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I realized there is something fundamental about equivalent resistance I don't understand in terms of application. I understand the part about mechanically simplifying resistors into one resistor, just not how it applies in electronics.

This is a MEMS accelerometer (Datasheet, MS3028) with equivalent circuit.

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My understanding is that the equivalent resistance is the resistance between two terminals, and in a real-life application, it is the resistance between the input and the output of a component.

My understanding is that, in order to get the equivalent resistance, we assume terminal 2 and terminal 1 are connected as they are tied to the same -ve supply. Then we calculate the resistance between terminal 3 and terminal 2/1.

The part I don't understand is why we picked terminals 2/1 as the "end point"? Why didn't we pick terminals 4 and 5? In most bridge circuit explanations, there is a connection between 4 and 5 which makes it not a viable "output", but in this case, they are meant to output electrical signals to an external component. Why does that not make them viable "end points" for the equivalent circuit?

Please let me know if more information is required.

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2 Answers 2

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The part I don't understand is why did we pick terminal 2/1 as the "end point"? Why have we not picked terminal 4 and 5?

If I was asked to calculate the equivalent resistance, I would ask the questioner between which two points. I would ask this because we need to know this in order to make a proper answer.

But, if this information wasn't provided and, I had to answer (on pain of torture or death), I would give several answers and, all would be right but, only the person originally asking the question would know which one specifically applied to their question.

Of course my "several answers" would be wrapped within a scenario. For instance, I would definitely consider a scenario from the point of view of the bridge excitation supply (pins 1, 2 and 3) and, I would make that clear. I would also consider the similar scenario from the point of view of the measurement circuit connected to pins 4 and 5.

Short-story: if it doesn't specifically state which terminals, then answer with both perspectives.

Why does that not make them viable "end points" for the equivalent circuit?

They are viable.

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  • \$\begingroup\$ Thank you very much. I thought I was losing my mind. \$\endgroup\$ Commented Dec 14, 2023 at 14:44
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As the other answer says, for a multi-terminal circuit, there can be several important equivalent resistances, and only the person asking the question knows which one is important to them.

  • If we want to know how much the circuit loads its power supply, we might want to know its equivalent as a load resistance.

  • If we want to know how much a two-port network will load the circuit that drives its input, we might want to know the equivalent input resistance. Mathematically this is \$\frac{dV_{in}}{dI_{in}}\$. This doesn't apply to your bridge circuit because the bridge circuit doesn't have an electrical input port.

  • If we want to know how stiff the load needs to be on the output of the circuit, we might want to know the equivalent output resistance. Mathematically this is \$\frac{dV_{out}}{dI_{out}}\$.

The equivalent input resistance doesn't apply to your bridge circuit because the bridge circuit doesn't have an electrical input port. The equivalent output resistance is a very important parameter for bridge circuits and is likely what you are being asked for, because we need to know this to know how much error we introduce in the measurement due to the load on the output.

But it's also possible that the equivalent load on the power supply is needed in this particular instance. Only by knowing how the answer is going to be used can you know which answer is actually wanted.

If you do want the output resistance, you can calculate it by tying terminals 3 and 2 together, then calculating or measuring the resistance from terminal 5 to terminal 3/2. Then multiply by 2 to account for the equal resistance on the other output.

Or, to give a measurement more directly tied to the definition of the output resistance, tie 3, 2, and 1 together and place a current source between terminals 4 and 5, then measure/calculate the voltage between terminals 4 and 5.

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