Electronics- Internal resistance of a voltage...

Internal resistance of a voltage source

ElectronicsInternal resistance of a voltage source

Terry Sturtevant

Wilfrid Laurier University

June 13, 2014

Terry Sturtevant Electronics Internal resistance of a voltage source

Internal resistance of a voltage source TheoryExample

Any device which produces a voltage output has a limit to thecurrent it can provide.This includes

batteriespower suppliesfunction generatorsamplifierslogic gatesand so on

This can be represented as a resistance in series with the idealvoltage output of the device.

Any device which produces a voltage output has a limit to thecurrent it can provide.

This includesbatteriespower suppliesfunction generatorsamplifierslogic gatesand so on

batteries

power suppliesfunction generatorsamplifierslogic gatesand so on

batteriespower supplies

function generatorsamplifierslogic gatesand so on

batteriespower suppliesfunction generators

amplifierslogic gatesand so on

batteriespower suppliesfunction generatorsamplifiers

logic gatesand so on

batteriespower suppliesfunction generatorsamplifierslogic gates

and so onThis can be represented as a resistance in series with the idealvoltage output of the device.

“ideal”voltage

riinternal resistance

V0 VTterminalvoltage

The potential difference between the terminals of a battery,VT , decreases as the current supplied by the battery increases.

This is due to the potential drop which appears across thebattery’s internal resistance ri .Thus the maximum potential difference is attained only withzero output current.

The potential difference between the terminals of a battery,VT , decreases as the current supplied by the battery increases.This is due to the potential drop which appears across thebattery’s internal resistance ri .

Thus the maximum potential difference is attained only withzero output current.

The potential difference between the terminals of a battery,VT , decreases as the current supplied by the battery increases.This is due to the potential drop which appears across thebattery’s internal resistance ri .Thus the maximum potential difference is attained only withzero output current.

We can see the effect of this by watching how the measuredvoltage changes if we put a resistance across the terminals of thedevice.

Notice that what we’ve done is to create a voltage divider with riand R.

VT = V0(

VT = V0

( RR + ri

(R + riR

= 1 +riR

V0VT− 1 =

( V0VT− 1

VT = V0

( RR + ri

(R + riR

= 1 +riR

V0VT− 1 =

( V0VT− 1

VT = V0

( RR + ri

(R + riR

= 1 +riR

V0VT− 1 =

( V0VT− 1

VT = V0

( RR + ri

(R + riR

= 1 +riR

V0VT− 1 =

( V0VT− 1

VT = V0

( RR + ri

(R + riR

= 1 +riR

V0VT− 1 =

( V0VT− 1

VT = V0

( RR + ri

(R + riR

= 1 +riR

V0VT− 1 =

( V0VT− 1

So if we can measure the voltage with no load; (i.e.essentially V0), and then the voltage VT with some resistanceR, then we can determine ri .

V0VT− 1

Note that the bigger the drop with load, the bigger ri is,which makes sense.Question: If R is chosen to equal ri , what will be the value ofVT ?

So if we can measure the voltage with no load; (i.e.essentially V0), and then the voltage VT with some resistanceR, then we can determine ri .ri =

(V0VT− 1

Note that the bigger the drop with load, the bigger ri is,which makes sense.

Question: If R is chosen to equal ri , what will be the value ofVT ?

(V0VT− 1

Graphical analysis

To graph this, it makes sense to linearize the equation:

= 1 +riR

Dividing both sides by V0 gives

Graphical analysis

= 1 +riR

Graphical analysis

= 1 +riR

Graphical analysis

= 1 +riR

Graphical analysis

= 1 +riR

Graphical analysis

= 1 +riR

Graphical analysis

= 1 +riR

If we plot 1VT

versus 1R then we will get a straight line with a slope

of riV0

and a y -intercept of 1V0

.We can get ri by dividing the slope by the y -intercept.

If we plot

of riV0

If we plot 1VT

of riV0

If we plot 1VT

versus

1R then we will get a straight line with a slope

of riV0

If we plot 1VT

versus 1R

then we will get a straight line with a slopeof ri

V0and a y -intercept of 1

We can get ri by dividing the slope by the y -intercept.

If we plot 1VT

versus 1R then we will get a straight line

with a slopeof ri

V0and a y -intercept of 1

If we plot 1VT

of riV0

If we plot 1VT

of riV0

and a y -intercept of

If we plot 1VT

of riV0

If we plot 1VT

of riV0

Here’s the setup:

Here are the data collected for a new 9 volt battery.

External resistance R Terminal voltage VT( Ω ) ( V )∞ 9.27

1000000 9.27680000 9.27

......

6800 9.243300 9.222200 9.201000 9.17680 9.14330 9.08220 9.01100 8.8047 8.40

1000000 9.27680000 9.27

......

6800 9.243300 9.222200 9.201000 9.17680 9.14330 9.08220 9.01100 8.8047 8.40

1000000 9.27680000 9.27

......

6800 9.243300 9.222200 9.201000 9.17680 9.14330 9.08220 9.01100 8.8047 8.40

1000000 9.27680000 9.27

......

6800 9.243300 9.222200 9.201000 9.17680 9.14330 9.08220 9.01100 8.8047 8.40

1000000 9.27680000 9.27

......

6800 9.243300 9.222200 9.201000 9.17680 9.14330 9.08220 9.01100 8.8047 8.40

1000000 9.27680000 9.27

......

6800 9.243300 9.222200 9.201000 9.17680 9.14330 9.08220 9.01100 8.8047 8.40

1000000 9.27680000 9.27

......

6800 9.243300 9.222200 9.201000 9.17680 9.14330 9.08220 9.01100 8.8047 8.40

1000000 9.27680000 9.27

......

6800 9.243300 9.222200 9.201000 9.17680 9.14330 9.08220 9.01100 8.8047 8.40

1000000 9.27680000 9.27

......

6800 9.243300 9.222200 9.201000 9.17680 9.14330 9.08220 9.01100 8.8047 8.40

1000000 9.27680000 9.27

......

6800 9.243300 9.222200 9.201000 9.17680 9.14330 9.08220 9.01100 8.8047 8.40

1000000 9.27680000 9.27

......

6800 9.243300 9.222200 9.201000 9.17680 9.14330 9.08220 9.01100 8.8047 8.40

1000000 9.27680000 9.27

......

6800 9.243300 9.222200 9.201000 9.17680 9.14330 9.08220 9.01100 8.8047 8.40

The graph of 1VT

versus 1R looks like this:

0.0000.005

0.0100.015

0.0200.025

0.0000.005

0.0100.015

0.0200.025

0.0000.005

0.0100.015

0.0200.025

0.0000.005

0.0100.015

0.0200.025

0.0000.005

0.0100.015

0.0200.025

0.0000.005

0.0100.015

0.0200.025

0.0000.005

0.0100.015

0.0200.025

0.0000.005

0.0100.015

0.0200.025

0.0000.005

0.0100.015

0.0200.025

0.0000.005

0.0100.015

0.0200.025

0.0000.005

0.0100.015

0.0200.025

0.0000.005

0.0100.015

0.0200.025

0.0000.005

0.0100.015

0.0200.025

0.0000.005

0.0100.015

0.0200.025

0.0000.005

0.0100.015

0.0200.025

A least squares fit to the data gives:

V0 = 9.246± 0.006V

Imax = 1.89± 0.05A

ri = 4.8± 0.1Ω

V0 = 9.246± 0.006V

Imax = 1.89± 0.05A

ri = 4.8± 0.1Ω

V0 = 9.246± 0.006V

Imax = 1.89± 0.05A

ri = 4.8± 0.1Ω

V0 = 9.246± 0.006V

Imax = 1.89± 0.05A

ri = 4.8± 0.1Ω

Here are the data collected for a used 9 volt battery.

1000000 7.32680000 7.32

......

6800 7.113300 7.012200 6.931000 6.71680 6.56330 6.10220 5.72100 4.7147 3.55

1000000 7.32680000 7.32

......

6800 7.113300 7.012200 6.931000 6.71680 6.56330 6.10220 5.72100 4.7147 3.55

1000000 7.32680000 7.32

......

6800 7.113300 7.012200 6.931000 6.71680 6.56330 6.10220 5.72100 4.7147 3.55

1000000 7.32680000 7.32

......

6800 7.113300 7.012200 6.931000 6.71680 6.56330 6.10220 5.72100 4.7147 3.55

1000000 7.32680000 7.32

......

6800 7.113300 7.012200 6.931000 6.71680 6.56330 6.10220 5.72100 4.7147 3.55

1000000 7.32680000 7.32

......

6800 7.113300 7.012200 6.931000 6.71680 6.56330 6.10220 5.72100 4.7147 3.55

1000000 7.32680000 7.32

......

6800 7.113300 7.012200 6.931000 6.71680 6.56330 6.10220 5.72100 4.7147 3.55

1000000 7.32680000 7.32

......

6800 7.113300 7.012200 6.931000 6.71680 6.56330 6.10220 5.72100 4.7147 3.55

1000000 7.32680000 7.32

......

6800 7.113300 7.012200 6.931000 6.71680 6.56330 6.10220 5.72100 4.7147 3.55

1000000 7.32680000 7.32

......

6800 7.113300 7.012200 6.931000 6.71680 6.56330 6.10220 5.72100 4.7147 3.55

1000000 7.32680000 7.32

......

6800 7.113300 7.012200 6.931000 6.71680 6.56330 6.10220 5.72100 4.7147 3.55

1000000 7.32680000 7.32

......

6800 7.113300 7.012200 6.931000 6.71680 6.56330 6.10220 5.72100 4.7147 3.55

The graph of 1VT

versus 1R looks like this:

0.0000.005

0.0100.015

0.0200.025

0.0000.005

0.0100.015

0.0200.025

0.0000.005

0.0100.015

0.0200.025

0.0000.005

0.0100.015

0.0200.025

0.0000.005

0.0100.015

0.0200.025

0.0000.005

0.0100.015

0.0200.025

0.0000.005

0.0100.015

0.0200.025

0.0000.005

0.0100.015

0.0200.025

0.0000.005

0.0100.015

0.0200.025

0.0000.005

0.0100.015

0.0200.025

0.0000.005

0.0100.015

0.0200.025

0.0000.005

0.0100.015

0.0200.025

0.0000.005

0.0100.015

0.0200.025

0.0000.005

0.0100.015

0.0200.025

0.0000.005

0.0100.015

0.0200.025

The steeper slope of this graph compared to the last one indicatesthat the internal resistance is much higher; in other words, theoutput voltage changes much more as current increases comparedto the new battery.

A least squares fit was produced, as before, but only the last fivepoints were used due to the curve at the left of the graph.

V0 = 6.78± 0.08V

Imax = 0.33± 0.02A

ri = 20± 1Ω

(Using all of the data points gives slightly different results, butones that are still much different than those for the new battery.)

V0 = 6.78± 0.08V

Imax = 0.33± 0.02A

ri = 20± 1Ω

V0 = 6.78± 0.08V

Imax = 0.33± 0.02A

ri = 20± 1Ω

V0 = 6.78± 0.08V

Imax = 0.33± 0.02A

ri = 20± 1Ω

V0 = 6.78± 0.08V

Imax = 0.33± 0.02A

ri = 20± 1Ω

Parameter New battery Old batteryV0(V ) 9.246± 0.006 6.78± 0.08Imax (A) 1.89± 0.05 0.33± 0.02ri (Ω) 4.8± 0.1 20± 1

Even though the nominal battery voltage is only 27% lower,the internal resistance is 5 times as high, and the maximumcurrent available is only 1/6 of what it is for the new battery.

Parameter New battery Old batteryV0(V ) 9.246± 0.006 6.78± 0.08Imax (A) 1.89± 0.05 0.33± 0.02ri (Ω) 4.8± 0.1 20± 1

Even though the nominal battery voltage is only 27% lower,the internal resistance is 5 times as high, and the maximumcurrent available is only 1/6 of what it is for the new battery.

Electronics- Internal resistance of a voltage...

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