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I'm studying the schematic for the constant current/constant voltage (CC/CV) indicator on an Agilent E3610A:

enter image description here

My question is: What purpose is served by the 10kΩ resistor (R26, R16) on the non-inverting input to the LM393 comparator?

The circuit operation is pretty straightforward. When the constant current error amp (U4B) is controlling the pass device, it's output voltage is around -1V. When the constant voltage error amp is controlling the pass device, the constant current error amp output jumps up to near the positive rail, say 14V in this case. The inverting input to the comparator (U5A) is set at around 12.3V. So the LED comes on when the constant current error amp is in control.

The thing I don't get is why a resistor is required between the error amp (LM442 op amp) and the comparator input. The input impedance of the LM393 is megohms, isn't it? What difference would an extra 10k make?

The only hypothesis I have is that I understand from the datasheet that if one of the inputs is driven more than 0.3V more negative than the negative supply, that as much as 50mA can flow out of the input pin. So I suppose this might be a safety precaution; but I can't work out what anticipated condition would drive that voltage down that far since it uses the same supply rails. So that possibility isn't making much sense to me.

Can anyone help me understand why the design engineer might have added these resistors?

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    \$\begingroup\$ I believe (remembering back to circuit design class circa 1999) this helps minimize offset errors due to offset current. Op-amp inputs do have some bias current flowing in or out, so the idea would be to match the impedance seen looking out of the other input. So putting a resistor there is a good idea, but 10k seems way to high. \$\endgroup\$ – mkeith Nov 10 '15 at 6:04
  • \$\begingroup\$ Yes, I should have added this hypothesis. I did consider equalizing offset voltage related to bias current as a possible reason, but then the value should be closer to 1kΩ, i.e. R25 in parallel with R41, wouldn't it? Also the input values are pretty sharply one way or the other, so I couldn't figure out why a mV or so one way or the other would matter. I notice that if I fiddle carefully with the setpoint controls, I can get both LEDs to light up at once. I'm figuring this is actually kind of a feature, indicating you're at the cusp of both setpoints; I wonder if it's related to that somehow. \$\endgroup\$ – scanny Nov 10 '15 at 6:43
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    \$\begingroup\$ Just doing the math, I see that 400nA Input Bias Current across a 100kΩ resistor develops 40mV. That might be enough to widen the trip point enough to get the "both on at once" behavior for a narrow window. This would actually make sense from a user interface standpoint, to avoid a "neither on at once" behavior. Maybe that's it. I notice the bias currents flow out of the input pins, so that would keep the non-inverting input say 40mV higher than the error amp output. \$\endgroup\$ – scanny Nov 10 '15 at 6:53
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    \$\begingroup\$ Many linear PSU get stung with Big RF fields .Try using the ptt of a CB or marine VHF and you may see the voltage move the current limit setting change or all sorts of strange behaviour .The resistors will stop RF mucking up the lamps if they are close to the chip. \$\endgroup\$ – Autistic Nov 10 '15 at 10:52
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    \$\begingroup\$ Scanny I guess that you are designing a linear psu . The resistors have more impedence than a choke so they actually work better and they are broadband .Because the circuit is high impedence the DCR is not a penalty .In fact some DCR can be of advantage as others have said.Please make your psu RF proof because otherwise it will be useless to many people . Remember that any circuit switchmode or linear can get stung ! I did a S TRAP buck convertor that flew through radiated and conducted EMC with no EMC components only to find that it got stung by a VHF transciever 1 meter away. \$\endgroup\$ – Autistic Nov 10 '15 at 19:12
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enter image description here

Here is the schematic of op Amp LM393.

You see the internal diodes on input stage. You have to use external resistors in series to the input for limiting the internal diode currents. Otherwise your accuracy will decrease.

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