[sdiy] breadboard woes

[Gordonjcp]

On Tue, Apr 04, 2017 at 01:43:09PM -0500, Elain Klopke wrote:
> So I was trying to get a reading on my breadboard to see how much voltage
> was lost between the input of three passive low pass filters in series and
> the output.
> 
> My measurements between ground and:
> positive supply: 5.06 V
> Output from chip: 2.4 V (was expecting 5 V because digital square wave)
> junction of first and second resistor: 2.4 V
> junction between second and third resistor: 2.4 V
> after third resistor: 2.4 V ....
> 
> All resistors measured at 5.07 kOhms. I'm not sure what's going on. Is my
> breadboard messed up somehow or am I missing something fundamental about
> electronics?

Two things - the square wave will spend half its time at 5V and half its time at 0V.  What does that average out to?  Well, 2.5V, which is just about what you're measuring.

How do you work out the voltage drop across a resistor?  Well, it's V = IR.  What's I in your case?  That rather depends on the load on the filter, doesn't it?  If the load is just your DMM then it's effectively infinitely high, so the current is basically nothing.  What's nothing times 5kOhms?  Nothing.  So you won't see any voltage drop across the resistors.

-- 
Gordonjcp MM0YEQ