non-inverting Schmitt-Trigger need other values than for the inverting if you want to achieve the same hysteresis.<\/p>\n For comparision we put the values of R2<\/em> and R3<\/em> for the non-inverting Schmitt-Trigger into the formulas for the inverting Schmitt-Trigger.<\/p>\n With that values the threshold voltages of V4<\/em> are 1.56\u00a0V and 3.44\u00a0V. In some cases one needs a comparator with a hysteresis. This curcuit suggestion shows further the possibility of a inverting or non-inverting operation. The ciruit above shows a typical comparator circuit with addional input voltage dividers. On the left side are two voltage dividers. R4 and R5 divide the input voltage V1 and generate the … Continue reading Comparator With Hysteresis(Schmitt-Trigger)<\/span>
\nR2<\/em> and R3<\/em> are defined to 10\u00a0k\u03a9. V3<\/em> will be set as reference voltage to a value of 2.5\u00a0V.<\/p>\n
\nThat leads to: the ouput of the OP has ground potential. So V+<\/em> can be calculated as follows:
\n.
\nV3<\/em> is 2.5\u00a0V. Thus V+<\/em>=2.5\u00a0V*(10\/20)=1.25\u00a0V.
\nFor the comparator to switch\u00a0 (when Va<\/em>=5V), V4<\/em> needs to be less than V+<\/em>, that means less than 1.25\u00a0V.<\/p>\n
\nThe output is on the level of the positve supply voltage(Va <\/em>= 5\u00a0V). Va<\/em> is greater than V3<\/em>. The voltage divider consisting of R2 <\/em>and R3 <\/em>has an offset to the ground wich is the value of V3<\/em>. That needs to be considered in the calculation. The voltage V+<\/em> is calculated as follows:<\/p>\n<\/div>\n
\nThe histeresis in this case is (3.75\u22121.25)\u00a0V=2.5\u00a0V.
\nThe comparator is symmetric to the voltage V3<\/em>.Observation of a non-inverting Schmitt-Trigger<\/strong><\/span><\/p>\n<\/a>
\n<\/a><\/div>\n
\nV+<\/em> is now connected to the inverting input (\u2212) of the OP.<\/p>\n
\nConclusion<\/strong>:
\nLet’s look on the requirements for the switch in case 1 and case 2: for the ouput voltage to go low (Va=<\/em>0V) V4<\/em><\/strong> must be less than 3.75\u00a0V<\/strong>. For Va=<\/em>5\u00a0V, V4<\/em><\/strong> must be higher than 1.25\u00a0V.<\/strong><\/p>\n
\nThen\u00a0 both<\/strong> contitions are true. That may lead to a constant switching of the output of the OP between\u00a0 0\u00a0V und 5\u00a0V: We have created an\u00a0 oszillator<\/strong>.
\nObviosly this approach does not work<\/strong>.<\/p>\n
\n<\/strong><\/p>\n<\/div>\n
\nLet’s take a look on that 2nd possibility and use the original circuit. On V4<\/em> is now a voltage of 2.5\u00a0V. On V3<\/em> we connect a variable voltage that can change between 0\u00a0V and 5\u00a0V.<\/p>\n
\nAs long as V4 <\/em>is higher than\u00a0 V+<\/em>, then Va=<\/em>0\u00a0V.<\/p>\n
\nVa=<\/em>0\u00a0V. For the comparator to switch V+<\/em> must be higher than V4<\/em> werden. V4 <\/em>is 2.5\u00a0V. V+<\/em> must be higher than 2.5\u00a0V.<\/p>\n
\nVa=<\/em>5\u00a0V. For the comparator to switch V+<\/em> must be less than V4<\/em> werden. V4=<\/em>2,5\u00a0V.<\/p>\n
\nThe resistors R2<\/em> and R3<\/em> need other values.<\/p>\n
\nThe hyteresis with these values of\u00a0 R2<\/em> and R3<\/em> in this case is 1.88\u00a0V for the inverting and 4\u00a0V for the non-inverting Schmitt-Trigger.<\/p>\n","protected":false},"excerpt":{"rendered":"
![<\/a>](\"http:\/\/transistorgrab.de\/figures\/komparator_falsch_nicht_invertierend.png\"){kind=link}