Hi
Say I have a device which has
3.3V -> 100 ohm res -> connector
In an automated test, I want to check that is correct. The test jig is run from an MCU. So we have:
3.3V rail -> 100 ohm res ----> cable to test jig ----> (some circuitry to be chosen) -> MCU input pin
The most obvious is simply to test the pin reads 3.3V, but that would not check the impedance (the 100 ohm res). So I was thinking something like adding a 100 ohm resistor to ground on the test jig, and then simply checking the voltage with an ADC (which should be about 1.65V ish). It does not need to be uber precise, but the test should fail if the resistor is 10k, or a short for example.
Any thoughts?
Thank you!
Simon
Anything from in-circuit tester to a hand held meter. So many variables that only you would know.
For something simple, measure the voltage with it open circuit then with it loaded with a known value, say another 100ohm. From there, calculate the value.
Based on your description I would do same. You could also make a switchable 100R load (with a low side mosfet), then you could measure both voltage without load and with load. As a bonus it would not drain current all the time, you could just pulse it for a short moment.
In the original question it is not clear if the 3.3 V is a given or if it needs to be verified along with the 100 Ohm resistance/impedance.
Basic algebra tells us if we want to solve for two variables (the source Voltage and impedance) then we need two equations. In this case that means we must take two readings. One way to do this is, as Manul says, is to read the Voltage with the output into two different loads, 100 Ohms and infinity. That gives us the two equations and we can easily solve for both values.
It also tells us that finding both values can not be done with only one measurement. But if we know one of the two values, say the source Voltage, then that reduces the problem to just one unknown and one measurement will suffice to determine the other, the source resistance.
There is a reason why algebra is taught in school. It is also necessary to clearly state the problem.
Based on your description I would do same. You could also make a switchable 100R load (with a low side mosfet), then you could measure both voltage without load and with load. As a bonus it would not drain current all the time, you could just pulse it for a short moment.
There are two methods:
1. Measure output Voltage with open and with load, where load has known resistance
2. More precise way is to measure output Voltage with two different load where each load resistance is known. Load resistance should be near expected impedance, more difference between them leads to better precision. For example, if you expect about 50 Ω impedance, you can use load with 25 Ω and 75 Ω. But be careful because some DUT can have load resistance limitations. Use load resistance which is compatible with DUT output. If your DUT is very sensitive to the load impedance you can use a small difference for test load, for example with 50 Ω DUT it can be 45 Ω and 55 Ω...
When you have measurement U1 for R1 and U2 for R2, then you can calculate impedance:
Z = R1*R2*(U1-U2) / (R1*U2 - R2*U1)
Awesome guys, all super helpful answers thank you very much!
@EPAIII: yep, so correct, I need two measurements indeed!
I do not need anything uber precise, and the voltage source will be either 3.3V or 0V (I am mainly testing for soldering issues so I want to check shorts).
So I think I'll have a switchable 100 ohm load and I'll measure the voltage with and without it.
For an arbitrary system, the #2 system in the above post is a better idea, since some systems are unhappy with an open-circuit at the output.
The two load values near the normal load are an obvious trade-off between being close to normal (especially if the actual voltage vs. current is non-linear) and being far enough apart to get a measurable difference.
In elementary textbooks, you will see the Thevenin equivalent output resistance defined in terms of open-circuit voltage and short-circuit current, but shorting an output is often dangerous.
short to ground and measure the current. you can measure the current delta in the 3v3 rail. if your device consumes 15mA , then your short now it should consume 33mA more... (3.3v/100)