PIC as unwanted resistive load

Bryan · Joined: 23 Apr 2005 Posts: 73

I have created a circuit using a PIC with another separate DC circuit powered by the same 5V power supply. I am running into a problem because the PIC is acting as a resistive load in parallel with my DC circuit that is separate from the PIC. If you cannot picture this think of one voltage supply hooked to 2 different resistors, one for my separate DC circuit and the other as the PIC with both grounds common, and it is easy to see how this is happening. I verified this using my multimeter. My question is, does anyone know how to separate these 2 circuits without using two power supplies? I know it must be possible, and I cannot proceed with my project until I learn how because my particular application is resistive sensitive.

kender · Joined: 09 Aug 2004 Posts: 768 Location: Silicon Valley

It's a little difficult to envision your problem. Does the +5V supply droop when you connect all the load? If it does, then your power supply doesn't supply enough current.

In your particular case, why is the PIC in parallel with the other load bad? Is it a power consumption issue? Is it a noise issue?

Nick

Ttelmah · Guest

I suspect the other circuit, has a varying resistance, that he wants to use as a 'measure'. Hence the parallel consumption of the PIC, represents an error 'term'.
If this is the case, then there is nothing whatsover that can be done to get rid of the 'problem'. The PIC draws power, and without a seperate supply, has to source it from the rail.
However there may be ways of making the term 'acceptable'. The current drawn by a PIC varies with frequency, so lowering the speed of the PIC, might reduce the error to an acceptable level.
The other thought, is to make the consumption of the PIC circuit constant. You can run most PICs happily off less than 5v. If a resistor is added from the supply to the PIC circuit, and then an op-amp is added to measure the voltage drop across this resistor, and adjust the base drive on a common-emitter transistor, to keep this drop constant, then the current drawn by the PIC, and this transistor, would remain a constant value, allowing any change caused by the other circuit, to still be detected and measured. Effectively you turn the seperate cicuit into a constant current system.

Best Wishes

sseidman · Joined: 14 Mar 2005 Posts: 159

Bryan · Joined: 23 Apr 2005 Posts: 73

Thank you for the insight guys. My problem is not one of sourcing enough current or voltage, the problem is that I need the resistance of the secondary circuit (one not involving to the PIC) to be unchanged by outside sources (aka the PIC circuit). Since both are connected to the positive rail this creates a parallel connection and the resistances I am trying to measure in the secondary circuit drop drastically (basic equivalent resistance principles). All I want is a separate voltage divider circuit that consists of 2 resistors in series so I can check the voltage drops across each of them powered from a 5V source (which I am currently sharing with the PIC which should be completely separate). If you think about this, the PIC acts as a resistor now that is in parallel with my voltage divider circuit and this throws off all my calculations for my particular application since I need those series resistances to be unchanged by another "resistor" in parallel (the PIC). I just cannot figure out how to deliver 5V to each circuit (PIC and voltage divider) without them interacting since they have a common negative rail. As long as the PIC's resistance stays constant durign operation I could determine this value and factor it in, but I would rather find a better way to resolve the issue. Does this clarify?

treitmey · Joined: 23 Jan 2004 Posts: 1094 Location: Appleton,WI USA

That is what Ttelmah was getting at(see above). As long as the resistor(PIC) is know, you can figure it into the voltage divider. To keep it know, (in a steady state), he is powering the PIC with a constant current source.

Right Ttelmah?

sseidman · Joined: 14 Mar 2005 Posts: 159

sseidman · Joined: 14 Mar 2005 Posts: 159

asmallri · Joined: 12 Aug 2004 Posts: 1634 Location: Perth, Australia

Hi Bryan,

I am obviously not understanding the true topology of your circuit. An ideal +5V supply is a very low impendance source. Ideally the output impedance of the souce = 0R.

Your two ciruits, as I understand your description of them, are in parallel with this 0R impedance. Your non PIC circuit should see the power rail inputs as a 0R source impedance. From the measurement circuits perspective, the PIC appears as a xxxOhm load in parallel with the 0R output impedance of the power supply. The resulting parallel impedance is 0R (unchanged). Therefore the "load" the messurement circuit sees is also unchanged. If this is not the case then the low impedance source for your 5 volt supply is not low impedance and this is where you need have a look at what is going on.
_________________
Regards, Andrew

http://www.brushelectronics.com/software
Home of Ethernet, SD card and Encrypted Serial Bootloaders for PICs!!

Bryan · Joined: 23 Apr 2005 Posts: 73

I just connected the circuit up again as described and everything seemed normal - I measured approximtely 10k across the 10k resistor (within tolerance). Then I turned on the power supply without changing anything else and as soon as power came on the resistance JUMPED up drastically to somewhere close to 100k. Then once power was turned off there was a 2-3 second delay before the resistance returned to its nominal value once again (a cap discharging?). The power down time seems consistent with how long it takes the PIC to power off I/O pins when an LED/LCD are connected to them so I can see when they power down. This makes me wonder what role it is playing in all this. Anyone have any clue how turning on a power supply could make resistance shoot through the roof like that?

asmallri · Joined: 12 Aug 2004 Posts: 1634 Location: Perth, Australia

You cannot measure resistance this way. Conceptually, a multimeter measures resistance by passing a (constant) current through the device under test (DUT) and measuring the voltage across the device. Knowing the current and resultant votage the instrument can calculate resistance. Alternatively the multimeter can put a constant voltage across the DUT, measure the current and then derive the resistance from these two values.

What you are doing is injecting a current from the power supply and the results you are recording are completely meaningless. If you want to measure the effective resistance in this scenario you need two multimeters, one measuring current through the DUT and the other measuring voltage across the DUT. Then you an calculate the resistance.

To summarize, the values you are reading are rubbish.
_________________
Regards, Andrew

http://www.brushelectronics.com/software
Home of Ethernet, SD card and Encrypted Serial Bootloaders for PICs!!

treitmey · Joined: 23 Jan 2004 Posts: 1094 Location: Appleton,WI USA

I think we are all "on the same page". You were just reading resistance while the circuit was powered.
I think your set now. Do you have more questions?

btw. you can indirectly read the resistance(while powered up) by reading the voltage drop and current through the resistor and using ohms law. R=V/I

Guest · Posted: Thu Jun 23, 2005 10:20 am

Thanks for the help everyone. Looks like everything is cleared up now. I didn't realize that you couldn't directly test the resistance of a resistor with power on due to how the multimeter is designed, but that is a good tidbit to know. Thanks again!

sseidman · Joined: 14 Mar 2005 Posts: 159