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# what is input impedance?

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#### PG1995

##### Full Member level 5
Hi :smile:

Please have a see Example 1-5 on the given link:
https://img854.imageshack.us/img854/8376/inputimpedance.jpg

The example mentions input impedance. What is it in simple words? The book says: Because this input impedance is across the measured terminals , a small current flows flows through the multimeter....

What is book saying? Please help me with it. Thank you very much for this.

If you know Ohms Law (V = I*R), then you know part of what makes up impedance.

The book is asking, what impedance (resistance) does the circuit look like, if you combine all of the elements together. That is the input impedance... the equivalent load of everything combined together. Example, you have a load circuit that consists of two resistors in parallel. R1=1000 ohms, R2=3000 ohms. The input impedance of this circuit is the effective resistance of both resistors connected in parallel. So, what would a multimeter read if you measured the resistance of that entire load circuit together? Answer, 1k || 3k = 1 / (1/1000 + 1/3000) = 750 ohms. So, the input impedance of the load would be 750 ohms.

Resistance (R) is a real number with no imaginary component. An impedance is a complex number, which can be written as a real number plus an imaginary number, such as 5+j*10 or 23-j*2. Impedance can be written as R+jX, where R is the resistance, and X is the reactance, both in Ohms. In the Ohms law equation written above (V=I*R), this only takes into about the real part of the impedance, R. To be complete, the equation would be written as V=I*Z, where Z is the component's impedence, and Z = R+jX. In the case of a purely resistive device, like a 50 ohm resistor.... Z = 50+j*0 ohms. In that case, we just drop the imaginary part, since it's zero and write Z = 50 ohms, or R = 50 ohms (no reactive/imaginary component).

Points: 2

### naavid

Points: 2
A multimeter have a battery source and a circuit built in it. Battery is used to power the display and virtually no power from the circuit under test. Multimeter's DC voltage ranges have a very high resistance between 1 M Ohm ~ 10 M Ohm. This Resistor is known as INPUT IMPEDANCE of the multimeter and has negligible affect on circuit under test apart from delivering a small current to activate the measuring probes.

Senthilkumar_rjpm

### Senthilkumar_rjpm

Points: 2
A multimeter have a battery source and a circuit built in it. Battery is used to power the display and virtually no power from the circuit under test. Multimeter's DC voltage ranges have a very high resistance between 1 M Ohm ~ 10 M Ohm. This Resistor is known as INPUT IMPEDANCE of the multimeter and has negligible affect on circuit under test apart from delivering a small current to activate the measuring probes.

This is true, unless the circuit you are measuring is high-impedance. Then, the impedance of the multimeter circuitry can adversely affect your measurements. This can be seen when trying to probe some op-amp circuits.... their input impedance runs ~100k to 1M ohm, which is within 1 order of magnitude of the input impedance of a standard handheld multimeter, and can cause significant error in your measured values.

I know, that the original poster is reading the exercise problems very critical. To assist you with your critics, related to the present DC network calculation problem, it would be more appropriate to talk about input resistance (the real part of impedance, as enjunear explained). This would avoid to introduce a new term, that's neither needed nor explained in the problem.

Strictly spoken, no instrument has an exact input impedance of 1 or 10 MOhm. It's always a resistance with a parallel capacitance. But the latter does't matter for DC problems. Talking about input resistance makes the statement better understandable and fully correct.

Points: 2

### Senthilkumar_rjpm

Points: 2
Thank you all of you, enunear, mishrashashi, FvM.

I think I understand it now. Say, voltmeter measures how much voltage is appearing across its terminals. In the Example 1-5 open circuit voltage is 9 volts and a Rth of 1.5 K. Measuring with a multimeter which has impedance of 200 K ohm will give a reading of 9 * ( 200 K / 201.5 K) or 8.933 volts (I've used voltage divider rule). Error is 0.067.

In the scanned example in my first post the author used the phrase "measured terminals", what does the author mean by it?

The author also says that moving coil has typical sensitivity of 20k ohm per volt. What does the part "20k ohm per volt" mean? That would mean for two volts it would be 40k ohm.

The author also suggests to use a field-effect transistor. What does this mean? If it's complicated, then you can ignore it because perhaps I won't be able to understand it.

Please help me with the above stuff. Thanks a lot. Things would have been really difficult without your help.

In the scanned example in my first post the author used the phrase "measured terminals", what does the author mean by it?

The author also says that moving coil has typical sensitivity of 20k ohm per volt. What does the part "20k ohm per volt" mean? That would mean for two volts it would be 40k ohm.

The author also suggests to use a field-effect transistor. What does this mean? .

As I think measured terminals in the scanned data refers to the terminals A & B. You can easily identify the Thevenin equivalent circuit and load here.

For moving coil sensitivity you can refer the link...
Voltmeter - Wikipedia, the free encyclopedia

Field-effect transistor(FET) is used in circuit to control the electron flow and it increases the resistance value. It is just a basic idea. For FET and meter basics you can refer some standard books.

In the scanned example in my first post the author used the phrase "measured terminals", what does the author mean by it?

They are simply the two terminals at which we try to find out what could exist behind them as an 'equivalent' circuit. For example, measuring the voltage at these terminal can give us a good idea about Vth (terminals a and b said open). If we like to also measure Rth, it can be done indirectly. We add a known load at the “measured terminals" and measure the new voltage (in such circuits, it will be less than Vth). The equivalent loop is formed by Vth, Rth and RL. We know Vth and I = Vab/RL so Rth=(Vth/I) - RL. You see, even if there are 100 loops behind a and b terminals, after just two measurements we were able to replace them by a simple equivalent circuit; in our case Vth and Rth only.

The author also says that moving coil has typical sensitivity of 20k ohm per volt. What does the part "20k ohm per volt" mean? That would mean for two volts it would be 40k ohm.

When you read a sensitivity as "20k ohm per volt" it means that the input resistance of the voltmeter is equal to ( 20K x Voltage full scale ). So if you set the voltmeter to measure from 0 - 250V then the full scale is 250V and its input impedance is 20K * 250 = 5 Mega Ohms (5000K) for this setting. If the voltmeter has also a range from 0-2V, the full scale is therefore 2V and as you said its input resistance would be 40K.

The author also suggests to use a field-effect transistor. What does this mean? If it's complicated, then you can ignore it because perhaps I won't be able to understand it.

When we like to measure a voltage we usually allow a very small current to flow into the meter (say Iin) and let it move a very sensitive needle (magnetic moving coil for example) or we amplify it first then display, in a way or another, the voltage that corresponds to it, using the formula V = Iin * Rin since Rin of the voltmeter should be known by design. Later, you will hear about two simple devices that help us do the amplification; bipolar and field effect transistors. Both have an input and output (and a third terminal as a common ground). Only their inputs differ. A bipolar transistor amplifies a current to a greater one, say 200 times (it gets its amplified output current from a power supply). On the other hand, a field effect transistor converts a voltage (not a current) at its input also to a current say 200 mA/V. As you see, using a field effect transistor at the input of a voltmeter makes its input resistance very high since it needs to sense the voltage only (actually there is a very small ‘leakage’ current which makes the input resistance very high but not infinity).

Kerim

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