Impedence: so am I getting it right?

[Herr Masel]

I've been reading about impedence but one thing never explained that always seemed strange to me is that lower input impedence is HARDER to drive. So tonight I was listening to Mahler I think it was with my grandfather and an explanation (inspiration?) came to me. Lower impedence requires less drive (voltage) in order for current to flow, therefore the current is not driven hard enough and is "lazy", whereas higher impedence requires more voltage or effort and therefore signal passes through whole, which explains why higer value resistors retain more treble than lower ones. Or is impedence like a gate that keeps the signal from oozing out and degrading, so the input impedence of something needs to be higher than the output impedence of the circuit driving it so that it keeps things boarded up and moving in the right direction (in a DC circuit). Am I right, or is everything I said total nonsense?

[gaussmarkov]

hi.  i have been trying to understand this, too.  here's some useful stuff that i have found on the web:

The input impedance or sometimes loading impedance of a circuit or electronic device is the impedance actually experienced by a signal which is connected to its input.

For example, an amplifier with 100,000 ohm input impedance looks equivalent to a 100,000 ohm resistor to the signal coming into it.

Their illustration helps clarify what "experienced" means:



Z_S is the source impedance and Z_L is the loading impedance.  if  Z_L is really low, then it looks like the signal drains to ground rather than going on to the circuit.

they go on to say

Generally in audio and hi-fi, the input impedance of components is several times higher than the output impedance connected to them. This is called voltage bridging or impedance bridging. In this case,

    Z_load >> Z_source

In general, this configuration will be more resistant to noise (particularly power line hum).

i have seen similar statements other places and i found this helpful: to view input impedance as a pull-down resistor at the input.  i hope it helps you, too. 

R.G. has lots of useful information about this on his site.  for example, look at his description of the input buffer stage of the tube screamer where "the 510K biasing resistor accounts for almost all of the signal loading at the input."

[SaBer]

Refering to the picture in the second post: Zs and ZL form a voltage divider, which works exactly like a volume control. The lower the input impedance, the less volume you have.

Also, the smaller ZL is, the more current you have to "push out" (actually "is pulled out") of the output. So, with lower impedances, the output has to "do more work".
You can actually make an op-amp clip, by loading it down enough (putting a resistor from output to ground).

[Joe]

It's easer to understand if you look at it from a different angle. Resistance is analogous to an object with mass. A larger resistance requires more force (voltage) to accelerate it to a particular velocity (current). In reality the electrons flow through the resistance, and not the other way around, but the effect is identical.

Impedance acts similarly to resistance, but is frequency-dependent. The signal voltage won't degrade with a high-Z input because it takes a huge voltage to "push" a million ohms. (Swing a 9-iron at a bowling ball and it won't go far.) A low-Z input is easily "pushed" by the signal voltage, which is mostly used up so less is available for amplification.

[Herr Masel]

Thanks, this helps, what I think I was seeing wrong was that I thought that resistors prevent from current to go through, "starving" the circuit etc. when in reality they help stabalize the flow so that too much current doesn't flood the circuit, which makes sense now that I think about it because resistors used in say a stompbox won't be strong enough to block a 9v battery.
I am still not sure why it is HARDER to push lower impedence.
This brings me into the world of biasing, which I don't know much about. For instance I was asking why my big daddy distortion had some subtle octave, and was answered that it is a biasing problem. I measured the voltage on the gate/source junction of the input transistor and it was -32, the datasheet for the transistor says it should be -25, is this a big difference? And I can change it by changing the resistor? The schematic calls for 4M7 and I used 2M2, I also changed the value of a resistor in the tone part of the circuit (which is not in the schemtaic). Is this the cause?

"Impedance acts similarly to resistance, but is frequency-dependent." Impedence and resistance are not the same thing??!

[bioroids]

"Impedance acts similarly to resistance, but is frequency-dependent." Impedence and resistance are not the same thing??!

I think of impedance as a particular resistance at a particular frequency.

Resistors for example have the same impedance at all frequencies, so they are pure resistances.
On the other hand, capacitors present a different impedance at different frequencies (from infinite at DC to zero at infinite frequency an ideal capacitor). Idem with inductors (with the reverse frequency response). They are called reactive components.

If you want to know what's the resistance of a capacitor C at frequency f you use the capacitive reactance (Xc) formula: Xc=1/(2*pi*f*C)

The output impedance of the guitar and the input impedance of the effect (Zout and Zin) form a voltage divider just like the standard output volume pot. If you lower Zin (or raise Zout) you are turning the pot towards ground ie reducing the volume of the signal entering the circuit. As Zout is higer at higer frequencies, you will loose first the higs (that's what happens in old tone sucking pedals), as if the pot were turned more towards ground at higer frequencies.

Hope this helps!

Miguel

[gaussmarkov]

i also found these responses helpful.  thanks to everyone.

"Impedance acts similarly to resistance, but is frequency-dependent." Impedence and resistance are not the same thing??! 

i will try to add to the answers about resistance and impedance.  if i make any mistakes, i hope someone will correct me. (corrections and additions in green. thanks, niftydog! )  there are 3 terms that are closely related:

• resistance
• reactance
• impedance

resistance and reactance are special cases of impedance, the general term for the ratio of voltage to current.  it seems pretty unusual to see anyone refer to reactance on this forum, but we could because reactance is the impedance of capacitors. 

distinguishing between these concepts also depends on understanding that alternating current and alternating voltage are not necessarily in phase.  two sine waves with the same frequency are in phase when their peaks and troughs occur at the same points in time.  a capacitor actually delays the voltage back 90 degrees (one quarter of a cycle) from the current. so the peak of the sine wave for voltage coincides with end of the "hump" in the sine wave for current.  there is a very nice graphical explanation of this by Joe Wolfe (UNSW) on this webpage: "AC circuits, AC electricity."

resistors do not change the voltage-current relationship.  so resistance refers to impedance that causes no delay.

reactance refers to impedance that delays (or shifts) the phase of the voltage by 90 degrees behind the current.  impedance does not restrict the amount of the phase shift.  i have not tried to understand phasers and flangers yet.  but i am guessing that this distinction becomes important there.   in resistor-capacitor (RC) filters, the phase shift depends on the frequency of the signal and it varies between 0 and 90 degrees.

[Herr Masel]

Ahh, so lower impedence is harder to drive because the battery has to pump a high enough voltage to get to the required current level, and a low value resistor would "waste" alot of the current, causing the battery to work more. Thanks!
What about the biasing?

[niftydog]

...has to pump a high enough voltage to get to the required current level

Other way around! Usually voltage is the important thing in most cases.

Lets say you have a signal source that is capable of putting out 1VAC at 50mA. Therefore, it's output power is P = VI = 1VAC x 50mA = 50mW.

Now hang a 10k resistor from it's output to ground. Ohms law tells us that 1VAC across 10kohms means that 100µA will flow. (I = V/R) So in this case, the load is only dissipating 100µW, far below what the signal source is capable of supplying.

Now, change that 10k ohm resistor for a 10 ohm. Trouble! Ohms law says 100mA should flow, but we know that the signal source is not capable of 100mA @ 1VAC, so it cannot maintain the voltage while it tries to source that current. Hence, the voltage drops and a consequence of that is that the current drops as well.

Resistance is a special case of impedance. Impedance is the combination of resistance (DC) and reactance (AC). If an impedance has zero reactance, then it is purely resistive ("real" not "imaginary" [long story, don't ask, google "imaginary algebra"!])


Without knowing the circuit you describe, I cannot accurately comment. Suffice to say that quite often a tiny difference in bias can result in very strange results. Also, keep in mind that impedances are rarely if ever 100% isolated from the next/previous stage. So, in some rare cases, changing a component down stream of a transistor might upset it's delicate biasing.

[H.Manback]

No offence or anything, but with the topic title being as it is I couldn't resist

It's Impedance

[Herr Masel]

bunch of stuff

Thanks, that was an excellent explanation! My ohm's law and symbols for unit's names still need to be whetted more, but I'll learn.

A few more questions, but I'm getting there: how do I know how much amperes a source can provide? Is it simply ohm's law - I=V/R ? If I am looking at a certain point in a circuit do I have to calculate from the input what the current or voltage is, and update that figure with every resistor or cap or whatever that is there until I reach the point I am looking at? That sounds way too tedious and there is probably a better way. (this question is unclear, so it is in reference to: "Lets say you have a signal source that is capable of putting out 1VAC at 50mA.", so I am asking what would happen at the second or third resistor. If the current is supposed to be equal in all the circuit then calculating with ohm's law would yield different results, depending where you are looking!).

And what is the bias? Is it the process of calculating how much current [voltage, sorry] should flow in a given place in the circuit? How do I know if my bias is good or not, if the place I got the schematic from (ROG) doesn't tell you what the voltages should be?

[gaussmarkov]

If I am looking at a certain point in a circuit do I have to calculate from the input what the current or voltage is, and update that figure with every resistor or cap or whatever that is there until I reach the point I am looking at? That sounds way too tedious and there is probably a better way.

it's worse than that!    in general ASAIK, you have to calculate based on everything in the circuit--from input to output.  sometimes you can make approximations but it takes some experience to know when those will be reliable.  even calculating from input to output involves approximations about the impedance of what precedes and follows your circuit.

you are absolutely right.  the calculations can be tedioius.      and there are alternatives.     some people use a SPICE program to make the calculations for them.  i have been using 5spice and superspice, programs that have free (but limited) versions.  these programs can handle much more complicated calculations than those involving Ohm's law using simulation methods.  still, these programs involve some approximations, too.  and given the variation in actual component behaviour relative to nominal characteristics, you still have to make adjustments in actual building/design.

[niftydog]

how do I know how much amperes a source can provide?

depends on the source, but op amps for instance will often give you some idea of their maximum output on the datasheet. You can, given the right info, use ohms law, but it requires intimate knowledge of the sources output impedance which is not often in the datasheets. Otherwise, it's kinda trial and error.

Now, you're question about current neglects the fact that current can be sourced by active components from the power supply rail(s). So, measuring input current and working forwards isn't necessarily the way to go, unless you want to account for current gain etc etc. Tedious is an understatement, and it's prone to massive errors. If you really want to get into this, you need to start understanding nodal and mesh analysis. These methods will help you analyse a circuit in smaller "chunks" or branches. However, it is not often necessary to do such detailed analysis - just some common sense and a little knowledge goes a long way!

Ideally you want everything in a circuit to be working well inside of it's maximum capabilities. So, in general we set impedances such that the maximum power outputs of active devices are at a mid-range level so as not to stress anything out. If you have to measure, analyse and calculate branch currents to find out if a device is working as intended, then you are not leaving yourself any wiggle room in your design and things are likely to fail from stress at some point in the future. Er on the side of caution, and everything will be cool (figuratively and literally!)

If the current is supposed to be equal in all the circuit

clarification; current is equal in a closed loop (branch, mesh etc)
(Don't wanna go down this path too far though! Confusion follows.)


The best thing for you to do to better understand biasing is to have a look around on the net for some beginner tutorials about transistors and simple class A amplifiers. I would recommend that you start by searching this forum for the words "beginner tutorial" and click on the links we have provided.