Electronics > Projects, Designs, and Technical Stuff
14.4V ground reference for ADC
Zero999:
--- Quote from: Simon on May 19, 2019, 07:08:15 pm ---
--- Quote from: Zero999 on May 19, 2019, 06:51:56 pm ---
--- Quote from: Simon on May 19, 2019, 04:37:33 pm ---That won't work. A µC is too slow to react to load changes plus the conversion and communication time of the ADC. Contrary to what youtube channels like Great Scott say using a microcontroller inside such a critically fast control loop does not work.
You could even use the LM358 to measure you current sense resistor now that you have it but specific chips for this job which are essentially a special purpose opamp cost pence/cents.
So what type of voltage regulator is this? what is doing the power control?
--- End quote ---
The problem with the LM358 is its common mode range does not extend to its positive rail.
An op-amp with inputs which do work up to the positive supply can be used with a transistor to amplify and level shift the signal to the 0V rail.
https://www.analog.com/en/analog-dialogue/articles/high-side-current-sensing-wide-dynamic-range.html
--- End quote ---
Yea, it can only go to within 1.5V of the positive rail so a voltage divider would be required each side. You loose a small bit of the 1V range but it can be made to work on the budget. there is already a current sense in the circuit that can be used for what the OP wants.
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Yes, but accuracy will be poor.
Another option would be to use an op-amp with P-JFET inputs, which will work up to the positive rail and be fine with a total supply voltage of over 15V, thus eliminating the zener diode. The trouble is cheap P-JFET op-amps tend to have a fairly high offset error: 3mV for the TL082, although that can be nulled out. Further cost cutting could be achieved by using a BJT, rather than a MOSFET, at the expense of reduced accuracy, although that could be mitigated somewhat by using one with a high hFE, such as the BC560C.
Simon:
Oh I am sure there are more suitable parts than the LM358, i was just trying to explain that what he is trying to do is far harder work that using what he already has to do something better. He wants to drop up to 1V on the shunt resistor, 1.5V on 18V does not require too much of a loss in that range. The circuit he has posted already has a current shunt amplifier. To be honest i am confused as to what he is trying to do here.
Zero999:
Oh, I think I know what you meant now. Were you talking about the standard differential amplifier configuration?
https://en.wikipedia.org/wiki/Differential_amplifier#Operational_amplifier_as_differential_amplifier
Yes, that should work well enough, as long as a the resistors are fairly well matched and the gain isn't too high, so the inputs remain within the common mode range of the amplifier. The voltage across R1 and R2 must be under 1.5V, which would limit the gain to a maximum of (18-1.5)/1.5 = 11, with a supply voltage of 18V, but at lower supply voltages, the maximum gain will be less.
Another issue with the LM358 is it doesn't have a true negative rail output. Its output is pulled low by a current sink and will not go near 0V, unless the output is sinking a tiny current, so R1 + RF will need to be quite large: over 1M for <18μA of output current sunk would be ideal.
Arjunan M R:
--- Quote from: wraper on May 19, 2019, 07:12:53 pm ---
--- Quote from: Arjunan M R on May 19, 2019, 05:15:30 pm ---Here is the circuit i am using to regulate voltage and controlling the current.Please have a look at it and if you spot any error please notify me.
The dumb thing about this is I don't need that precise 12 bit DAC .But i just used it because it is under my budget. :-//
Because Dave used it in his microsupply(REV A).
So i was sure it will work and I don't haveto respin the pcb.
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It is non functional. For example because all opamps are powered from 5V. EDIT: Nope, not all (point stands) powered from 5V but why in the hell would you note +18 V as VCC? :-//.
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+18V is regulated to +5V using L7805.Because +18V is the common supply voltage for everything in this schematic I noted +18V as VCC as I said this is not the full schematic.
Arjunan M R:
--- Quote from: Simon on May 19, 2019, 06:06:42 pm ---Well there is already a current sense circuit there, you just need to connect your ADC input to that. You have an independant voltage regulator LT3080 so I'm not sure what your trying to do.
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Yes you are right.I will change the circuit.I will connect the ADC input to the current sense amp output so i can get rid of that 14.4V GND REF.It was a differential ADC so I thought i can just measure across that shunt :-DD |O.
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