### Why is high input impedance good?

Naive perhaps, but

- Why is high input impedance a good thing?
- Is high input impedance always a good thing?

Oli Glaser Correct answer

9 years agoIt is a good thing for a

*voltage*input, as if the input impedance is high compared to the source impedance then the voltage level will not drop too much due to the divider effect.For example, say we have a \$10V\$ signal with \$1k\Omega\$ impedance.

We connect this to a \$1M\Omega\$ input, the input voltage will be \$ 10V\cdot\frac{1M\Omega}{1M\Omega+1k\Omega} = 9.99V \$.

If we reduce the input impedance to \$10k\Omega\$, we get \$10V \cdot \frac{10k\Omega}{10k\Omega + 1k\Omega} = 9.09V\$

Reduce it to 1k and we get \$ 10V \cdot \frac{1k\Omega}{1k\Omega + 1k\Omega} = 5V\$

Hopefully you get the picture - generally an input impedance of at least 10 times the source impedance is a good idea to prevent significant loading.

High input impedance is not always a good thing though, for example if you want to transfer as much power as possible then the source and load impedance should be equal. So in the above example the 1k input impedance would be the best choice.

For a current input a low input impedance (ideally zero) is desired, for example in a transimpedance (current to voltage) amplifier.People always use the matched load for max power transfer point. None of my high power equipment does that. You do not want to be dissipating a ton of power in your source, instead they use a higher voltage and you design for a high load impedance. Not to say I do not understand your point, but just to note for others.

For maximum power transfer the source should have as low output impedance as possible. However, if the source has a relatively high output impedance and you can't modify it then the load should have the same impedance for maximum power. If the load impedance is higher, the power will be lower, if load impedance is lower, more power will be dissipated by the load. That's why vacuum tube amplifiers use output transformers to match the high impedance of the amp and low impedance of the speakers.

The "best" value of Impedance depends on the situation and application.

When it is appropriate to have or need a high impedance it is because it is an approximation to an infinite impedance.

An input applied to a signal source acts as a voltage divider.

Vout = Vsignal x Zinput / (Zsource + Zinput)

To get no loading either Zsiganl is zero (low or no impeadance output) and / or Zinput = infinite.

"Suitably high" is the practical version of infinite would be nice."How large "suitably" is depends on the application.

AC mains has an impedance well under 1 ohm (usually). A test meter with 1000 ohms impednace woul draw about 100 mA !!!! from 110 VAC mains but would only load it down my under 0.1 of a Volt in the process. A test meter of 1 megohm input impedance would draw about 100 uAmp which would be much more acceptable.

For high impedance sources "suitably) needs to be quite large.

A high impedance input places very little load on a signal that is applied to it.

It thus does not reduce it in level (or not much). A unity gain buffer usually has very high impedance and is often used as an input stage to an amplifier chain. A pH probe, used for measuring acidity and alkalinity of a solution, mat have an output impedance of 10's to 100's of megohms. It's voltage level is a direct measure of pH. So anything that seeks to measure the voltage must try not to alter it in the process. A voltage measuring probe will effectively act like a voltage divider. The probe impedance needs to be >> the measured impedance if loading is not to occur.A probe which is 256 times the impedance of a circuit being measured will cause 1 bit error in an 8 bit system.

A probe which is 4096 times the impedance of a circuit being measured will cause 1 bit error in a 12 bit system.So to measure with 1 bit in 256 = 1 bit in an 8 bit system with a 1 megohm source impedance you need a 256 Megohm input impedance. For a 10 Megohm source you need a 2.6 Gigohn input impedance. And for a 100 Megohm ource you need ... !!!

As per the formula above, for outputs, LOW impedance is good, with the ideal being zero impedance (a perfect voltage source).

Then there is the special case of matched impedances where source and input are the same. Half the signal is dissipated in the INPUT and half in the output (assuming otherwise lossless connection) BUT there are no reflections due to impedance mismatch. A whole new subject for another time.

Infinite input impedance would allow one to feed any amount of voltage into a load without it absorbing any power. Zero input impedance would allow one to feed any amount of current into a load without it absorbing any power. In cases where one wants to sense voltage without absorbing power, infinite impedance is thus the ideal; conversely, if one wants to sense current, zero impedance is the ideal.

Although sometimes one wants a load that doesn't absorb any power, there are times one wants to feed power into the load. The amount of power fed into a load will be maximized when the input impedance of the load matches the output impedance of whatever is driving it. This situation does not imply maximal energy efficiency, however. Depending upon what's driving the load, a higher or lower input impedance may cause the driving device to waste more or less power internally.

The word "high input impedance" is always related to the amplifier (audio intermediate frequency power amplifier... etc.)

So let's consider the following circuit:

The input voltage \$V_{in}\$ has an internal impedance (\$Z_{in}\$) this voltage injected to the base of transistor to amplify the signal. We calculate the voltage across \$Z_{in}\$, called \$v\$ as follows:

$$v = \frac{V_{in} Z_{in}}{Z_{in} + Z.V_{in}}$$

If we take \$V_{in}=5V\$, \$Z.V_{in}=2,000Ω\$, \$Z_{in}=10Ω\$ we get:

$$V=\dfrac{5 \cdot 10}{2,000+10} = 0.02V$$

That's a very low voltage compared with the input voltage.

If we take \$V_{in}=5V\$ , \$Z.V_{in}=2000Ω\$ , \$Z_{in}=1,000,000Ω=1MΩ\$ we get:

$$V=\dfrac{5 \cdot 1,000,000}{2,000 + 1,000,000}=4.99V$$

That's a good voltage compared with the input voltage.

Let's see some value of the input impedance in the table below.

The answer is the high input impedance is good for the amplifier circuit to have a good amplification of the input signal other wise we get low voltage in, so low amplification.

I hope this can help, thank you.

To get all the voltage from a source to a target without loss.

you need high input impedance. This principle is called "voltage bridging" or "Impedance Bridging".That is a relative low output impedance to a higher input impedance.

Usualy the input impedance is at least ten times higher then the output impedance.**Voltage Bridging**

one which maximizes transfer of a voltage signal to the load.

The other typical configuration is an "Impedance matching connection",

which maximizes power delivered to the load.The high impedance is not always good but it varies from application to application. In order for impedance matching with other circuits the designer will select the high input impedance using the theorem "Maximum Power transfer Thoerem"

linkAt high frequency impedance matching reduces reflected power ( look up transmission lines for more ).

An electrical signal has two components: (a) a voltage component (b) a current component.

To build a POWER amplifier requires equal amplification of both components and the

i.e. a Load impedance must equal (the purely theoretical) Source impedance.*"Maximum Power-transfer Theorem applies:*NOte that a soure impedance is not a true impedance - it cannot be measured but only calculated.

To drive an active component (valve or FET which has a high input impedance - large V/small I) a

*voltage amplifier*must be driven from a low Source-impedance but deliver from a relatively low-impedance. (Thevenin's Theorem.)To drive an active component (bipolar tansistor) which has a low input-impedance - small V/large I) a "current amplifier" must be driven from a high Source-impedance but deliver from a relatively high impedance. (Norton's Theorem.)

High Input means you only need the SIGNAL. Or lets call it the message of voltage. In this case low current is fine to drive the stuff.

High Input is NOT always a good thing. In case of not using the signal but driving an electronical part (for example for LED light) you need to calculate the current and you need to decrease the output resistance.

If you are using too high resistance while working with a signal message, the only point of view is the capacity to other parts.

If you are working in HF range of frequency modulation, it becomes more difficult. In any other case, yes, high input is a good thing to use to have less power consumption.

Regards

High impedance is not always good when a current must flow to achieve the desired result. For example, large area electrodes and conducting jelly are used to lower the impedance in Edison’s great invention, the electric chair.

License under CC-BY-SA with attribution

Content dated before 6/26/2020 9:53 AM

Kortuk 9 years ago

People always use the matched load for max power transfer point. None of my high power equipment does that. You do not want to be dissipating a ton of power in your source, instead they use a higher voltage and you design for a high load impedance. Not to say I do not understand your point, but just to note for others.