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I have a resistor 1MOhm soldered in a PCB that looks like that enter image description here And the schematic is looking like that: enter image description here When I measure the resistance by touching the a multimeter on the two ends of the resistor the measured value is around 500KOhm. When the resistor is not soldered the value is the correct one 1MOhm measured by the same multimeter in the same way. Why is that and what is the resistance value seen from the resistor "input"?

If we make the equivalent circuit of the measurement it looks like that, doens't it?:

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Instead of asking for examples how this could look like you should rather use the example you already have. That is, show us the schematic \$\endgroup\$
    – PlasmaHH
    Commented Sep 1, 2015 at 7:58
  • \$\begingroup\$ I'll do that, but it won't help. \$\endgroup\$
    – judoka_acl
    Commented Sep 1, 2015 at 8:06
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    \$\begingroup\$ Your schematic won't help because you do not include everything. You assume it is the part shown but if the circuit was only the part you show then this cannot explain measuring 500 kohms instead of 1 Mohms. Are you measuring R205 of 910 kohm ?? It looks like it's floating doesn't it ? Is it ? No it is not, there's a node called Iref, where does that go ? I do not see it in the schematic, thus the schematic is incomplete. \$\endgroup\$ Commented Sep 1, 2015 at 9:04
  • \$\begingroup\$ It won't help, not because I haven't included everything, but because the resistor is connected to a display pin. The display is not shown since it's not part of the design. Inside the display there is an integrated controller SSD1322 for which I don't have schematic. You can look into the datasheet if you wan't but this question is already finished. \$\endgroup\$
    – judoka_acl
    Commented Sep 1, 2015 at 9:16
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    \$\begingroup\$ It will always help since the schematics is all what is needed to analyse it at the level needed. But nobody forces you to do it, so nobody can exactly explain your case but only speculate about the general mechanism, I.e. "some resistances or similar things in parallel" \$\endgroup\$
    – PlasmaHH
    Commented Sep 1, 2015 at 10:02

2 Answers 2

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  • If the resistor is placed in a circuit. Then you'll measure the equivalent resistance value of the resistor parallel with the rest of the connected circuit. Just like here every component has it's own static or dynamic resistance, a diode, transistor, ic... If you now measure the resistance of a component none of them will show it's own resistance but only the circuit resistance. enter image description here

  • It can also be that you've made a bad soldering connection and short the other end also to the ground-plane.

  • Another option is maybe a residue of soldering flux around your component shorting it's terminals. You can remove this with a cotton-tip and the right product. enter image description here
  • If you can upload the whole schematic and provide a picture of your soldering connection, maybe we can help you better.
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  • \$\begingroup\$ I'm ok with the first, but I need a example to be convinced. The second can't be because I'm still measuring 500KOhm too big for a short circuit. For the third - I'm not using soldering flux for soldering resistors and capacitors, I've done this only with solder wire. \$\endgroup\$
    – judoka_acl
    Commented Sep 1, 2015 at 7:50
  • \$\begingroup\$ in solder wire is already some solder flux. It found out from my own experiences it also has a resistance and can be a short. With a short I mean an unwanted connection from high or low resistance. \$\endgroup\$
    – Bruce
    Commented Sep 1, 2015 at 8:00
  • \$\begingroup\$ @lalamer I've added a picture of an in circuit resistor. \$\endgroup\$
    – Bruce
    Commented Sep 1, 2015 at 8:06
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    \$\begingroup\$ thank you for the circuit. Now let's discuss it. We remove the battery, because I'm measuring the circuit off. If we put now the multimeter the test current will split around the other resistors and the measured value will be really the parallel resistance of the 100Ohm and 50Ohm. So far so good. I'm accepting your answer, thank you the efforts ;) \$\endgroup\$
    – judoka_acl
    Commented Sep 1, 2015 at 8:13
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The answer is simple: when a resistor is soldered on a PCB it will be connected to other components, this is the purpose of using a PCB ;-)

These other components will disturb the measurement. It depends on the circuit on the PCB how bad this disturbance is. Sometimes there is no change, you would measure the resistor's original value, sometimes you measure something completely different. This is to be expected and perfectly normal.

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  • \$\begingroup\$ Could you give a particular example with a schematic, because I tried to realize how is that possible and I don't see it happen. \$\endgroup\$
    – judoka_acl
    Commented Sep 1, 2015 at 7:40
  • \$\begingroup\$ You have a 1M resistor, you place it on a PCB, now you measure it to be 0.5 Mohms. What has happened ? Suppose there was another 1 Mohm resistor on the PCB in parallel with the first 1 Mohm resistor. 2 x 1 Mohm in parallel gives ... ? Yes, 500 kohms. Your example may be different but the principle is the same. Other components on the PCB influence the measurement ! \$\endgroup\$ Commented Sep 1, 2015 at 8:57
  • \$\begingroup\$ It was already answered by Bruce. In general you are measuring the Thevenin equivalent resistance. \$\endgroup\$
    – judoka_acl
    Commented Sep 1, 2015 at 9:00
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    \$\begingroup\$ That's the simple answer for Bruce's example. It is correct but only applicable to the circuit he showed. It might be a different effect on your PCB. What if there's a diode at the input of a chip (there almost always is!) in parallel with your resistor. You may have learned about Thevenin but be critical in applying it. There are many traps "for young players" to fall in in electronics. The trick is to be critical and check if the explanation applies to your problem. \$\endgroup\$ Commented Sep 1, 2015 at 9:10

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