Impedance ?'s

[keninverse]

I understand the basics of impedance such as bridging but how do you actually determine the  impedance of a circuit?  In many synth modules I always see resistors connected to the inputs what's that about...

[Rob Strand]

Generally it's not a simple thing, you can have a big messy circuit and a good of the parts can be affect the input impedance.

A typical input circuit might look like input jack -> resistor to ground (R1) -> series resistor (R2) and cap (C1) -> resistor to Vcc/2 (4.5V) (R3) -> input to opamp or transistor buffer.  Typical values might be R1=2.2MEG, R2 = 1k, R3 = 1MEG.

You basically have to calculate the effective resistance looking into the input jack.  This looks like R1 in parallel with the series combination of R2 and R3.

So first calculate the series resistance of R2 and R3,

Rseries = R2 + R3

Then calculate the input impedance, which is the parallel combination of  Rseries and R1,

Rin =   R1 parallel with Rseries =   R1 * Rseries /(R1 + Rseries)

You can make practical approximations to simplify the calculations.  Go back to the original circuit and notice that 1k in series with 1MEG is pretty much 1MEG, so the 1k in series part can be omitted without affecting the accuracy.  The result is the input impedance is largely determined by R1 and R3 in parallel.

The above only works because the opamp input impedance itself is really high and has no bearing on the input impedance of the unit.  The input impedance of a transistor in a transistor buffer stage is pretty high too and can often be ignored.  When you have other transistor circuits it's a messy job, something you would have to study quite a bit to understand - for example the input impedance of a fuzz face isn't a simple thing to work out.

Lastly, the input impedance doesn't have to be a single number, when you have capacitors and inductance the impedance can vary depending on the frequency.

[petemoore]

I had to study that, I think' I got a lot out of it!!
 Ibeen ponderin such quations            OK..that helps

[R.G.]

Rob's explanation of the complexity of determining the impedance is correct if you take an analytical approach.

The experimental approach is to hook a large pot in series with the input in question, hooked up as a variable resistor (two wire), but turned to the minimum resistance.  Apply a signal, make sure that the signal goes through **linearly**, not clipped, and set the output to some convenient value that you read on a scope or a meter.

Turn the pot resistance up until the output is exactly half the original output. Ohm's law now sez that the pot resistance is the same as the input impedance **at that frequency**.  If you've picked a good, midband frequency, you're done.

R.G.

[keninverse]

Thanks gentlemen...I've always wondered about this and it seems fairly intuitive; however I was a little confused and needed someone to just "put it to me straight."  It's fairly easy to just estimate the impedence from just glancing at a schematic...

[Andy]

what I dont get is what does having a "high input" impedance or a high output impedance do?  I've noticed that having a high input imped seems to be desirable.  That has SOMETHING to do with loading the circuit down that FEEDS the other, right?  exactly how does this "loading" happen? and how do you overcome this when you didn't design the circuits?  I read somewhere a "buffer" can be used?  Where are some plans for a buffer (if that's right)?

[Andy]

what I dont get is what does having a "high input" impedance or a high output impedance do? I've noticed that having a high input imped seems to be desirable. That has SOMETHING to do with loading the circuit down that FEEDS the other, right? exactly how does this "loading" happen? and how do you overcome this when you didn't design the circuits? I read somewhere a "buffer" can be used? Where are some plans for a buffer (if that's right)?

[Paul Marossy]

What RG describes sounds exactly like what Anderton describes on how to measure impedance in his book entitled "Do-It-Yourself Projects for Guitarists". I've been thinking about building that little gadget... I could come in real handy.

[R.G.]

Andy says:

what I dont get is what does having a "high input" impedance or a high output impedance do?
I've noticed that having a high input imped seems to be desirable. That has SOMETHING to do with loading the circuit down that FEEDS the other, right? exactly how does this "loading" happen?and how do you overcome this when you didn't design the circuits? I read somewhere a "buffer" can be used? Where are some plans for a buffer (if that's right)?

1. Think of "resistance" for the word "impedance". It's not strictly the same, but for low audio frequencies, and especially this question, it's an OK simplification.
2. "high resistance" inputs and outputs have exactly the effect you'd think - very little current flows per voltage applied. Every input and output has *some* resistance. Think about how this works: a pickup on a guitar has an output resistance of about 4K to 18K for common pickups. You can think of this in the EE's lumped model - the pickup coil is (in our minds) a perfect source of voltage, but there's this 4K to 18K resistor in series with it.

A perfect voltage source will supply any amount of current, no matter how large, to keep the voltage where it's supposed to be. This is impossible in the real world, so all perfect sources are imaginary, but let's not let that stop us. The imaginary version leads to some useful results for us.

We have this perfect source of , for instance, a tenth of a volt of signal, in series with a 10K resistor, just picking reasonable values.  If we measure the voltage at the perfect voltage source (which we can't, but let's forge ahead as though we could), we would measure 0.1V. If we measure the voltage at the end of the 10K resistor, we measure the same 0.1V, because the resistor is open. No current flows through it, so no voltage is developed across it, and the same voltage is at both ends.

If we now connect a load resistor of 10K to ground at the outer end of the 10K pickup resistance, we have a complete circuit and current can flow. The perfect signal voltage source sees a load equal to its own internal impedance/resistance of 10K plus a series load outside the pickup of 10K, total of 20K, and so Ohm's law tells us that 0.1V/(10K +10K) = 5uA of current flows. This flows through both resistors, so 5uA*10K = 0.05V of signal appears across each resistor. More importantly, the only place we here in the real world can reach, the junction of the internal 10K source resistance and the 10K load resistor,  measures 50 mV (0.05V), half of the signal voltage back at the perfect signal voltage generator.

The external load resistor being equal to the internal source resistor has divided the internal signal voltage by the ratio of the two resistors, and that's the only voltage we can get at. The imaginary perfect signal voltage inside the pickup is hidden from the real world by its source impedance. We call this internal resistance "output impedance" for signal sources.

If we're interested in measuring (or amplifying) that internal perfect signal voltage very accurately and not losing any, we need a high **INPUT** resistance on whatever the pickup is connected to. The input impedance/resistance of whatever we connect to the pickup acts like our 10K load resistor and reduces the signal voltage we have to work with.

We lose half the signal voltage with a 10K load resistor; if we have a 100K load resistor, we get 100K/(10K +100K) = 91% of the signal. If we have a 1M load, we keep 1M/ (10K+1M)=99% of the signal voltage.

That's what's important about impedances for voltage sources: you want low source impedances and high impedances loading them, because those impedances determine how much of the signal you get to keep. Otherwise, it's lost in the voltage divider ratio of the source and load resistances.

Here's where our "resistance" versus "impedance" part company. A real pickup output impedance is not really resistive. It's resistive plus inductive. If we load down a real pickup too much, the inductive part makes you lose more treble than bass, and the guitar signal sounds dull and muffled because the treble is lost more than the bass.

What happens if the source/output impedance is very big? Let's say that we have an imaginary pickup that has a huge voltage source - a thousand volts, as well as a very high source impedance, 1M ohm. If we short the output to ground, then the amount of current that flows is 1000V/1M = 1uA. If we put in any resistor less than about a tenth of the source resistance of 1M, then very close to the the same 1uA flows. The high output souce/output impedance acts like a constant current generator. you won't need to know this much, but it's the other side of the coin.

On input impedances: The input impedance of anything which reads or amplifies another signal acts like an impedance (which we simplify to resistor for a minute) to ground.  We imagine that the amplifying device that reads the signal is perfect and "eats" no signal current to work, and we personify the signal losses that will inevitably happen in the real world as a resistor load to ground that's impossible to remove from the inputs.

So if we have an amplifier with an input impedance/resistance of 10K  that we want to amplify the signal from our imaginary 0.1V through 10K pickup, we find that we start out with a signal loss of half. Worse, we start out with a really, really muffled sound because even if our 10K input impedance is a real resistor, the pickup still has an output impedance that goes up with frequency, so we lose proportionately more treble.

We found out pretty quickly with a few calculations that for a 10K output impedance on that example pickup, we needed at least a 100K, preferably a 1M load resistance to avoid signal loss. That translates directly to the INPUT resistance we need on an amplifier to amplify it. We need at least a 100K, preferably 1M input resistance/impedance amplifier.  Any thing lower will lose signal, and in the real world, lose signal unevenly over the frequency band.

For voltage amplifying circuits (which is most of what effects circuits do), you want LOW source (output) impedances and HIGH input impedances to keep from losing signal, and you also want the HIGH impedances to be at least 10:1, preferably 100:1 bigger than the LOW output impedances. This is the ideal setup for the best voltage signal transfer.

Now we're down to buffers.

A buffer is a circuit, usually a simple one compared to the rest of the junk you might use with it, that has a relatively HIGH input impedance and a relatively LOW output impedance. It may or may not have voltage gain. It can, but it's usually OK if it has a voltage gain of 1.

If you think about it, buffers have lots of available current gain. They take tiny little whiffs of signal current and can drive bulldozer loads of current out to keep the signal voltages from being lost.

You can usually insert a buffer between any two sections of a circuit that depend only on voltages. For instance, if we take our imaginary 0.1V and 10K pickup driving a 10K input amplifier, and we insert between the two a buffer circuit with a 1M input and 10 ohm output impedance (and those are quite reasonable, real world numbers for the buffer) then the pickup cna only see the 1M input impedance of the buffer, and the amplifier can only see the 10 ohm output impedance of the buffer; both pickup and amplifier think the other is ideal, and the buffer has made things work between them. In fact, that's how they came to be called buffers.

You can only insert buffers in an intelligent way - that is, you cannot ignore the biasing and DC circumstances of the place where you insert a buffer. You must know the buffer's needs for DC bias as well as the needs of the two stages you insert the buffer between and make sure that the DC conditions at all of the output of the signal source, input of the buffer, output of the buffer, and input of the following signal load are satisfied. This may mean you have to design biasing and signal coupling networks out of resistors, caps, and sometimes inductors and/or transformers, to make all of the circuits happy with each other in the DC sense.

So to put in buffers where you have a circuit you did not design, you have to find a place where you understand the DC and biasing well enough not to mess it up. The easiest places for this are the inputs and outputs, because these places usually have a DC level of flatly zero, and if you can just put in a buffer with 0Vdc on its input and output, everybody's happy. Mucking around inside the circuit where there are bias voltages mean you have to understand the circuit well enough to get the biases right with the buffer inserted.

Buffer circuits are so useful, they're found everywhere. They're a stock item in every circuits cookbook ever written, and have been devised from every amplifying technology every known. The EE world is practically littered with buffer circuits reflecting the zillions of slightly different circuits that are useful in the zillions of slightly different circumstances. The epanorama site should have a full load of them. I think Jack Orman collected up a lot of the simple buffer from the cookbooks in a  page on his site.

In the larger sense, any circuit that is non-loading on its input and relatively non-loadable on its output can act as a buffer, whether it amplifies or not. Even circuit not specifically intended to be used as buffers can be used.

What you're really missing is understanding of Ohm's law, plus likely some understanding of the terminology. I cannot tell you how useful Ohm's law is in electronics. You can have a complete professional career as an EE if you really understand Ohm's law well, and very little else. On the other hand, you cannot design electronic circuits (in the sense of understanding what you're doing and getting there by means other than easter-egging parts) without knowing Ohm's law. I highly recommend that if you want to go any further with electronics, you dig into that.

Between meetings, I've been writing a beginners book in electronics aimed specifically at effects for this community. You're just the latest in a long line of people with the same questions. The FAQs don't seem to be all that much help to beginners, so I'm going to try to get a little deeper into the subjects for beginners. Just what they need, none of the fancy stuff they don't.

Yell if I confused you more than helped, or if there was a place I glossed over that you need more on.