Again with the impedance confusion

[Vitrolin]

So, input impedance i set by the resistor to ground, also called pulldown resistor, should this resistor be before or after the input cap?
Looking at GGG's LPB2 (which i build) theres a 1M resistor (R6) to ground before the input cap (C1), after the input cap theres a 43K resistor (R2) to ground, I've seen similar things in other designs, but wouldn't these be in parallel and there for produce a input impedance of 41kZ?
Where is the improvement?


Is the input resistor to ground the only thing that sets in put impedance?
How does one know the out put impedance of a circuit?

[brett]

Hi

So, input impedance i set by the resistor to ground

No.  The input impedance is usually set by the circuit.  A few high-impedance circuits have their input impedance set by the pull-down resistor, but this is rare.  The input impedance of bipolar junction transistors (the usual ones) is usually much lower (typically 50k to 300k) than anything to do with the pulldown resistor (usually 470k to 2 M).

The pull-down resistor "bleeds" charge (to ground) from the input side of the input cap.  Otherwise, it can sit there and wait for the circuit to be re-connected by the stompswitch and cause a "pop" by adding something extra to the signal.

Don't hesitate to ask more questions.  None of us were born with this knowledge.
cheers

[Vitrolin]

i found this article which agrees with me that the pull-down resistor set input impedance,
http://www.muzique.com/lab/imp.htm

and the ROG omega booster has a pot to ground that they claim is a variable impedance, though to me it looks like a low-cut filter.
http://www.runoffgroove.com/omega.html

i'm now even more confused, i understand the importance of impedance matching, being a sound technician,

there for i would like to know how out put as well as the in put impedance are given

[R.G.]

i found this article which agrees with me that the pull-down resistor set input impedance,
http://www.muzique.com/lab/imp.htm

I agree, that article is confusing. Here's the scoop.

Think about the input to a pedal, just the metal contact on the input jack. What is the impedance which a guitar "sees" when it tries to drive that point?

Let's be simplistic and say that there is no circuit board in there, just a resistor to ground of 10K. The guitar plugged into the jack obviously has to drive a 10K load.

Now let's add a 1K resistor in series with the 10K. What load does the guitar see? Yes, 11K, the sum of the two.

Next illustration, leave the 10K, but put the 1K in parallel with it. What does the guitar see as a load? It sees both the 1K and 10K in parallel, or a load of 1K*10K/(10K+1K) = 10/11 K = 0.9090909K = 909.1 ohms. If it was a 10K we put in parallel with the original 10K, that would instead be 5K - the two loads in parallel.

Next illustration. 10K to ground, plus a 1uf cap and another 10K in series. At some very high frequency, the cap looks negligibly small compared to the 10K it's in series with, so the load is 10K paralleled with 10K. At DC, the cap is an open circuit, so the load the guitar sees is just the 10K to ground (yes, I know guitars don't put out DC. This is for illustration!  )

Where does it go from 10K (the DC value) to 5K (the high frequency value)? Well, it depends on the impedance of that cap. The impedance of a cap is Xc = 1/2*pi*F*C, where F is the frequency and C is the capacitance in farads. So for a 1uF, the impedance is Xc = 1.(2*pi*F*1E-6) = 159,236*1/F. So the cap looks like a 159K impedance at 1Hz. It's 15.9k at 10Hz, 1.59K at 100Hz, and 159ohms at 1000Hz. So it looks like we can ignore it for DC at 1Hz, and for AC at 100Hz, and we have to take it into account at frequencies around 10-20Hz.

But what if we glom on another 10K resistor in parallel with the one behind the 1uF cap? No problem, we just calculate the parallel combination of the two resistors behind the cap (now 5K), recalculate where the paralleled resistors interact with the capacitor, and the 10K load at DC stays the same.

If you're with me so far, I'm going to generalize: the load on the input jack is equal to the equivalent series/parallel combination of **everything that is connected to the input jack**. That includes every resistor, capacitor, inductor, and semiconductor. All but the semiconductors are easy. What's the input load of a transistor? Or an opamp?

Those are complicated. An opamp will have an input impedance that is often - not always! - negligible compared to the resistors and caps connected to it. The (+) input is often many megohms, and so that leads people to say "the input biasing resistors set the input impedance" , leaving off the very important "because we can neglect the impedance of the (+) input because it's so high." Without that last qualifier, beginners like you are led down the garden path until they run into a situation where they simply cannot use the approximation. For instance, an inverting opamp input pin has an impedance of just about zero ohms, fixed by the feedback network. In this situation, the input resistor DOES set the input impedance because the semiconductor and the feedback resistor are forcing the input to zero ohms.

Unfortunately for you, you simply must know the input impedances of the devices you connect to any point or be able to approximate them somehow to get any kind of accurate estimate of the impedance to signal at that point.

and the ROG omega booster has a pot to ground that they claim is a variable impedance, though to me it looks like a low-cut filter.
http://www.runoffgroove.com/omega.html

Think about what I just typed above. It may well be a low cut filter. However, it is also a variable resistor, which is by definition a variable impedance, and so it varies the impedance of the nodes it's connected to as you turn the pot. It's possibly BOTH things at once.

i'm now even more confused, i understand the importance of impedance matching, being a sound technician,
there for i would like to know how out put as well as the in put impedance are given

Actually, you don't understand the importance of impedance matching. There are only very limited situations where you want impedances to match. In many situations it's important that the NOT match. Matched output impedances are important for maximum power transfer situations. We want that in tube amplifier output stages and some microphone setups. But usually in audio we want impedances mis-matched in a specific way.

In most cases, we want the input impedance of a piece of equipment to be much, much higher than the output impedance that drives it. That's because we want maximum signal voltage transfer, not maximum power transfer. Matched impedance necessarily mean that half the signal power is lost in each of the driver and the load. Also half the signal voltage. We usually want all of the signal voltage, or as much as we can get.

This leads rapidly to the run of 10 to 1: in general, you can neglect things that are ten to one greater/lesser than other things. If you're carrying a 10 poung grocery bag, putting in another pound or removing a pound usually doesn't make a huge difference. If you have 1V of signal, generally you're OK with 0.9V or 1.1V. If your input impedance is 10K, a driving impedance of 1K won't change the signal voltage you get by very much. Obviously, this is just a rule of thumb, and sometimes ten to one isn't good enough. But it's a good start!

[drk]

just a little question for RG, about this part:
"That's because we want maximum signal voltage transfer, not maximum power transfer."

by maximum power do you mean maximum current? when you match the impedance.

[Vitrolin]

ok i think i've got it,

so looking at the GGG LPB resulting input impedance by C1, R1 and R6 are:

   30 Hz   90K
100 Hz   58K
    1KHz   46K
  10KHz   43K
  20KHz   43K

this is of course ignoring the impedance of the rest of the transitor and the rest of the circuit
according to this to get a the 10:1 thumb rule across the hole audible spectre it should have a 4K impedance at higher frequencies.

but i hope ive gor this right
thanks R.G.

[R.G.]

by maximum power do you mean maximum current? when you match the impedance.

No, actually, maximum *power* transfer.
You get maximum voltage transfer when the load is an infinite resistance.
You get maximum current transfer when the load is a zero resistance. That way the current is limited only by the internal resistance of the source.

You get the maximum power transfer when the source and load impedances are matched. Any variation lessens the amount of power you get out, at least instantaneous power. Batteries may provide more energy (power *time) when discharged more slowly, but the instantaneous power out is maximum with matched impedances.

This is why you hear about matched impedances where it really, really matters - in RF ciruits where the incoming power may be -160dbm. Squander any of your incoming antenna power and you just get noise. Well, there's that speaker thing for tube amps, and 600 ohms balanced lines for pro audio. That's more needing to keep the impedance low than max power transfer though.

[R O Tiree]

You can't ignore the transistor because the emitter resistor is so small and also the "intrinsic emitter resistance", re, is of the order of 63 ohms, because the emitter current is also small (around 400µA, so 25mV/0.4mA = 62.5).

The signal looking into the base "sees" an impedance of Re (390R) plus re (63R) times hfe. Call hfe 600 for a 2n5088 and that gives you 271k. (From the Philips datasheet, min hfe is 300, max 900)

Go back one step... the signal looking into the junction of the 2 bias resistors towards the transistor's base sees those 2 bias resistors paralleled (39.1k) and that result in parallel with the base impedance. That total is 34.2k.

Now go back one more step, so we're looking into the input cap - what we've effectively done is reduced a complex situation into a simple one - the 0.1µF cap connected to a "fictitious" 34.2k resistor which is, in turn, connected to ground. Now we can work out some impedances:

You can't just add Zc and Zr, BTW. You have to work out SQRT(Zr^2 + Zc^2) to get the final figure for Zrc.

 30Hz : 63.1k
 82Hz : 39.3k (low E string)
100Hz : 37.7k
 1kHz : 34.2k
10kHz : 34.2k

That seems a little low for the value quoted on the AMZ page (42k). Maybe the original circuit had bias resistors of 56k and 560k? That works out around about 42k... Or maybe someone figured that 271k //el 43k was as close to 42k as made no difference?

Or maybe I've missed something? I don't think I have, because I just ran the GGG circuit in the simulator and fed it an AC source in series with a variable resistor. With the VR at 0 ohms, I measured the amplitude of the waveform at the base of the transistor, then wound the VR up until I got half the measured value. At that stage, the resistance of the pot = input impedance. Repeating this for all the calculated values above, the sim gave values within 1k or so in each case (except the 30Hz case - the sim wanted 67k to balance).

[drk]

i was forgetting about the possibility of zero resistance load, was only considering the matched impedance, and the big input impedance. you're right as usual

So matched impedance is kinda of having a balanced voltage/current ratio?