Volt Ampere Meter

I'm trying to get real time data off of a rectifier by collecting it on a microcontroler and passing it to a computer. I am using a voltage divider to get voltage (A0 PIN) and the drop across the shunt resistor for current (A1 PIN). The problem is that when I apply rectifier I only get signal of the A1 PIN and no signal on the A0 Pin.

What am I doing wrong?

You speak of a rectifier, but the circuit does not show a diode. It seems to me that you may have simplified your schematic with the generator symbol and obscured the actual connections.
If you put a voltmeter or scope on the A0 and A1 locations, what do you read? They should be about the same, since the R1/R2 voltage divider is about 95%. What is R4 doing?

95% of what? The "rectifier" is actually a power supply. It will regulate voltage and current.

A0 reads -0.30V. A1 reads 1.3V. R4 is a buffer between the current and microcontroler.

_OK, it's a bit clearer now.

There's clearly something odd going on with those voltages. If A1=1.3volts, the current through R3 will be 1.3/0.47=2.77A. R4 should not contribute any voltage drop if the input resistance of the micro's ADC is high enough (as it usually is), though 87K is an odd sort of value.

If A0=-0.3 volts, then how do you suppose a negative voltage is getting created there?
The R1-R2 divider is 95% of the applied voltage from the power supply, so A0 should be 0.95*1.3=1.235 volts. If you are getting a negative voltage there (notify Kirchoff!) you probably have a bad ground somewhere, or your connections are not what you believe. And of course your micro most likely won't read a negative voltage.

Disconnect your micro from A0 and A1 and read the voltages again. If they are the same as above, check your wiring. Then check the response of A0 and A1 to voltage independently of this monitoring circuit. Maybe try a 1-1/2 volt battery on each pin. It may be that one of the inputs is defective.

Good luck!

It still isn't clear, to me, what you're trying to achieve. You have a "shunt" [R3] that is connected directly across the supply (or "rectifier" as you call it). Is this, also, intended to be the load? The R1, R2 voltage divider seems redundant, because you're already getting a read of that voltage at the A1 point.

Also, if you are connecting A1 to a uController input, then you probably don't need R4, unless there is a possibility that the voltage across R3 will exceed the max input voltage of the uController -- but, I'm talking in a vacuum because there is so much missing information.

So, which of these components are part of the intrinsic circuit, and which are for measurement purposes? It would help if you also supplied the original circuit (without measurement components). But, it looks like the original circuit will just be the supply?!? But a schematic of the supply would help. Also, if you could tell us what the supply voltage and max current are, that would help, too. Otherwise its very difficult to diagnose it, because the intention is unclear.

Yeah, the more I think about it, the only thing that makes sense is that R3 is the original load, and that R1 and R2 are unnecessary.

OR, if the R1, R2 divider is needed because the voltage across R3 can go slightly above the max input on the uController, then you can get the voltage and the current from A0 [and A1 is redundant]. Also, I'm guessing that you are reading A0 and/or A1 with an analog input, so make sure the impedance is high enough to not swamp your divider [R1 & R2] and/or series protection resistor [R3]. For example, the input resistance of an analog input on a PIC is something like 10K (if memory serves)!


So, using the divider [R1 & R2] to read the voltage, us the following formula:

Vs = (R1+R2)*Vm/R2

where
* "Vs" = the source voltage [i.e. the voltage across R3 and the voltage supplied by your "rectifier"]
* "Vm" = the measured voltage [i.e. the voltage seen by the uController]

Then the current will be:

I = Vs/R3 = Vs/0.47

That's how you get the voltage and the current from the A0 reading alone.


BUT, because you are not seeing the voltages you expect to see, I suspect that it's because that "shunt" really shouldn't be across the source (i.e. "rectifier") and if so, then I need to know what the intended load is. In other words, the shunt is shorting out the supply [aka "source" aka "rectifier"].

So, more info please

The trouble with Colin's circuit is that the current reading ['A'] will be a negative voltage and I doubt you will be able to read that with a uController. You could use an op-amp to invert the voltage, but that would be a lot of overhead [more than likely, including a separate dual power supply].

And, actually, come to think of it, Colin's circuit doesn't make sense unless the voltage divider is also the load -- otherwise what current are you measuring? Or to put it differently, why are we interested in the current flowing through the voltage divider-- it's merely for measurement purposes. And, the current through the divider can be calculated from the voltage read across the divider resistor -- because those values will be known (once they are established), so the current resistor (at point 'A') is not even needed.

Perhaps this is more like what you are looking for...

Here is a schematic that shows the whole arrangement of what it sounds like you are trying to do, David. Let us know if this hits the mark. And if not, please fill in the blanks ;)

The diodes [D1 & D2] are to protect the MCU inputs from voltages that rise higher than the maximum allowed on the MCU inputs. D2 is probably not needed, but included for completeness.

...and because the voltage across R4 should be as low as possible, so it doesn't alter what is happening at the load, there are the following considerations/concerns:

* Unless there is an internal reference in the MCU that can be set at or near the maximum voltage at R4, there won't be much resolution when converting this signal [in the MCU]. Say the maximum voltage at R4 is 0.6V and the MCU's ADC has a reference set to, say, 5V and the ADC is 8bits, then that is 5V/255 = 196mv/bit. If you divide 06V by 196mv you get 3! Which means there will be only 3 steps of resolution when reading the current. In other words, here is what the output of the ADC will look like:
* ADC: 0 Interpreted as 0V
* ADC: 1 Interpreted as 196mV
* ADC: 2 Interpreted as 392mV
* ADC: 3 Interpreted as 588mV
* ADC: 4 Interpreted as 784mV [which will never be reached, because the voltage across R4 will never go that high!

Some MCUs have an internal reference that can be used with the ADC. Typically 1.2V

With a 1.2V reference, the resolution is better: 1.2/255 = 4.71mv, which amounts to 127 steps (or 7 bits of resolution).

So, if the MCU you are using has such an internal reference, or has a way to use an external reference, then the above circuit might do the trick [though there are possible noise considerations].

But, if not, then you'll probably need to add an Op Amp as seen in the circuit, below. Use a "single supply" op amp with an "input voltage range" that includes ground (such as an LM358 or LT1013 -- but the choice of op amp will be determined by the voltage levels and the frequency involved and also by the need for precision, and if the amount of current the op amp uses is an issue).

To determine the values of R5 and R6, you need to know the maximum voltage across R4 [lets call it VR4] and the maximum voltage at the Op Amp output [call it Vout]. Then use the following formula:

VR4/R5 = Vout/(R5 + R6)

I'll leave the algrebra to you

Now, notice that the op amp, in my last circuit, has a separate supply. I did that because the output of most op amps will not go all the way to the output high "rail". In other words, if V+ on the op amp is set to Vdd, and lets say Vdd is 5V, to make this discussion more "real world", then in most cases the output of the op amp will only go as high as 5-1.5 or 3.5V. But, in order to get the full resolution from the MCU's internal ADC, that ADC would need to be run with a 3.5V reference. In most cases this would have to be supplied externally. And, to insure stability, a reference other than Vdd should be used. But, if Vdd is deemed stable enough for your situation, then to get the full resolution of the ADC, the output of the op amp needs to be able to go all the way to Vdd. And, unless you use an op amp that can swing all the way to the uppper rail [they do exist], you will need to supply a voltage that is at least 1.5V [adjust this for the particular op amp you are using] higher than Vdd. In other words, the op amps V+ line needs to be high enough for the op amps output to swing to Vdd.

And, when using an op amp with a V+ set higher than Vdd, it might be wise to include the input protection circuit shown [R7 and D2].

But, if you are using an op-amp with "rail to rail" output capability, or if you are using a reference on the ADC that is equal to the maximum voltage the op amp will output, then the following circuit is the choice [again, assuming I have the right idea of what it is you are trying to achieve].

Heh heh -- I suspect poor David's eyes are rolling back in his head from these in-depth answers, that, rightly, bring up more issues than he probably ever thought of! I wonder if we'll hear from him again.
We engineers can turn even an apparently-simple question into an elaborate discussion, of the sort that drives our spouses and friends crazy. In our world, simple questions are usually the worst kind, since they often lack boundaries or context, or contain unspoken and unspecified assumptions, all of which must be put in place before we can make an informed answer.

Your are so right!

-snip-