Difference between revisions of "Input and output impedance"

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(Battery example)
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=Principle for loading circuits=
 
=Principle for loading circuits=
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After observing the previous example, you can see that loading a source where the input impedance of the load is similar in magnitude to the output impedance of the source causes the output voltage to drop. How can you load a source such that you maintain the output voltage?
  
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If '''<math>Z_{in} >> Z_{out}</math> (input impedance of load >> output impedance of source),''' then the source can output close enough to it's open circuit voltage that we can ignore the changes induced by the load. This is a good principle to follow when connecting different components together.
  
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=Practice examples=
  
 
=OpAmp example?=
 
=OpAmp example?=

Revision as of 20:43, 18 August 2016

20.309: Biological Instrumentation and Measurement

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Overview

What happens when we connect one circuit component to another? Sometimes the circuit component doesn't behave in the same way when it is by itself versus when it is connected to another component. To understand how the circuit will behave, we must consider the input and output impedances of the different parts. The output impedance refers to the impedance, or opposition to current flow, of the component bearing the electrical source. Meanwhile, the input impedance refers to the load component's opposition to current flowing in from the electrical source. In general, you will want to have a high input impedance relative to output impedance, and you will see why in the following sections.

The source component has an output impedance, and the load component has an input impedance. What happens when we connect Thing 1 with Thing 2?

Battery example

Let's look at an example of connecting a battery to a resistor. When we see a 9V battery, we often think it should output 9V from its terminals. It truly does, but only when it's not sourcing any current. It turns out that there is some internal impedance inside the battery that prevents the output voltage from remaining at 9V once current starts to flow. We can thus model the battery as a pure voltage source in series with a resistor, where the resistance value $ R_{out} $ is equal to the output impedance of the battery. Here is an example of a 9V battery's I-V curve (current-voltage relationship).

Model of a 9V battery, showing the internal impedance as a resistor with value $ R_{out} $. $ I_{SC} $ refers to short-circuit current (if a wire was placed between the output terminals), and $ V_{OC} $ refers to the open-circuit voltage (the voltage measured between the output terminals when no current is flowing)

Now let's connect load to this source. For simplicity, we will connect a resistor with resistance $ R_{in} $. Hence, the input impedance of this load is equal to $ R_{in} $. When we plot the I-V characteristics of the load with the source, the intersection of the two lines is the operating point.

The load has an output impedance equal to $ R_{out} $. When the load is connected to the source, the operating point is given by the intersection of the two lines on the I-V plot. The operating point gives the voltage applied across the load, as well as the amount of current flowing through it.

Here we can see that once the load is connected to the source, the output voltage is no longer 9V; rather it is given by the operating point. The example drawn above shows the 9V battery having an output resistance of 1.5Ω and the load having an input resistance of 3Ω. If you use voltage divider relations to calculate the voltage being applied across the load, and the current flowing through it, you will find it to be 6V and 2Ω, respectively, as given by the operating point graphically. This shows us that even though the 9V battery outputs 9V when there's no current flowing, once we apply the load, the output voltage drops down to 6V to account for the increased current flow.

What would have happened if the load resistance were very high?

Principle for loading circuits

After observing the previous example, you can see that loading a source where the input impedance of the load is similar in magnitude to the output impedance of the source causes the output voltage to drop. How can you load a source such that you maintain the output voltage?

If $ Z_{in} >> Z_{out} $ (input impedance of load >> output impedance of source), then the source can output close enough to it's open circuit voltage that we can ignore the changes induced by the load. This is a good principle to follow when connecting different components together.


Practice examples

OpAmp example?