# Using PCscope with a voltage divider

i have been using PCscope as a 'rough and ready' oscilloscope for a few days and only just realized that i might be straining the Arduino if the voltage pops above 5V - the power supply on the circuit i am testing is supposed to be 5V but i've seen it go above that.

so, i thought i should just play it safe and rig the circuit to measure up to 9V instead.

to do that, i should make a voltage divider, right ?

could you please confirm if i have the circuit correct ?
(it's just adding the two RVD1 and RVD2 into the original circuit specified by PCscope. - was not sure if i had it correct, a bit thrown off by the extra (clamping?) diodes in there.)

the values of both resistors should be the same to make it measure (2x 5V= ) 10V, right ?

thanks !

EDIT:
changed topic heading to better reflect new question on the subject.

The schematic doesn't look right to me...

[u]This is a voltage divider[/u]. Yes, if the resistors are equal the voltage will be cut in half.

The diodes make a [u]protection circuit[/u]. Normally, you need a current-limiting resistor. But, you already have resistors forming the voltage divider and a 3rd resistor will mess up your voltage divider calculations.

DVDdoug:
The schematic doesn't look right to me...

no, it doesn't - without totally understanding the setup, the two RVDs look like they won't change anything with respect to the 'existing ratio'.

the other thing throwing me off is that i'm used to seeing voltage dividers with Vout being the result. in this case, it's an input...

DVDdoug:
The diodes make a [u]protection circuit[/u].

ahh, that is what the "clamping" is for - so it is already protecting from surges above 5V ?
but it's dependent on;

if you have a 5 V rail that can sink enough current,

i don't get this - usually a 5V rail sources current, and the GND is what sinks current ?
can the Arduino 5V rail "sink current" then ?

DVDdoug:
Normally, you need a current-limiting resistor. But, you already have resistors forming the voltage divider and a 3rd resistor will mess up your voltage divider calculations.

yes, that 10K is infact a current-limiting resistor according to the PCscope specs.

BabyGeezer:
the other thing throwing me off is that i'm used to seeing voltage dividers with Vout being the result. in this case, it's an input...

okay - so i spread the circuit out, and i think i've solved it this time.

anybody think that this is the correct way now ?

EDIT:
is the diode protection section superfluous now ?
and the current-limiting 10K now has to be included in the voltage divider as part of Z1, right ?

Make Z1 0.0 ohms ( easiest way to say remove it) and then make Z2 10K and then I think it may do what you want.
The diodes will keep the voltage from going below -0.7 ( D2 turns on if any lower then that) and to keep it from going above. +5.7V (D1 turn on if it is any higher then that.) ( the 0.7 could be 0.2 depending on what kind of diodes you are using.)

1steve:
Make Z1 0.0 ohms ( easiest way to say remove it) ...

LOL

OK, i think it's settled now.

i've decided to keep the Protection Diodes, probably best practice - i've shifted it "to the inside" which is where it's needed anyway.

okay - so am now using the PC scope with the voltage divider in effect.

have used it to analyze a simple op-amp oscillator circuit with the LM386 as per this schematic;

the 5V has since dropped to half as expected.

but comparing the frequencies and the calculated theoretical values, it also seems to be half as well.

is this also because of the voltage divider ?

1N4148's won't protect Ardiuno pins as the internal protection diodes in the chip are more sensitive and
will burn out just the same as if the 1N4148's weren't there. Use schottky diodes to bolster the internal
protection diodes as schottky's conduct much sooner.

The 10k input resistance will protect the Ardiuno pin quite well by itself for moderate voltages.

A BAT64-04 is ideal for this job.

Allan

BabyGeezer:
LOL

OK, i think it's settled now.

i've decided to keep the Protection Diodes, probably best practice - i've shifted it "to the inside" which is where it's needed anyway.

The diode protection = nice name in your case,
Remove the diodes and you are ok.

allanhurst:
A BAT64-04 is ideal for this job.

yes, i saw the datasheet - a nice pair for clean layout.
but those are SMD, and i was thinking of incorporating the components into the "probe cabling" itself, so i just have to plug it into the Arduino power (5V/GND) and A0 pins, nice clean setup with no mini-board or PCB needed.

ted:
The diode protection = nice name in your case,

do you mean it's just "in name only" - and really has no effect, as mentioned by MarkT ?

ted:
Remove the diodes and you are ok.

so, they are infact superfluous in the original circuit too ?
(or only if involving 5V levels)

Post #5
R2 is shorted by diode D2 = no more than 0.72V on A1.
If you have only 2 resistor divider max. voltage from 10V = 5V on A1, so what you want to protect ?

ted:
If you have only 2 resistor divider max. voltage from 10V = 5V on A1, so what you want to protect ?

ok, confirmed then - the diodes are infact superfluous if using the voltage divider.

ted:
Post #5
R2 is shorted by diode D2 = no more than 0.72V on A1.
If you have only 2 resistor divider max. voltage from 10V = 5V on A1, so what you want to protect ?

R2 is only shorted ( if we are going to use that term) by D2 with neg voltages on A1 and are more neg then -0.6 to -1.5 volts depending on current. I guess we could say if the voltage goes above +100 volts the diode will fail and may short, but D1 should stop that.
The voltage on A1 with the D1 and D2 should be keeping the voltage in the range of about -0.7 to +5.7 voltage. Someone said there are diode build in to the board that are better than this, in that case these diode are redundant.

If you like protecting diodes you can use 2 of them,

1. 5V zener diode A0 to ground
2. Reverse Polarity Protection

http://sound.whsites.net/appnotes/an013.htm