Measuring voltages without common ground - Page 1
Forum section: electronics / general

author: krish2487
author_rank: Frequent Contributor
timestamp: February 05, 2013, 05:48:49 am
content: Hello.I have a peculiar problem that I need some help with..I need to make a A2D measurement that uses a low side switch between a power supply and a 12V lead acid battery.The catch is that the measurement needs to be made with the switch off thereby removing the common ground between the power supply an the battery.The battery powers a 7805 regulator that powers a PIC16F676 and the pic makes the measurement.I am attaching a png file that contains the essential representation of the circuit in question. The low side switch is actually a N channel mosfet.I have simulated the circuit and made a DC sweep of the power supply.The readings in the excel file more or less represent what i have measured on a breadboarded circuit.I am unable to understand the reason.Without a common ground, where is a path for the current to flow that results in the measured/simulated readings of the power supply??I would be grateful if anyone here took the time to explain.Thanks.

author: mswhin63
author_rank: Frequent Contributor
timestamp: February 05, 2013, 05:55:04 am
content: The standard approach that I used many years ago mostly for analog circuits was to use an isolation transformer for the test gear. This may be easy for me as I have an isolation transformer though.

author: krish2487
author_rank: Frequent Contributor
timestamp: February 05, 2013, 05:57:53 am
content: Well..I should have mentioned it earlier.It is an existing product and I am just trying to understand how as the engineer who designed it at the company where i workis not with the company anymore and it seemed like a opportunity to learn something new..:-)

author: Psi
author_rank: Super Contributor
timestamp: February 05, 2013, 06:53:39 am
content: Can't you just throw a linear opto-isolator at it?

author: krish2487
author_rank: Frequent Contributor
timestamp: February 05, 2013, 07:13:04 am
content: As I mentioned earlier..It is a working product out on the field..It is not a project I am working on presently..I was going through the design papers and saw that particular circuit configuration.I tried to understand how it works by working it out on pen and paper.. then a multimeter..The rest of the circuit is pretty easy to understand..When the switch is on, no matter how much the power supply the A2D in point is pretty much at half of 5V i.e 2.5V.My interest lies in how the power supply voltage is being measured when the switch is off..( I do know for a fact that the power supply measurement is taken when the low side switch is off.)Besides, when the switch is on, the power supply (if it is greater than battery voltage) bucks to about 0.1 ~ 0.2V above the battery voltage.Hence the question...It was just an itch I needed to scratch...

author: codeboy2k
author_rank: Super Contributor
timestamp: February 05, 2013, 10:30:42 am
content: What is this ?? home work?Come on. It's basic KVL / KCL and node analysis.There is no need for there to be a common ground. There are voltages between points in every circuit.Write the KVL voltage loops, you will see it. That's all I'm going to give you.

author: lewis
author_rank: Frequent Contributor
timestamp: February 05, 2013, 10:58:36 am
content: At first glance this circuit looks totally weird, but it actually appears to be quite clever.There is a common ground! Let's work it through:Assume the switch in the circuit is actually a switch, not an N-MOSFET. SW1 is open. I'm assuming the ADC negative reference is ground.V1=20V. The input to the 5V regulator is 12.7V relative to ground because the battery sits between the 7805 input and ground. That means that relative to ground, the positive terminal of V1 is also at 12.7V. Since there is 20V between the + and - terminals of V1, relative to ground the negative terminal of V1 is at -7.3V, or 20V below the 12.7V voltage. Again, relative to ground, the top terminal of R2 is at 5V, and the left terminal of R1 is at -7.3V. R1 and R2 form a potential divider, and given these inputs it just happens to produce -1.15V at its output. (5-(-7.3)/2)+(-7.3) = -1.15VLet's do the same for V1=10V: V1+ = 12.7V relative to ground, V1- = 2.7V relative to ground, potential divider output = ((5-2.7)/2)+2.7 = 3.85V.Let's do the same for V1 = 1V: V1+ = 12.7V relative to ground, V1- = 11.7V relative to ground, potential divider output = ((5-11.7)/2)+11.7 = 8.35V. WRONG! Why? D1 conducts and limits the voltage at the potential divider output to about 5.6V. That's why V1 voltages from about 4V downwards in the excel file tail off to 5.56V. As V1 approaches zero, the potential divider allows more current to flow through the diode increasing it's forward voltage slightly which is why there are small differences the further down you go.I might be way off here but this circuit looks like a PWM lead-acid battery charger. If it wasn't, when the switch was closed and V1 was at 20V, an enormous current would flow into the battery and boil it's bollocks off. So it's gotta be. Or, the power supply isn't really a power supply at all, its a load for the battery and it works the other way.With that in mind, let's go for how I think the circuit really works:Firstly, we assumed that SW1 was a switch. It isn't. It's an N-FET. So that means there is the FET body diode across the switch. It's cathode faces the battery and anode faces V1-. (Although that means it's gate drive circuitry would have to be referenced to V1- which seems weird). This means that when the V1 voltage is less than the battery voltage minus a diode drop, the battery back-feeds the power supply. Clearly this circuit is designed for a power supply that is OK with that, and most of them are. But that does mean the simulation is not quite right - essentially V1 can never go below the battery voltage (minus a diode drop). So the data in the table from about 12V downwards is meaningless.(If i've got the direction of the FET wrong, then the body diode will allow the power supply to massively charge the battery all the time when it is more than about 15V, so that seems unlikely. However, if this thing is not a charger at all and is designed to allow the battery to power external circuitry, then the FET would go the other way, but V1 would never go above the battery voltage. I doesn't make a lot of sense.)Secondly, when the 'switch' is open, the ADC measures the difference between the power supply voltage and the battery voltage. It cannot measure the absolute value of either. This makes sense for a PWM charger, the input voltage needs to be a volt or so higher than the maximum charging voltage. However, what doesn't make sense is the need to measure the absolute battery voltage to determine state of charge which is not done on this circuit unless the supply is well regulated, defeating the whole point of everything.Thirdly, when the switch is closed, it's possible for the ADC to primitively measure the charging current. The charge current will create a voltage drop across the NFET which in turn means the left-hand input of R1 will be slightly below ground. The output of the potential divider will then vary very slightly with charge current, but it will be very non-linear and inaccurate, but might be enough to detect any gross short-circuits.If the ADC is built into a microcontroller, it is extremely unlikely it is bipolar, so it won't be able to measure negative voltages. That means everything in the table from 18V upwards is meaningless too. I would guess the ADC just makes a rough measurement of the difference between the charging voltage and the battery voltage when the switch is open, and measures approximate charging current when the switch is closed.I said at the start that there is a common ground. Remember that 'ground' is just the name for the reference point for all the voltages in a circuit. We can make anything we like ground. We could make the output of the potential divider 'ground' if we want, and all the other voltages in the circuit would move about this point. So let's make the V1+ terminal 'ground'. Suddenly there is a common connection between the power supply and the battery, and consequently voltages begin to appear when we probe about relative to this point whether the switch is open or not. Just because the switch is open, it does not mean the battery and PSU are isolated. As long as there is a path for current to flow about, stuff happens.You're probably going to tell me that its nothing to do with charging whatsoever, in which case ignore everything I've said!

author: krish2487
author_rank: Frequent Contributor
timestamp: February 05, 2013, 04:18:19 pm
content: @codeboyNo it is not homework...I am just weak at analog circuits...I do know it is an application of KCL/KVLbut did not get my head around how to solve it..@lewisThank you!!!The explanaion you gave makes perfect sense...You are right, it is a battery charger and we are talking about 12V 180Ah battery here.It is a Solar charger, not a PWM one and yes the FETS are not operating in PWM mode. just binary on or off.Regarding the switch, it is not a single switch it is a pair of back-to-back N FETS in D-S-S-D with the gates tied such that a signal switches on both the FETS but no reverse flow from battery to panel because of two body diodes back-to-back.The FETS have a really low Rds on allowing us to achieve charging current of 45 A or so without even a heatsink.Of course it is not a single fet but 3-4 pairs of fets in parallel.