Thevenin's Theorem, Output Impedance, and Input Impedance

today I'm going to explain thevenin's theorem and the interaction between output impedance and input impedance as I often do I'm going to express thevenin's theorem a little differently from what you might see in textbooks and that is thevenin's theorem states that any circuit that produces an output will act exactly like a circuit that has a single voltage source with some impedance in series with that voltage source since we're talking about DC circuits we're going to reduce that voltage source to just a battery with a resistor in series with it so thevenin's theorem says that any circuit no matter how complex that produces an output will act like a battery with a resistor in series with it and so how do we determine what those are well let's take a look at how we can do that first we start by putting a voltmeter across the output this will give us what's called the open circuit voltage remember that the voltmeter has a very high impedance so for all intents and purposes it has no current flowing through it so we have an open circuit here so we're going to measure a voltage let's say we measure 5 volts so what does that tell us well we know there's no current flowing through this resistor because it's an open circuit and to get a voltage differential across that resistor we must have resistance and current there's no current so there's no difference in the voltage from here to here so whatever voltage we have here is the same voltage we have over here so by measuring the open circuit voltage we've just determined that this is a 5 volt battery now how do we figure out what the resistance is well now we replace this voltmeter with a current meter and measure the current so let's say I measure that and I have 100 milliamps so what does that tell us well we use Ohm's law divide the 100 milliamps into the 5 volts and that gives us 50 ohms for the resistor so that's the basic way we find out what the thevenin equivalent circuit is now this is a circuit let's say it's in a black box or something we don't really know what is there but it acts like a battery with a single resistor in series with it let's take a circuit and see how this works I'm going to make a fairly simple circuit up here I'm going to write it up here so I have room to do the thevenin equivalent below so here's our battery make it a simple circuit so we don't have to get to thinking too hard about it let's make this a 20 volt battery make this a 1K resistor and another 1K resistor and we'll put a couple of terminals here which will constitute the output of that circuit so what we're going to do now is measure the open circuit voltage by putting a voltmeter here now that's the open circuit voltage this is a closed circuit here but over here we have an open circuit so there's no current flowing through the voltmeter let's draw the of an equivalent circuit here and find out what we find out by doing this so we measure that voltage what do we get well we have 20 volts here and we have two equal resistors so we know the voltage across these resistors will be equal and they must add up because it's a series circuit here they must add up to this 20 volts so that's going to be 10 volts across this resistor and 10 volts across that resistor so we measure 10 volts so our voltmeter measured 10 volts and as we explained before that must be the battery voltage that's a 10 volt battery how's that work well stay with me we'll see that it does act like this how do we find out what that resistor is well now we take the current so we remove the voltmeter replace it with a current meter and measure how much current we have how much is that going to be well the current meter is essentially a short circuit so basically it eliminates that resistor because we have a 1K resistor in parallel with a zero Ohm resistor and we know that we have two resistors in parallel or any number of resistors in parallel the total resistance will be lower than the lowest resistor well the lowest resistor is zero ohms and we can't get any lower than that so together they are a total of zero ohms so basically that's no longer part of the circuit so now we measure our current and we find that with 20 volts and 1K we're going to get a current of 20 milliamps so now our current meter 20 milliamps so what does that tell us well we have a simple series circuit 10 volts 20 milliamps and some resistance to find that resistance we divide our current into our voltage if you know your voltage you divide into it so 20 milliamps goes into 10 volts that gives us a resistance of 500 ohms so the thevenin equivalent Circuit of this circuit equals that circuit so this circuit looking at the output here will act exactly like that circuit well if that's true then if we put some kind of a load out here we should get the same results here as we do here so let's try that let's put a 1K resistor out here as a load and find out what we get so we'll do the same thing here and if this circuit acts like that circuit we should get the same voltage and current of course through the load here because that's another 1K resistor let me pull out my calculator because I certainly won't be able to do these calculations in my head so let's find out what we have here okay so let's find out what the voltage across here is that's going to be we have two 1K resistors together they are how much resistance equal resistors in parallel will give you half the resistance so the total resistance of two 1K resistors in parallel will be half of either one or five hundred ohms we put that in series with this so we have 1K plus 500 ohms that gives us 1500 ohms of resistance total so that's going to be divided into the 20 volts 20. divided by 1500 gives us a total of 13.3 milliamps through that 1K resistor so how much voltage are we losing there so that's going to be 13 .3 milliamps times 1000 and that gives us 13.3 volts across here foreign leaving us take 20 volts and subtract 13.3 volts and we're going to get left over here six point six volts and of course 6.66 volts with 1K is going to give us a current let's try that again 6.66 volts divided by 1000 equals going to give us 6.66 milliamps Okay so we put this load out here let's break this right here just to remind us this is the load we're putting on the circuits we put a 1K load across that circuit we got 6.66 volts across it with 6.66 milliamps okay now if this circuit really does act exactly like that circuit we should also get 6.66 volts across this resistor by the way I did this by Ohm's law but it could have also done it as a voltage divider here we have 500 ohms and 1000 ohms we have two times the resistance we have here so two-thirds of our voltage would be here and one third of our voltage could be there we could just take our 20 volts and divide it by three and that gives us 6.66 volts and we do that Division and that gives us the voltage across the lower resistance this is 500 ohms we'd have 66 6.66 volts so I did that by Ohm's law but don't forget these things can be a lot easier than they look once again 500 ohms 1000 ohms twice the resistance as here so twice the voltage so 20 volts divide that by three one third goes here two-thirds goes there and could have figured that out a little easier but I did the Ohm's law just because I could now we're going to do the Ohm's law and find out if this circuit gives us the same voltage here as that circuit well what do we have we have 10 volts and we have what 1500 ohms divide that by and we get lo and behold 6.66 volts so indeed this circuit acted exactly like that circuit if we look at it as there's that circuit and there's that circuit so we took this complex circuit slightly more complex and reduced it down to this thevenin equivalent circuit and no matter how complex this is it would have worked I could have used a more complex circuit but that may have caused us to lose focus on what we were doing so we did see that this circuit did act like that circuit so I have a 20 volt battery two 1K resistors taking the output across the second one k resistor acts exactly like a circuit that has a 10 volt battery and a 500 ohm resistor in series so what does that do for us well understanding thevenin's theorem is a big step to understanding what output impedance is because the thevenin equivalent resistance or in the case of an AC circuit the thevenin equivalent impedance is the output impedance of that circuit so let's erase this and draw a circuit we know that any circuit well let's do this with an AC circuit just because we can so there's an ac voltage source and some kind of impedance so some AC circuit that produces an output will act exactly like a circuit that has a single ac voltage source and a single impedance in series with it and this thevenin equivalent impedance is the output impedance so something you'll hear a lot about is output and input impedance we're going to talk about that in just a moment but that's what your output impedance is is your thevenin equivalent impedance we're going to go back to working with a battery and resistor because we're doing DC and we don't want to get into AC for quite a while so let's take a look at a typical circuit reduced to its definition equivalent so we have a battery and a resistor now this is going to be connecting to some other circuit even if it's connecting to some kind of device that's not a circuit for example if this is a radio transmitter it's going to attach to an antenna well that's another circuit so we always have one circuit going to another does the antenna go to another circuit yes it couples to The luminiferous Ether to create radio waves so everything is coupling to somebody else and so that other circuit does not produce an output it's an input so it's passive and it takes the energy from the previous circuit and when we do that we have that represented by a single resistor so here's our output impedance just a resistor here not the complex impedance we have in AC circuits and over here we have the input impedance so every circuit will act like that so we have a circuit and then the input impedance of course is the load that we put across there so how do these interact with each other let's get the Clutter away here for just a minute so I can write on the board and a couple of things to remember a little Mantra if you will is that a high impedance like a high resistance is associated with a low current and a low impedance I'm using the Z for impedance since this is a a DC circuit we could use ohms in Z represents a complex impedance you have in AC circuits but this could be ohms I'm just using Z because we're talking about impedance I'm going to use the symbol for impedance but that could be an Omega if we were talking about DC circuits and simple resistors so high impedance is associated with low current low impedance is associated with what guess what high current so basically if you have a high resistance you're going to have a low current if you have a low resistance you're going to have a high current so low impedance High current High impedance low current now let's see how that works when they interact with each other I'm going to leave that up there for now so let's say we have two circuits coming together let's just show a little connection here 's our output impedance and here's our input impedance and let's say we have a low output impedance and a low input impedance so let's uh let's draw that this way low impedance low impedance what's that saying the load is saying or the input impedance is saying I'm a low impedance I need a lot of current low impedance High current this is saying I'm a low impedance too I can give you a lot of current so they work well together now let's see what happens if we have a high impedance this High impedance says I cannot give you much current I'm a high impedance I can only give a small amount of current this is a high impedance it says I'm high impedance too I don't need much current I can't give much current I don't need much current they work together now let's see what happens if we have a low impedance here and a high impedance here now we have a mismatch let's see if that's a problem this says I'm a high impedance I don't need much current this says I'm a low impedance I can give you all the current you want and more no problem so a low impedance going to a high impedance is usually no problem now it can be sometimes sometimes um well what happens is when we have a low impedance going to a high impedance we pretty much maintain the voltage because since we have a low resistance here our low impedance we don't have much of a voltage drop so the voltage here is pretty close to the voltage over here so we maintain that voltage so if we want the output voltage to remain high well we have a high impedance input to go to and if we have a low impedance output that just helps that voltage be maintained the only problem is sometimes you might want that voltage to be lower one particular case is with audio systems where you have microphones and if you have a low impedance microphone going into a high impedance input you're going to develop too much voltage there and that causes Distortion you get the AC wave tries to get too big and it gets chopped off and you get what's called clipping Distortion so a low impedance microphone to a high impedance amplifier input is not good but let's see what happens if we take a high impedance output to a low impedance input now this says I need a lot of current this says sorry boss I'm a high impedance I can't give you much current so what's going to happen we're going to get a voltage sag here and we're not going to develop much voltage across here let's take a look at a example of how that might be let's say this is 20 volts and let's say this is oh how about um 9k and this is 1K so a high impedance going to a low impedance what's going to happen here well I'm going to lose about 18 volts going across here having only about 2 volts left over we can just look at this as a voltage divider I have nine tenths of my resistance here one tenth of my resistance there so I get one tenth of my voltage nine tenths of my voltage so my 20 volts I lose 9 10 of it that's 18 volts leaving one tenth of it 2 volts left over so if I go from a high impedance to a low impedance that's going to be problematic I'm not going to get much voltage left over here there might be some rare cases where that's desirable but usually it's a problem for example if I have a PA system and I take a high impedance microphone to a low impedance input I'm not going to get the amplification I expect because the input voltage is going to be too low so just like if I take a low impedance microphone to a high impedance input I get too much voltage the signal tries to go too high and too low it gets clipped off I get Distortion a low impedance microphone or should I say a high impedance microphone to a low impedance input I don't get a enough of a signal to amplify so that's usually a problem so let's look at that again just a quick review low impedance to low impedance that's good low impedance delivers a high current this needs a high current they work together High impedance to high impedance is good this doesn't need much current this can't deliver much current so they work together a low impedance to a high impedance is usually good because I maintain my voltage I don't have much voltage drop across it but if I have a high output impedance to a low input impedance I'm going to get a major voltage sag here and I'm not going to get my voltage transferred over so that's the problem so low to low high to high low to high is almost always good but high output impedance low input impedance is usually problematic now let's say I have a situation where I want to get the maximum current across my load this is uh desirable in a couple of places I can think of off the top of my head for one radio transmitters you want to develop the most power you can across the antenna to transfer as much energy into The luminiferous Ether and create the strongest radio waves you can and when it comes to audio amplifiers you want to deliver the most power to the speaker you can because you want that power turned into sound waves and so in both cases you want to develop the maximum power on the output in those cases you want the impedances to match so I want for example let's say this is a 50 ohm impedance I want this to be a 50 ohm impedance so this is a common situation in radio because a lot of antennas just by the nature of the way they are built happen to have a 50 ohm impedance so then we have to design the transmitter to also have a 50 ohm impedance to get the maximum power across that antenna let's see how this works so we have 20 volts there 50 ohms 50 ohms I'll use my trusty calculator to get these numbers let's see how much power we're going to develop across this resistor so we have 50 ohms that's 100 ohms and we're going to divide that into our 20 volts that's going to give us 200 milliamps so what's the power 200 milliamps I squared times R that's going to be that's going to give us 2 Watts so at 50 ohms I get 2 watts of power let's see what happens if I drop this down to let's say oh let's make it 10 ohms let's see what happens to our power so now we have 50 we have 60 ohms goes into 20 volts that gives us a current of 330 milliamps so what's our power going to be that's going to be I squared r so 330 . that's going to give us 1.1 Watt so we reduce this to 10 ohms so at 10 ohms we only get 1.1 Watt so let's go back up to 50 ohms just momentarily so at 50 ohms we had 2 Watts but we reduced this to 10 ohms we went down to 1.1 watt so if we increase the resistance we should get more power right let's see what happens let's put that up to um nice even number how about 75 ohms so now we're above 50 ohms so now we have 50 and 75 that's going to be 120 ohms 20 divided by 120 gives us a current of of 166 milliamps we square that multiply it by the 75 ohms so we do that and we come out with at 75 ohms we have a higher power no we have 1.9 Watts so notice the power dropped again so to get the maximum power across our load we want the load impedance to match our output impedance so for a quick review the evidence theorem says that a circuit no matter how complex will act like a single voltage source with a single impedance in series with that and we've proved that by a little bit of mathematic on the board here and the thevenin equivalent impedance is the output impedance of the circuit and so we have an output impedance and an input impedance and those two interact with each other High impedance is associated with low current low impedance is associated with high current and so if we have a low impedance and a low impedance that's good we get a good match there if we have a high impedance to a high impedance that's a good match a low impedance to a high impedance is usually a good match uh some exceptions that I talked about earlier but when we have a low impedance here and a high impedance here we preserve our voltage across there which is usually desirable but if we have a high output impedance and a low input impedance we have a voltage drop and we have a significant loss in the voltage so that's usually undesirable so the one thing that's undesirable is to have a high output impedance and a low input impedance they don't work well together and finally if we want to maximize the power develop the most power across our load we want the load impedance to match the output impedances as I showed here we get our maximum power across our input impedance if it matches the the output impedance of the circuit driving it if you found this video useful and informative please give me a thumbs up down below it really helps the channel and subscribe because that not only informs you when I put new videos up but it really helps the 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Channel: Vocademy - Electronics Technology

Views: 9,988

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Id: 0ulZjUbdQJE

Length: 25min 39sec (1539 seconds)

Published: Thu May 18 2023