One Circuit to Rule Them All

[sirdowny]

Hi everyone. First off, forgive me if this question is totally that of a newbie, but here's what I'm trying to do:

I have a power supply and an electronic load loading the power supply. The pwr supply outputs say 12VDC. Now I also have a data logger (namely Picotech's ADC-11) with an input voltage range of 0 - 2.5V. SO, I need to bring the voltage entering the data logger down to within 2.5V. I'm thinking I have to use voltage dividers somehow... But here's where I'm confused, how should the circuit be laid out so that I can accurately read in voltages to the logger without really affecting the power supply's output going into the electronic load? What does this circuit look like?

Any help is GREATLY appreciated!!

-sirdowny

 

[BrianG]

Power supply + load (or is that multiple loads?) What exactly are you trying to measure?

 

[Skogsgurra]

I assume that you want to read power supply voltage when loaded by the electronic load.

It is also possible that you want to read the load current as well, but that is a bit more complex and I will not say anything about that for this time. Later, if you ask for it.

Yes, you need a voltage divider. The input impedance of the ADC-11 is >1 Mohm. The output impedance of your voltage divider should be low in comparison so that you do not introduce extra errors. I think that 2.5 kohms is a safe and convenient value. It will keep errors from the internal impedance below .25 percent.

If you connect the 2.5 kohm resistor to PSU ground (or minus, as you may call it) and add a 10 kohms resistor between other end of 2.5 and PSU output (or plus), then you have got a voltage divider that has a total resistance = 12.5 kohms. And you vill have one fifth of the PSU voltage between minus and divider output. That will be 2.5 V at 12.5 V PSU voltage. So your scale factor is 5.00. That means that you multiply the result from the ADC-11 with five to get the real value.

If you think that this divider loads your PSU, then consider this; a PSU usually delivers more than 1 A. The voltage divider consumes about 1 mA. That is 0.1 percent (or less) and doesn't influence your measurements at all. But if you are worried about that, you can correct for the extra load by adding actual voltage divided by 12.5 kohms to the total load of the PSU.

Oh, yes. Current measurement... We get back to that when/if you need it.

Gunnar Englund

http://www.gke.org

 

[sirdowny]

Thanks for the replies. I'm actually measuring a typical computer PSU (6 outputs: 3.3V, 5V, 12V1, 12V2, -12V, & 5Vsb). So this setup would be replicated 6 times. So given the circuit you described, the load would be inserted between the PSUs + terminal and the 10Kohm resistor followed by the 2.5Kohm resistor and finally the PSUs - terminal? And the data logger terminals would be measuring across the 2.5Kohm resistor?

As for current measurement, I do want to measure that as well.. Originally I'd thought of putting a shunt resistor (like a .01 ohm resistor w/ a high power rating) in series before the load and measuring the voltage drop across that. From there I'd apply Ohms Law and just divide the measured voltage by .01 ohms to get a current reading. BUT, then i'd need another 6 inputs to my data logger (and the ADC-11 has 11 inputs).

Can i do this another way without needing another 6 inputs, or should I just suck it up, lose the current measurement of the relatively insignicant -12V, and use the setup I desribed?

Thanks again for your reply! I truly appreciate it.

 

[nbucska]

Sir downy:
Gunnar slightly oversimplified it by neglecting the effect of the resistor tolerances.

If you want/need to use higher resistances, you can buffer
the output of the voltage divider with an op.amp.

You can ose an op.amp. to amplify the current sensor
voltage to the desired value.Of course an op.amp.
has its own errors you have to take into consideration.






Plesae read FAQ240-1032
WEB: <

http://geocities.com/nbucska/>

 

[Skogsgurra]

No, I always use 0.001 percent resistors ;-)

Gunnar Englund

http://www.gke.org

 

[sirdowny]

Indeed. I'll use resistors with low tolerances. So what do y'all think of using the described shunt resistor method to measure current - given the circuit I described? If you don't see any problems with it let me know, so I can move forward on this. I don't exactly trust my knowledge on all this :-\

thanks again

 

[FrankTech]

Dear Sir

Regarding the output of this type's of PSU's
----------------------------------------------
Please take notice that most switching power supplies need a minimum load on the main output. ( Which is allmost allways the +5 Volt ) If you do not load this main output for at least 10% you cannot load the other outputs !

Using very high resolution resistors is OK, but in practical measurements not needed. When you are loading for instance the +5V output with a load that draws say 20 Amps
you need a resistor value 5/20 = 0.25 Ohms
With this kind of low resistor values you will have to be very carefull with the wiring you use. Or else the whole measurement is giving misreadings. Using a high tolerance resistor say 0.1% is not relevant if the wiring that is used has a resistance of 0.2 Ohms.....

Adding a current-measuring-shunt to the wiring will complicate this further.
You will have to add the value of the shunt to the resistive value of the wiring. Again : result = mismeasurements.

Hope this is of any help ?...