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### Topic: Unexpected voltage drop (Read 788 times)previous topic - next topic

#### smhall05

##### Dec 14, 2011, 10:16 pm
I have a 100k resistor connected to pin 2 and driving it HIGH. When I connect my meter (one end to ground the other to the end of the resistor) I see 4.64V. The voltage out of pin 2 is measured as 5.09V, a difference of 0.45 Volts.

Can anyone explain this? Why am I not seeing something closer to the input voltage?

#### pwillard

#1
##### Dec 14, 2011, 11:04 pm
Ohms Law can explain it.

#### smhall05

#2
##### Dec 14, 2011, 11:06 pm
Would you mind showing the steps to obtain that value. Ohms law tells me the voltage should be the same at that point.

#### smhall05

#3
##### Dec 14, 2011, 11:20 pm
It looks to me that my meter looks like a 1M resistor. Does that make sense?

I have pasted a  LTspiceIV netlist here that shows a 4.64V reading at the R1 and R2 connection labeled Drop.

"ExpressPCB Netlist"
1
0
0
""
""
""
"Part IDs Table"
"V1" "5.09" ""
"R1" "100k" ""
"R2" "1031111" ""

"Net Names Table"
"N001" 1
"0" 3
"Drop" 5

"Net Connections Table"
1 1 1 2
1 2 2 0
2 1 2 4
2 3 2 0
3 2 1 6
3 3 1 0

#4
##### Dec 15, 2011, 03:58 am
If your voltmeter has an internal resistance of 1M it is time to get a new one and throw this one away.

Did you set the pin as an output?
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#### James C4S

#5
##### Dec 15, 2011, 04:37 amLast Edit: Dec 16, 2011, 05:56 am by James C4S Reason: 1

Would you mind showing the steps to obtain that value. Ohms law tells me the voltage should be the same at that point.

You have created a voltage divider of 100k and 1M.  I calculate you should get 4.627V.  Looks like it is working exactly as expected.

Vout = (R2 / (R1 + R2)) * Vin = (1M / 1.1M) * 5.09 = 4.627

Edit: Fixed math error.
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#### smhall05

#6
##### Dec 15, 2011, 06:54 am
Problem solved. madworm put in words what my simulation showed. The voltmeter I am using, Radioshack 22-810 has an input impedance of 1 M OHM (DC), this explains why I am falsely seeing this drop. If the input impedance was say greater than 10Meg, which is the case with better voltmeters I would have seen what I expected, 5.09V.

Thank you everyone.

#### James C4S

#7
##### Dec 16, 2011, 05:54 am

input impedance of 1 M OHM (DC), this explains why I am falsely seeing this drop.

The drop you are seeing isn't false. It is very real and correct.

If the input impedance was say greater than 10Meg, which is the case with better voltmeters I would have seen what I expected, 5.09V.

No, you would have seen 5.03V, not 5.09.
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#### MarkT

#8
##### Dec 17, 2011, 06:14 pm
1M ohm is a pretty poor spec for a multimeter - mine is 10M on higher voltage ranges and >100M on low voltage ranges.

Just be aware it's reading wrong on high impedance circuits...
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