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I'm analyzing the circuit in the picture below. I've read a little about unity gain amplifiers/buffers and think I understand the use of a single buffer. But I don't understand the purpose of having two buffers, one after the other.

Here are some more info about the circuit:

  • OPAMPS: MCP6002, A and B are in the same package
  • Input is from a 4 .. 20 mA signal (4 mA = 0.88 V at OpAmp-A's input, 20 mA = 4.4 V at OpAmp-A's input)

My questions:

  1. Why have two buffers or what is the designer trying to avoid by using two?
  2. What is the purpose of the 4.7k and 1k resistors? I.e. what will happen if I change, lets say the 4.7k to a 1k?
  3. Same as question 2 but regarding the capacitances 100 nF and 10 nF

enter image description here

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    \$\begingroup\$ Obviously, the designer wanted two stages of filtering, but with no gain required and the low impedances involved, it isn't clear why any buffering at all was required. Perhaps he thought the circuit didn't consume enough power, or that it needed more noise, offset errors, etc. \$\endgroup\$ – Dave Tweed Apr 11 '18 at 13:53
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The RC dividers (4.7kΩ and 100 nF, 1kΩ and 10 nF) are filters, with shoulder frequencies of 1/(2πRC) each, giving you two 1-pole low-pass filters cascaded into a 2-pole LPF. They're buffered so that each filter doesn't load the prior stage. The resistor divider out front sets the gain.

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  • \$\begingroup\$ It would make a whole lot more sense to put the buffers after the filters, not before. \$\endgroup\$ – Dave Tweed Apr 11 '18 at 13:54
  • \$\begingroup\$ If the ADC input is high-impedance, which it almost certainly is, then the buffers are after the filters, considering the resistor divider is a filter than simply adjusts gain. \$\endgroup\$ – schadjo Apr 11 '18 at 14:08

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