Energia Domestica y Salud/ Fuel for Life: Combustibles Para by Eva Rehfuess

By Eva Rehfuess

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However, the bad news is that while unity gain buffer chips are available, we are not recommending that you use one. In fact, unity gain buffers are usually constructed from operational amplifiers, and hopefully you have some of these. Operational amplifiers, or op amps for short, are the work horses of the analog circuit world. They measure the voltage difference between two nodes, and output a V+ V- A Vout = A (V+ - V-) voltage which is A times greater. Or at least that is the ideal. The gain, A, is so big, generally 106 or more, that even very small input differences quickly overwhelm the output voltage range.

In this case, instead of feeding back the output to the negative input, we Vout = A (V+ - V-) V+ A V- Rf Vin Rs first reduce the output with a voltage divider, and then feed back the reduced voltage. This has a rather curious effect. From the voltage divider equation, we can write: Rs V - = V out -----------------Rs + R f 42 A Pragmatic Introduction to the Art of Electrical Engineering What is an Amplifier? Plugging this into the equation for Vout, we get: Rs V out = A ( V + – V - ) = A  V in – V out ------------------  Rs + R f  Collecting the terms yields: AR s  - = AV in V out  1 + ---------------- Rs + R f  Solving for Vout and approximating for large A gives: AV in Rs + R f R V out = ---------------------------- ≅ V in ------------------ = V in  1 + ------f   AR s Rs Rs  1 + -----------------Rs + R f Oddly, the result is an amplifier with gain equal to the inverse of the voltage divider’s gain.

I 4kΩ VR1 5V 6kΩ VR2 Vout We begin by noting that the resistors are in series. Applying our result from before, we know that resistors in series effectively add together. 5mA. This current flows through each resistor. The voltage across the first resistor is given by VR1 = I R1 = 2V. For the second resistor, VR2 = I R2 = 3V. This is Vout. We could also have calculated this directly from the voltage divider equation: R2 6kΩ V out = V -------------------- = 5V ---------------------------- = 3V R1 + R2 6kΩ + 4kΩ It is also good to verify that KVL works.

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