Input/output impedance question

[nosrej]

Hi! here's my confusion. V= IR so it is evident that voltage is directly proportional to current. But why do they say that as voltage increases, current decreases or vice versa. Does it have something to do if I'm considering the source side or load side?

Another question is, in op-amps they say that ideally the input impedance is very high while the output impedance must be very low. based from V=IR we need a high amount of resistance so that output voltage is increased compared to the input or amplified which is not what they say.

Please clarify and explain things to me.

 

[LvW]

Hi! here's my confusion. V= IR so it is evident that voltage is directly proportional to current. But why do they say that as voltage increases, current decreases or vice versa. Does it have something to do if I'm considering the source side or load side?

Another question is, in op-amps they say that ideally the input impedance is very high while the output impedance must be very low. based from V=IR we need a high amount of resistance so that output voltage is increased compared to the input or amplified which is not what they say.

Please clarify and explain things to me.

But why do they say.....

Who is "they"? And where do "they" say this?

I cannot understand the content of the second part of your contribution.

 

[capcas]

when we say "hey say that as voltage increases, current decreases or vice versa" means they are saying in terms of power. P = VI. Usually in the systems you have a fix maximum power so if you increase voltage in P = VI then current ought to go down to keep the power fix and vice versa.

The input resistance is usually very high to avoid the loading and so that maximum voltage appear at the input of the device. (see voltage division rule for details). The output resistance is kept low so that maximum voltage appears at the input of the next stage.

 

[LvW]

..................
The input resistance is usually very high to avoid the loading and so that maximum voltage appear at the input of the device. (see voltage division rule for details). The output resistance is kept low so that maximum voltage appears at the input of the next stage.

OK, that's correct.
However, the most important arguments for high (low) input (output) impedances are as follows:

It is the aim of an operational amplifier to
*allow a variety of mathematical operations (multiplication (amplification), summation, subtraction,integration, differentiation),
*enable the realization of complex transfer functions.

In all cases, the particular operation shall be determined by external feedback components only - which means that the opamp should be a voltage amplifier with nearly IDEAL properties:
*open-loop gain infinity,
*input impedance infinity,
*output impedance zero.

Of course, this cannot be achieved, however - if "infinity" is replaced by "very large" and "zero" by "very small" the contribution of these real parameters to the desired transfer function is sufficient small so that the idealized formulas can be applied.

 

[FvM]

when we say "they say that as voltage increases, current decreases or vice versa" means they are saying in terms of power. P = VI. Usually in the systems you have a fix maximum power so if you increase voltage in P = VI then current ought to go down to keep the power fix and vice versa.

This is for the category "How to read a sense into an unclear question?". To achieve this, you e.g. need to change "current decreases" into "current ought to go down" etc. I prefer to ask for an unequivocal question.

 

[nosrej]

Ok thanks guys for explaining you made it really clear to me and sorry if I wasn't able to use the correct terms

 

[enjunear]

To help you in future analysis, you may need to start looking at the world of electronics through the framework of impedance (Z), rather than simply resistance (R). When you introduce capacitors and inductors, they do not (ideally) have resistance, but they do present an impedance (also measured in ohms) to the circuit they are in, at some defined frequency. It just so happens that a resistor's impedance is identical to it's resistance.

Resistor impedance: \[{Z}_{R} = R + j*0\] Ω
Inductor impedance: \[{Z}_{L} = 0 + j*\omega*L\] Ω
Capacitor impedance: \[{Z}_{C} = 0 + \frac{1}{j*\omega*C}\] Ω
where ω is the frequency in radians/sec, and ω = 2*pi*frequency(in Hz).

The imaginary term [includes i or j... which are both equal to sqrt(-1) ] is called the reactance. The impedance of a device is the sum of the resistance and reactance, so in basic form, Z = R + X.

Read more here.
This is kind of a neat primer to the concept of reactance: video

 

[7@rB]

hi nosrej,
this case you are talking about is related to transformers domain and their applications, and also to the transmission of electrical power. you should keep in mind that in trasformer (Vout/Vin) = (Iin/Iout) . with transformer that transforms form high voltage to low one, the voltage that you are losing here in some words you can say that it is transformed into current, high voltage with a low currrent in this case taken as input will give us at the output a low voltage and a high current.

 

[fightshan]

but why the nearly ideal properties is "input impedance infinity"and "output impedance zero"?

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OK, that's correct.
However, the most important arguments for high (low) input (output) impedances are as follows:

It is the aim of an operational amplifier to
*allow a variety of mathematical operations (multiplication (amplification), summation, subtraction,integration, differentiation),
*enable the realization of complex transfer functions.

In all cases, the particular operation shall be determined by external feedback components only - which means that the opamp should be a voltage amplifier with nearly IDEAL properties:
*open-loop gain infinity,
*input impedance infinity,
*output impedance zero.

Of course, this cannot be achieved, however - if "infinity" is replaced by "very large" and "zero" by "very small" the contribution of these real parameters to the desired transfer function is sufficient small so that the idealized formulas can be applied.

but why the nearly ideal properties is "input impedance infinity"and "output impedance zero"?

 

[vyomax]

but why the nearly ideal properties is "input impedance infinity"and "output impedance zero"?

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but why the nearly ideal properties is "input impedance infinity"and "output impedance zero"?

for getting any functional output from OPAMP we need to provide some source voltage....
If input impedance is lower then it will draw more current and will overload our source....this is undesirable.
such situation will never happens if there is high input impedance....means OPAMP drawing less current.

for output case: any output will drive some load device (resister, LED, speaker,etc)....means load device will draw current from OPAMPs output...If output impedance is higher it is restricting this once again this is undesirable.
so output impedance should be as small as possible so that OPAMP would provide its service to maximum number of load devices.