![]() ![]() This stands as a stark contrast to single-transistor amplifier circuit designs, where the Beta of the individual transistor greatly influenced the overall gains of the amplifier. The voltages and currents in this circuit would hardly change at all if the op-amp’s voltage gain were 250,000 instead of 200,000. Just as with the voltage follower, we see that the differential gain of the op-amp is irrelevant, so long as its very high. For this reason, this circuit is referred to as a noninverting amplifier. A positive input voltage results in a positive output voltage, and vice versa (with respect to ground). However, the gain can be increased far beyond 1, by increasing R 2 in proportion to R 1.Īlso note that the polarity of the output matches that of the input, just as with a voltage follower. Since the voltage follower has a gain of 1, this sets the lower gain limit of the noninverting amplifier. If we were to lower R 2 to a value of zero ohms, our circuit would be essentially identical to the voltage follower, with the output directly connected to the inverting input. Note that the voltage gain for this design of amplifier circuit can never be less than 1. Gain can be calculated by the following formula: We can change the voltage gain of this circuit, overall, just by adjusting the values of R 1 and R 2 (changing the ratio of output voltage that is fed back to the inverting input). The 6 volt signal source does not have to supply any current for the circuit: it merely commands the op-amp to balance voltage between the inverting (-) and noninverting (+) input pins, and in so doing produce an output voltage that is twice the input due to the dividing effect of the two 1 kΩ resistors. Using the null detector/potentiometer model of the op-amp, the current path looks like this: Upon examining the last illustration, one might wonder, “where does that 6 mA of current go?” The last illustration doesn’t show the entire current path, but in reality it comes from the negative side of the DC power supply, through ground, through R 1, through R 2, through the output pin of the op-amp, and then back to the positive side of the DC power supply through the output transistor(s) of the op-amp. ![]() Counting up voltages from ground (0 volts) to the right-hand side of R 2, we arrive at 12 volts on the output. Knowing the current through R 2 and the resistance of R 2, we can calculate the voltage across R 2 (6 volts), and its polarity. In other words, we can treat R 1 and R 2 as being in series with each other: all of the electrons flowing through R 1 must flow through R 2. Because we know that both inputs of the op-amp have extremely high impedance, we can safely assume they won’t add or subtract any current through the divider. This gives us 6 mA of current through R 1 from left to right. Since the left-hand side of R 1 is connected to ground (0 volts) and the right-hand side is at a potential of 6 volts (due to the negative feedback holding that point equal to V in), we can see that we have 6 volts across R 1. With the 2:1 voltage divider of R 1 and R 2, this will take 12 volts at the output of the op-amp to accomplish.Īnother way of analyzing this circuit is to start by calculating the magnitude and direction of current through R 1, knowing the voltage on either side (and therefore, by subtraction, the voltage across R 1), and R 1‘s resistance. If R 1 and R 2 are both equal and V in is 6 volts, the op-amp will output whatever voltage is needed to drop 6 volts across R 1 (to make the inverting input voltage equal to 6 volts, as well, keeping the voltage difference between the two inputs equal to zero).
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