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I have faced a few instances in which I need to measure resistor or capacitor value which is soldered to the PCB board (SMD component). These are very small and difficult to look the value written on it. When I measure using a multimeter I am not getting the correct value (maybe due to other components connected to it).

Is there a way by which I can measure the component's value without desoldering it?

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    \$\begingroup\$ Generally not easy to achieve. A magnifying glass will help with components that are marked but SMD caps tend not to be marked. \$\endgroup\$ – Andy aka Oct 25 '15 at 10:27
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No, this is generally not possible because of the rest of the circuit. If you have a schematic you may be able to analyze the circuit and figure out what the value is from the 'incorrect' reading, but that is not always possible, even under ideal conditions. By ideal, I mean that you have a perfect understanding of the circuit, how you measuring instruments work (for example the waveforms used to measure capacitance including the voltages on each range) and how parts such as the chips Nick mentions behave when not powered and subjected to those waveforms). If a much smaller and/or less accurate part is effectively in parallel it may be impossible to get an accurate number.

The most reliable way is to remove the parts with a tweezer type desoldering tool and measure it, then replace. Some larger (0603 and up) resistors are still marked but there are many smaller resistors and most SMT capacitors that are not marked at all (photo from here).

enter image description here

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  • \$\begingroup\$ in-circuit resistance measurement is a top topic in electronic. You should not say it's not possible! \$\endgroup\$ – HOPE Oct 25 '15 at 10:41
  • \$\begingroup\$ "Not always possible" is what I said, and it is correct. If th OP is reverse engineering and does not have access to a complete schematic it's particularly fraught with difficulty. \$\endgroup\$ – Spehro Pefhany Oct 25 '15 at 10:42
  • \$\begingroup\$ Difficulty, of course. But not impossible. I mentioned a way if you consider it. \$\endgroup\$ – HOPE Oct 25 '15 at 10:56
  • \$\begingroup\$ @HOPE If you have access to a perfect schematic and perfect understanding of the behavior of all parts behavior under weird conditions, access to all all nodes, can put arbitrary voltages and waveforms on each node and use information from previous measurements it should be possible (to some, perhaps acceptable, degree of accuracy) to measure many values, but that is not a simple thing. For example measuring the load capacitors on a crystal with the chip inout/outputs in parallel, or measuring an individual bypass capacitor with 100 others in parallel would likely not be feasible. \$\endgroup\$ – Spehro Pefhany Oct 25 '15 at 11:08
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    \$\begingroup\$ @HOPE - "we oscope on DC performance on a circuit and assume its in off-state with no power attached to the circuit." There's an old saying which applies to this statement. "When you assume, you make an ass out of u and me." In other words, while you can occasionally make that assumption, in a very large proportion of the time you will get into trouble if you do so. \$\endgroup\$ – WhatRoughBeast Oct 25 '15 at 16:27
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In general, no. Consider, for example, a component that's shorted out by PCB traces. Obviously it's going to be impossible to measure it while on the PCB. Conversely, a component that's not connected to anything isn't going to pose any problems at all.

If you're trying to make an in-circuit measurement like this, you have to look at the schematic and consider what other paths could be formed between your probes besides the one you want. Consider not just the marked parts, but also ESD protection diodes in any ICs.

If the only current paths other than the one you want go via semiconductors, you may be able to measure your component accurately if your meter uses a test voltage lower than one diode drop. Otherwise, you're stuck with unsoldering the component in order to measure it.

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What are you asking called "In-circuit resistor measurement" and is discussed in different ways. Regard copy right low, I attach the exact file of a known method in this field.

'in-circuit' resistance measurement

The original link is :

ioScience- "in-Circuit" resistance measurement

Consider you have a equivalent circuit as below :

enter image description here

that R1e and R2e are input and output impedance respectly. Now assume a circuit as below :

enter image description here

That node 1 and node 2 are your probe signal. In this Signal VD is equal to :

enter image description here

Which delta-V is op-amp DC offset voltage and K is inverter circuit gain. When we adjust circuit so that VD=0 then we have :

enter image description here

It will be satisfied by using an variable resistor for R0 or Rs.

Suggested values for R0, Rs and op-amp model is described in original document that is attached.

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  • \$\begingroup\$ Please don't put your answer in a link to a filehosting service. They're unreliable, so your answer is likely to become useless in short order. \$\endgroup\$ – Nick Johnson Oct 25 '15 at 10:34
  • \$\begingroup\$ Ok Got it,I'll fix it now... \$\endgroup\$ – HOPE Oct 25 '15 at 10:35
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    \$\begingroup\$ You might consider linking to the original source instead. \$\endgroup\$ – Nick Johnson Oct 25 '15 at 10:37
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This is a difficult problem. Components tend to be connected on a board, and these connections corrupt the readings.

Many manufacturers, starting in the 1960s when automatic test equipment was just starting to be computer controlled, like Teradyne and Marconi Instruments to mention a couple, came up with a way of making some of these measurements.

In the best case, these connections can be 'guarded out'. This is essentially an extension of a 4 wire measurement.

Let's say you have a pi attenuator, and you want to measure the series resistor R2 (from that link). You would apply a known voltage to one port, so the current down R1 is irrelevant, and measure the current through R2 using a virtual ground amplifier across R3. To do this, you also need a connection to the ground, this is called 'guarding out' the ground connetion. As there is no voltage on R3, no current flows in it, so all the flowing current comes through R2. This allows calculation of R2.

Unfortunately, some components are just not feasible to guard out. The guard node might not be accessible. You might not have enough dynamic range to control the guarded components.

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