Load/Source Impedance

** Warning Will, danger, danger ....

** Basically - yes

** Do not isolate words from their context - this is where YOU are going wrong all the time. ** NEVER use the letter " I " to mean other than current.

This is where YOU are also going wrong all the time.

Saying a source is "low impedance" indicates that it has the *capacity * to deliver significant current into a load, a current level that is significant

*in the circumstances*.

With high frequencies, the main worry with loading is not the input impedance of the next device - but stray capacitance in connecting cables and PCB tracks.

This is a formula you need to know:

I = C dv/dt

If a signal voltage jumps by 5 volts in 10 nanoseconds,

then dv/dt = 500 exp6 V/s

If the stray load capacitance is say, 50 picofarads, then

I = 0.025 amps or 25 milliamps.

...... Phil

Uriah,

maybe you want to have a look at FEEE which explains impedance in various contexts. For a starting point you could use the index of the book, looking for impedance:

More specifically, there's a chapter on coupling which deals with impedance issues, too:

Regards,

Hans

Uriah, It sounds to me like you don't understand what impedance means. Impedance can be loosely defined as a resistance to current flow. Resistance and impedance are very similar in that they are both used to describe something which limits current flow. I imagine you know what resistance is. In DC circuits, you are usually dealing with resistances.

Example: If you take a 9V battery and connect a 100 k-ohm resistor across the terminals, the battery will supply 9V/100000 ohms of current. Given the current capacity of the battery, this is a high resistance (impedance), meaning it allows a small amount of current relative to the capacity of the battery. On the other hand, if you place a 10 ohm resistor across the same terminals, you will then be drawing 0.9A from the battery. This is a lot of current for the battery, and it is considered to be a large load (low resistance/impedance).

Low resistance or impedance >> large load High resistance or impedance >> small load

Now the difference between impedance and resistance is in a sense, just semantics. Impedance sort of means a resistance which varies with frequency. Resistance doesn't vary with frequency. A capacitor becomes highly resistive at low frequencies, and it is minimally resistive at high frequencies. An inductor does the opposite. It ideally has zero resistance in DC circuits (where the frequency is 0), and it becomes highly resistive at high frequencies.

Now the idea behind a low output impedance and a high input impedance is based on the idea that your input signal is going into an amplifier, not a speaker. High input impedance will prevent your output signal from having to produce more current than it is capable of. In other words, a high input impedance will retain the integrity of your signal.

I have no idea if this is going to help. I hope so.

flank

Hi, Uriah. Look at this circuit (view in fixed font or M$ Notepad):

| | Signal | Source Load | .----------------. .------------. | | | | | | |VCC-o Vo ___ | | | | | __--o-|___|-o----------o---o----o | | |GND-o Zo | | | + | | | | | | .-. | | | | | |Zi| | | | | | | | | | Vi | | | | | | '-' | | | | | | | | | | | | | | | | | | | | | - | | | '-----------o----------o---o----o | | | | | | | '----------------' '------------' | (created by AACircuit v1.28.6 beta 04/19/05

There's a signal source and a load. In this diagram, for the sake of discussion, the signal source Vo (Vout, the output signal itself) switches between Vcc and GND like a digital signal. This works for analog signals, too. All of your questions, in one form or another, all have to do with transmitting the signal Vo, and the question of what's happening at Vi (Vin, the input or load voltage). Zo (Zout, the output impedance) is a characteristic of the signal source, and Zi (Zin, the input or load impedance) is a characteristic of the load.

There are a couple of points worth mentioning about this circuit:

• If you want maximum Vi for a given Vo, you want Zo (output Impedance) to be as low as possible, and the load impedance Zi to be as high as possible. It's just a voltage divider question. It's that simple.

• The current flowing from the signal source to the load is just the voltage divided by the sum of the impedances. Ohm's law works for AC, too. It's that simple.

• If your signal source is a microprocessor pin, and your load is a bus with many CMOS inputs, you'll usually find the load impedance is primarily capacitive. This means that Vi will rise as the capacitance is charged up by the signal source, using Mr. Allison's equation. The speed with which Vi rises to Vcc or falls to GND (very important for many signals) is dependent on the source impedance Zo. The higher the impedance, the slower the cap will get charged up. This could mess up a clock signal, or another signal with specified maximum rise/fall time.

But the load impedance Zi is also important in determining rise and fall times. The more you increase the load capacitance Zi for a given Zo and signal, the slower Vi will rise and fall. So you want to endeavor to keep the bus impedance (load impedance Zi) as high as possible (in other words, try to keep the bus capacitance to a minimum).

I hope this has answered your immediate questions. There's a lot more really basic stuff to learn here. Your question also indicates you might not be entirely clear on what impedance is. It actually sounds like you need to back up the truck. I'd suggest you get back to the textbooks (assuming you bought them here

The *lower* the impedance (more load capacitance), the slower the cap will get charged up by the source.

Oops. Sorry.

Chris

Simple example.

Suppose you are driving a 9-volt, 0.1-amp light bulb. That is a load with a load impedance of 90 ohms.

And you're doing it with a 9-volt battery with a 1-ohm resistor in series with it. That is a source with a source impedance of 1 ohm.

The 90-ohm resistance of the light bulb and the 1-ohm resistor attached to the battery (or inside it) form a voltage divider. The bulb gets 90/91 of the voltage.

If that 1-ohm resistor were 90 ohms, the bulb would only get 90/180 of the voltage.

A "low impedance" source is "powerful" and has plenty of "drive". A "high-impendance" source is weaker with less "drive".

e.g. Case A A battery lights a bulb: then you can say "The battery has enough drive to light the bulb."

Case B But some time later the bulb gets dimmer. What has happened? Evidently the battery is getting flat but what does that mean? It means that the battery has *less* drive than it before or "less power".

The new battery has "low output impedance" while the flat battery has "higher output impedance". Let's measure the that output impedance in both cases.

1) Take the new battery and measure its voltage (with nothing except your volt meter connected to it). Say it is 1.50 Volts. 2) Repeat 1) while the bulb is connected. Say it is 1.40 Volts. 3) And measure the current. Say it is 100 mA.

Now we can calculate the output "impedance" (it is resistance really). You can imagine the battery when nothing is connected, it is having an easy time, doing no work and so it can hold its output at maximum of

1.5 volts. But once the bulb is connected, then it struggles to force current through the hot filament and its output is loaded down to say 1.4 volts.

We say "this drop of 0.1 Volts inside the battery is happening across the batteries output impedance" and we can say that it equals 1 Ohm because Ohms law says V/I = R and so 0.1 Volts divided by 100 mA is equal to 1 Ohm.

Evidently this thing called "output impedance" is not a resistor as such but is a sort of build of waste matter while the battery is doing work and this build up causes the loss of drive and the fall in output voltage. If you loaded the battery more with say two bulbs in parallel then the above calculation might produce an output impedance of 1.1 Ohms i.e. if all depends on the load e.g. if you short the battery out, then evidently the output voltage will be 0 Volts and if the current was 1 Amps then the output impedance would be

1.5 / 1 = 1.5 Ohms

In case B the voltage might be 0.4 volts (under load) and the current might be 50 mA. If the unloaded voltage was 1.4 volts then

1 / 0.05 = 20 Ohms

is the output impedance.

********************************************************************** The load impedance (resistance really)

1) In Case A and B the load impedance is (mainly) the restance of the bulb's filament (when hot) which is lowish.

2) In Case A and B without the bulb connected, the load was resistance of your voltmeter only and this will be very high maybe 10,000,000 Ohms if you are using a digital multimeter or 200,000 Ohms if you are using an old fashioned AVO. I.e. both very high.

Voltmeters are designed to have high resistances... so that they don't load the thing they are measuring... and give a falsely low reading :)

Cheers Robin

You say you teach electronics, then you give someone this lousy example? A light bulb is a non linear device, so you will not get what you stated. It will confuse the hell out of anyone who tries it as a hands on experiment. Now, if you said, We can use a 90 ohm resistor to simulate...

--
Service to my country? Been there, Done that, and I\'ve got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida

Admittedly, I oversimplified. But I wanted to keep things simple. Supplying power to a light bulb is a common, ordinary thing to do; supplying power to a resistor is not. To teach effectively, I like to avoid abstruseness.

Any experiment that can't be done on the bench should be avoided. At the very least you could have pointed out that the device is non-linear rather than post pure BS.

I was on a committee a few years ago to choose new textbooks for an electronics course and they were all full of this crap. We ended up choosing the one with the least mistakes and sending a list of errors to the publisher. It wasn't the best overall book, but it doesn't instill confidence when a teacher has to hand out thick errata sheets with the textbooks.

--
Service to my country? Been there, Done that, and I\'ve got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida

In general, I agree with you, and it was an omission on my part -- I should have added "in practice, the resistance of a light bulb changes with temperature, so this calculation doesn't describe an actual one."

I agree that experiments should be able to be done on the bench, and in fact I'm drafting some text materials in which every circuit, without exception, has all the component values indicated and can be breadboarded. It used to drive me crazy that all the circuits in the books lacked most of the component values.

Actually, I'd like to see some of your critiques. Did you find any in The Art of Electronics? (I have no connection with that book other than that I like it... but it's probably too hard for beginners.) Under "Bad circuits" in one place they list a circuit that is not one of their students' mistakes, but is taken from an earlier textbook by Diefenderfer.

I want to thank everyone for taking the time to reply to my post. I have been reading them over and will need to go over them a few times. There is allot there. It is starting to clear up. I don't know calculus so when I see the I=C dv/dt I am clueless but that is for another time. I think I am on the right track and once I go through this a few more times I think it will be clear. Once again thanks to everyone for taking the time to explain. I really appreciate it. Uriah

These books were all for a two year vocational electronics course, and the state didn't have AOE on the list. I may be ale to contact the man who was the teacher and see what he remembers. I turned all of the materials back in to the school board's book committee and I don't even have a list of the titles. They replaced the electronics course with a computer course the following semester and tossed all of the books, test equipment and supplies into a construction dumpster. They also left a dozen, new 4 channel analog scopes sitting out in the rain. The bastards.

--
Service to my country? Been there, Done that, and I\'ve got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida

Things have not changed much then, since Feynmann's chapter (probably titled) "judge a book by its cover" in his (definitely titled) "Surely you're joking Mr Feynmann"

Cheers Robin

(One of the choosen books was blank. All the pages were blank. The only print was on the front and back covers. No one had noticed because no one had read it, except Feynmann.)

Was it a state-funded institution? Most states have an office that solicits suggestions of ways to save money and reports of egregious waste. That would be worth reporting.

It's an absolutely shocking waste of money.

Graham

I don't think the explanation was difficult to understand. As far as a benchtop experiment is concerned, the light bulb would heat up fast enough (and reach its steady state characteristics) that the experiment would appear to have a linear load. If the experiment was done with an oscilloscope, designed to capture the first second or two, then the experiment would appear to be non-linear.

You sound quite knowledgeable. You should consider writing a better textbook. I also teach electronics, and I use the Hayt, Kemmerly, Durbin book, which is by far the best I have found, and it has many errors and truly horrible homework problems. Good luck with your book.

supplying

Sorry, but when you have the 90 ohm resistor in series with the lamp in the second part, there isn't enough current flow to allow the filament to come to full operating temperature, so it won't be 90 ohms, unless you double the supply voltage.

from the problem that was posted:

• If that 1-ohm resistor were 90 ohms, the bulb would only get 90/180 of the

• voltage.

I've also taught vocational electronics, back in the '60s. I ws offered a full time job teaching at another vocational school in the mid '70s, but they wanted me to move into their district, and they money just wasn't there. in the late '70s I was offered a job teaching at one of those TV and industrial tech "mills" where they tried to hammer the basics into their heads in a very short course. Their equipment and materials were 30 years out of date and I was not impressed by any of their graduates that I had to show how things really worked.

As far as writing a textbook, the teacher and I were were negotiating with a publisher during the book selection process, but the deal fell through. I was working about 60 hours a week, and the teacher had outside interests, so they deadline they wanted along with an insultingly small price it just didn't happen. It would have cost me money to write the book by their deadline because I would have had to turn down all the overtime to be able to complete it on time. Now, I am on a 100% disability. The VA tells me that I am allowed some small income, but I still have now answer as to what I can earn and not lose my pension and medical care.

I think I would rather see it done as an interactive CDROM book with a large section of experiments and test questions to encourage additional self study. I learned the basics by reading every electronics book I could find, starting when I was 10 years old. The ARRL handbooks, used college textbooks and the then common monthly electronics magazines and assorted tools took up every cent I earned. The authors had a way of teaching you electronics while keeping it interesting, yet leaving you hungry for more knowledge. I will see if I can find a copy of the book you are talking about. There is still one vocational electronics course in the area that might have some newer books. I lost my collection of electronics textbooks when I was laid off five years ago, and the company closed before I could get my personal property out. The company hired a number of wet behind the ears techs every year, and some needed refreshing fairly often.

--
Service to my country? Been there, Done that, and I\'ve got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida

They claimed that it was the contractors fault, the contractor claimed that they were told everything they wanted had been moved. A lot of finger pointing and no results, as usual. At least the moron that closed the course is gone. They claimed that the older items had been offered to other schools in the state, but no one wanted any of it. They also claimed that the water damaged equipment was an accident, then they had the nerve to auction it off without telling anyone about it sitting in the rain for days. You have no idea about the amount of money wasted by school systems while they cry for more money. I've seen complete school intercom systems at auction that were replaced when they were only a few years old. When I serviced that type of equipment through my commercial sound and industrial electronics business some systems were 50 years old and still in daily use.

--
Service to my country? Been there, Done that, and I\'ve got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida