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I am analyzing the following circuit used as an analogue input protection for an ADC ( image source: http://www.analog.com/library/analogDialogue/archives/43-04/process_control.html ):

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I don't understand the purpose of c5, c6, and D3. So my questions:

1- I'm guessing c5 can reduce spikes on the input. Is this correct?

2- I have no idea what D3 (1n4148) does. If a negative voltage is placed at the input of the circuit this diode would immediately fry (defeats the purpose of the circuit).

3- I also have no idea what c6 does?

4- Also the ref input is connected to 0.5 volts. Why? The op-amp is rail-to-rail, why is there any need to shift the output by 0.5 volts?

P.S. Anti aliasing filter is placed on the ADC input and is not shown in this circuit.

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A partial answer

1 - Usually a low value capacitance is placed as close as possible to the connector for better EMI and RFI.

2 - Me neither

3 - Also, don't know. Although I guess it is a very slow filter, since the signals are industrial and relatively slow. What confuses me is that it was placed upstream of the resistor divider.

4 - The offset is done because the input ranges are symmetrical and the ADC input is single ended.

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  • \$\begingroup\$ Thanks for the response. Regarding point 4: this circuit was developed for unipolar inputs (0-5,10 v) not symmetrical inputs. \$\endgroup\$ – hadez Apr 9 '16 at 20:50
  • \$\begingroup\$ " In this case, the input signal is attenuated, amplified, and level shifted to provide a single-ended input to the ADC" - page 3. Some of the analog inputs are single ended, but others are bipolar. \$\endgroup\$ – Claudio Avi Chami Apr 9 '16 at 20:57
  • \$\begingroup\$ Yes but it continues on the next paragraph that "This channel is designed to accept 4 mA to 20 mA, 0 V to 5 V, and 0 V to 10 V analog inputs. Other channels in the input module have been designed for bipolar operation to accept ±5 V and ±10 V input signals." The current circuit is for unipolar channels. As I stated earlier, applying a negative voltage to the current circuit e.g. 5 volts would probably fry the 1n4148 diode :( \$\endgroup\$ – hadez Apr 9 '16 at 21:04
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Others have addressed most of the points. I just wish to address R5, S2 and D3.

The 250 Ω switched resistor is a common feature on industrial analog circuits such as PLC (programmable logic controller) inputs. It converts a 4 - 20 mA (or 0 - 20 mA) signal to 1 - 5 V (or 0 to 5 V) for processing by the ADC. Leaving switch S2 open makes the input a voltage input. Closing S2 converts it to a current input. (A resistor is, after all, a current to voltage converter.)

D3 will protect the circuit in the event of a reverse connection of a 4 - 20 mA signal. There is little danger of the diode burning out as the current is externally limited.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Typical industrial 4 - 20 mA input. Without R1 the circuit becomes a voltage input. With R1 a 20 mA signal will produce a 5 V input to the ADC.

schematic

simulate this circuit

Figure 2. D3 is more likely to protect against reverse connection in this case.

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  • \$\begingroup\$ Thank you for the answer. But I believe D4-a and D4-b are supposed to provide the protection you speak of. And considering the fact that that D3 (1n4148) is always in the circuit and does not get disconnected by the switch, I think it should have some other purpose? \$\endgroup\$ – hadez Apr 10 '16 at 0:41
  • \$\begingroup\$ Maybe the people who designed this circuit simply copy-pasted it from elsewhere without thinking about the purpose of the diode and capacitor?! See Fig. 2 here: sound.westhost.com/appnotes/an011.htm . It is stated that "D3 and C2 help protect the MOSFET from external nasties, and C2 also prevents the MOSFET from oscillating with long leads attached." Designers used an old design but accidentally kept some unnecessary parts? \$\endgroup\$ – hadez Apr 10 '16 at 0:51
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The use of so many elements seems to partially be for redundancy, though the various protection techniques will apply to slightly different voltages/frequencies.

  1. The small-value C1 input capacitor helps filter out RF signals, and is likely high-breakdown voltage so it can handle ESD events. It probably also helps shunt the high frequency energy in ESD.

    The TVS will mostly help with ESD, but these devices can't dissipate much power, so larger diodes are also used.

  2. D3 is a fairly big diode. It can handle about 0.5 W continuous power. I can only imagine it as being for protecting against negative input voltages. Perhaps it is used to shunt the current away from the TVS, so that the circuit can handle a higher negative voltage continuously.

  3. C6 looks like it serves as a low-pass filter when the input is a current signal. It, along with the resistors, will set the low-pass cutoff frequency
  4. An ADC will only digitize signals above some minimum voltage limit. Shifting up the voltage would allow a slightly negative voltage to be properly digitized, or a zero input to be in the range of an ADC requiring a some minimum input voltage.
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  • \$\begingroup\$ 1- ok. 2- I disagree. The TVS is usually rated for at least 24 volts. Assuming a mere negative voltage of -5 volts on the input, the TVS will not conduct, but the 1N4148 will pass 7A of current (and fry). 3- I doubt the capacitor is being used for filtering when reading current because it is also in the circuit when reading voltage, and the resistors are voltage dividers reducing the voltage to an amount suitable for the ADC). 4- I don't know. The ADC used reads voltages from zero and upwards and the circuit only reads positive voltages. Doesn't make sense to shift the voltage. \$\endgroup\$ – hadez Apr 10 '16 at 0:35

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