Input and Output_ Resistance

8/8/2019 Input and Output_ Resistance

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Input resistance.

1. The input resistance of the following diode is 26 ohms. (It is also its incremental

resistance!!)

In a two terminal device, Rin = Rout. A diode is a two terminal device.

See the first circuit. V3 is chosen as 0.603V more than 10 so that current is exactly 1mA.

This is seen in second circuit. The ³input resistance´ Rin can be measured as shown and is 26.37

ohms, since current through R2 is 100uA.

In the last section we pretend to measure Rout. A current of 100uA is injected and voltage is

measured as 2.637mV, yielding 26.37 ohms as Rout. Recall that Rin = rout = incremental resistance.

This is 26 ohms for a diode at 1mA. At 100uA, this is 260 ohms as shown in fig 2.

Note that the injected current should be a small fraction of quiescent current. In this case the

quiescent current is 100uA. Injected current is reduced to 10uA. You could also choose it s 1uA.

2. Input resistance of a resistor R is obviously R, and so is its output resistance.

3. Input resistance of an inverting amplifier is Ri.

One method of measuring Rin is to increase Rs the source resistance till output drops to

half its value. This is shown below:

R4100k

R510k

C2

10u

10 Hz

V4-14.1/14.1V

+V

V510.603

AC V

2.637mV

DC~A

1.000mA

Q3

2N2222A

R310k

+V

V310.603

Q22N2222A

Fig 1

Q12N2222A

DC~A

1.000mA

AC V

2.637mV

+V

V210.603

10 Hz

V1-14.1/14.1V

C1

10uR2100k

R110k

C2

10u

10 Hz

V4-14.1/14.1V

+V

V510.542

AC V

2.602mV

DC~A100.0uA

Q3

2N2222A

+V

V3

10.542

Q22N2222A

Fig 2

Q12N2222A

DC~A

100.0uA

AC V

2.602mV

+V

V210.542

10 Hz

V1-14.1/14.1V

C1

10u

R41meg

R5100k

R3100k

R21meg

R1100k



In the second part of the above, note that R4 acts as source resistance, and it has been

increased till output drops to half the earlier value of 99.98mV. Thus Rin of amplifier is =

R4 = 1k.

Another method is to identify the input current and calculate Rin therefrom. This is shown

in fig 4.

Here Iin = 20uA and since applied voltage is 100mV, Rin = 100mV/20uA = 5k.

4. Input resistance of ac amplifier.

5. The current in the transistor is 1mA since 5V drops to 4.002V due to drop in R1. AC

output voltage will be lesser when Rs is not zero. If Rs is 1k, output will be

38.39mV*Rin/ (Rin+1k). Since this is 27.71mV, Rin can be calculated as follows.

38.39*Rin = 27.71 Rin + 27.71, or Rin = 27.71/(38.39 ± 27.71) = 27.71/10.68 = 2.59k

This could also be measured as Vin / Iin as below.

AC V

49.99mVAC V

99.98mV

+

U3UA741

+

V1210V

+V

V510

10 Hz

Fig 3

V4-141m/141mV

+

U2UA741

+

V910V

+V

V1010

10 Hz

V11-141m/141mV R11

1k

R51k

R41k

R81k

R91k

AC A

20.00uA

+

U1UA741

+

V610V

+V

V710

10 Hz

V8-141m/141mV

Fig 4

R65k

R710k

10 Hz

Fig 5

V3-1.41m/1.41mV

Q2NPN

+V

V45V

C2100u

AC V

27.71mV

DC~V4.002 VDC~V

4.002 V

AC V

38.39mV

10 Hz

V2-1.41m/1.41mV C1

100u

+V

V15V

Q1NPN

R51k

R41k

R3423k

R2423k

R11k



8. Active load. When the collector load is replaced by a current source we have an active

load. Note that two current sources are involved here and getting the voltage at collector

as about 1V or 2V, so that neither Q1 nor Q3 is in saturation.

The results show Rout as 63.24u/1nA = 63.24k. One might estimate the Rout as

VA1/0.98mA || VA3/0.98mA . VA1 is 139V and VA3 = 114V. This gives 141.8k||116.3k =63.9k, in close agreement with measured (simulated) values.

9. What if current is 10uA?? R4 can be calculated as follows: Vbe of Q4 is 600mV at 1mA

and hence 480mBV at 10uA. Hence R4 = (5-.48)/0.01k = 452k. {Later it is modified to

475k to get about 2V across Q6}.

Here the collector current is reduced to about 10uA. It is actually 9.641uA. Rout is seen to

be 3.662meg. The voltage gain is seen from the second part as -2112, at 1Hz. (Inverting..)

What would the gain be at 1 kHz?? [The gain drops to -1273!!!]

10. When a differential amplifier is used, it much easier to use a bipolar current source. See

the differential amplifier using 2 pnp transistors.

DC~A980.8uA

Q22N2907A

Q1

2N2907A

C1100u

10 Hz

Fig 8

V3-14.1m/14.1mV

DC~V1.098 V

AC V63.24uV

+V

V15V

Q3

2N2222A

R14.6k

R310meg

R2729k

DC~A9.641uA

Q22N2907A

Q12N2907A

DC~V1.931 V

AC V

21.12mV

+V

V15V

Q32N2222A

1 Hz

V3-14.1u/14.1uV

C1100u

Q62N2222A

+V

V45V

AC V

3.662mV

DC~V1.931 V

1 Hz

Fig 9

V2-14.1m/14.1mVC2

100u

Q5

2N2907AQ4

2N2907A

DC~A9.641uA

R1475k

R228megR6

28megR5

10megR4475k



resistance at collector of Q3 will be Rout1|| Rin2, and is 288.6/0.09825 k = 2.937meg. This

can be ascribed to an Rin2 of 5.554meg, since 6.232||5.554 = 2.937. Thus one has to

identify and differentiate between resistances ³before´ connecting a load and ³after´

connecting a load.

12. Another simpler example is shown below. In the first case, Rout = 1meg. However, when

a load of 100k is connected, the equivalent resistance is 100k||1meg = 90.9k

13. It is obvious that the Rs of first circuit is 1 meg. When a 100k is connected as shown, the

³resistance´ looking at junction of R5 and R6 is 100k|| 1meg = 90.9k. The voltage has

certainly dropped to 1/11th

of its earlier value as is to be expected. This is a method used

sometimes to match a transmission line to a source. If the cable resistance is 100 ohms

and source resistance is 1k, one can place a resistor of 111 ohms to form a potentialdivider. After this is done, the voltage drops to 1/10

thits earlier value but the source

resistance is now 100 ohms and matches that of the cable.

See fig 13.1 and b.

14. The circuit shows a transmission line as consisting of three sections of a T network. In

fig 13.1 the source has a resistance of 1k, while the characteristic resistance of the line is

100 ohms. In fig 13.b, a 111 ohms shunts the source and ensures that the µsource´

resistance is equal to the characteristics resistance of the line. The advantages obtained

in step response is as in graph below.

R6100k1kHz

Fig 12

V8-1/1V R5

1meg

R31meg

1kHz

V7-1/1V

166.kHz

V30/5V

C31n

L610u

L510u

166.kHz

V10/5V

Fig 13 a

C21n

L410u

L310u

C11n

L210u

L110u

L1210u

L1110u

C61n

L1010u

L910u

C51n

L810u

L710u

C41n

R17111

R1310meg

R7100m

R61k

R5

10meg

R4100m

R310meg

R2100m

R110

R1

R15100m

R1410meg

R12100m

R11

10meg

R101k

R9100m

Fig 13.b

R810meg



The bold line corresponds to ³matched´ conditions. In this case the terminating load

resistance is 100 ohms. See what happens when the termination is increased to 1k!!

It is clear that sever distortion can be expected when there is no matching, particularly

when load termination is not matched. In practice, there could be several loads, and one

might expect proper performance when 1 or upto 32 loads are connected, as in RS 485.

Hence one may not always ensure that the termination is always equal to characteristic

resistance of the line. It is essential that one ensures matching at source. Matching at load

is not that important as matching at source ensures that there are no reflections.. It is

preferable to have a termination, but the problem is the drive requirements from the

source. It is however preferable to have a termination of say 5*characteristics resistance

0 2.24u 4.49u 6.73u 8.98u 11.2u 13.5u-200m

0

200m

400m

600m

800m

1

Xa: 0.000 Xb: 0.000

Yc: 706.7m Yd:-200.0m

a-b: 0.000

c-d: 906.7m

fr q: 0.000

X: 0.000Offs ts Y: 0.000Offs ts

R f=Ground X=2.24u/Div Y=voltag 297%

d

c

baA

B

0 2.24u 4.49u 6.73u 8.98u 11.2u 13.5u-5

0

5

10

15

20

25

Xa: 0.000 Xb: 0.000

Yc: 25.00 Yd:-5.000

a-b: 0.000

c-d: 30.00

fr q: 0.000

X: 0.000Offs ts Y: 0.000Offs ts

R f=Ground X=2.24u/Div Y=voltag 297%

d

cba

A

B

C



of the line. It is mandatory that if you have several loads, only one should have the

desired load of R. Otherwise, with 32 parallel lines, one can end up with R/32 as the

effective load, a feature that should will attenuate the signal highly and result

additionally in improper termination.

Input and Output_ Resistance

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