Electronics > Beginners
Measure reversible voltage without PD drop?
David Hess:
--- Quote from: coppice on August 10, 2019, 04:32:35 pm ---It would be interesting to see how they do the input protection in that device. Its only single ended, though. Do you know what causes the common mode limits you referred to?
--- End quote ---
The first problem is that if a common junction isolated semiconductor process is used, then most or all semiconductor nodes are isolated from the substrate by a junction diode which is why no pin can be pulled negative. Older ICs sometimes had a separate substrate connection but modern parts tie the substrate to the negative most input which is usually the negative supply pin. For instance the LM317 has the substrate tied to the adjust pin and the LM337 has the substrate tied to the negative input pin. This is why these regulators have different pinouts; the center pins connect to the tabs and that is where the substrate connects.
But for an input that goes to an inverting amplifier, the junction at the input after the series resistor is held at ground. Linear Technology makes some switching regulator controllers which use this method to allow feedback from negative voltages. I think some ADCs and DACs do this also.
Dielectric isolated semiconductor processes avoid this problem entirely.
12407622:
--- Quote from: langwadt on August 11, 2019, 01:51:04 am ---
if everything is on the same ground you will short half the diodes. How about two diodes one from each terminal of the motor, when measuring turn on low side fet on the low voltage side
--- End quote ---
Hi langwadt,
I drew it out, but can't get my head around the diode shorting you've alerted me to. Yes, I think the ground is common, at least while the H-Bridge is energised. I need to be cautious, both when the motor is driven, and disconnected and freewheeling, it sounds like?
Not sure I can isolate any one leg of the H-Bridge chip I'm using, but low PD drop through 'active' rectification, by switching of an additional 'measurement' network of FETs sounds interesting.
Thanks for the tips,
12407622 :D
digsys:
I've used these in the past, for HV battery packs - https://datasheets.maximintegrated.com/en/ds/MAX9928-MAX9929.pdf
Ran multiples on isolated supplies, up to 300V. What I like about them, single FS ADC output, with a sign pin for direction. Pretty fast as well. May be useful info
coppice:
--- Quote from: David Hess on August 11, 2019, 03:43:23 am ---
--- Quote from: coppice on August 10, 2019, 04:32:35 pm ---It would be interesting to see how they do the input protection in that device. Its only single ended, though. Do you know what causes the common mode limits you referred to?
--- End quote ---
The first problem is that if a common junction isolated semiconductor process is used, then most or all semiconductor nodes are isolated from the substrate by a junction diode which is why no pin can be pulled negative. Older ICs sometimes had a separate substrate connection but modern parts tie the substrate to the negative most input which is usually the negative supply pin. For instance the LM317 has the substrate tied to the adjust pin and the LM337 has the substrate tied to the negative input pin. This is why these regulators have different pinouts; the center pins connect to the tabs and that is where the substrate connects.
But for an input that goes to an inverting amplifier, the junction at the input after the series resistor is held at ground. Linear Technology makes some switching regulator controllers which use this method to allow feedback from negative voltages. I think some ADCs and DACs do this also.
Dielectric isolated semiconductor processes avoid this problem entirely.
--- End quote ---
Its rare for an ADC to have an actual amplifier at its input, despite them appearing in many block diagrams. Those diagrams only represent the functionality, not the implementation details.
Zero999:
Why not simply use a differential amplifier?
http://www.eetimes.com/document.asp?doc_id=1272328
For a single supply application, the output can be biased, so it's centred around half the power supply voltage, say 2.5V, for a 5V supply. If the input voltage is +1V, then the output will be 2.5+1 = 3.5V and if the input voltage is -1V, then the output will be 2.5+(-1) = 1.5V.
If you need to handle +/-5V, then the circuit will need to have a gain of 0.5 and the op-amp a rail-to rail output.
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