Understanding Resonant Converters: Inverter and Rectifier Modeling Explained

hey everyone welcome back to power electronics i'm tim um i hope you're all having a great day

today i wanted to talk about resonating converters so last time we were talking about dabs the dual

active bridge and i thought i was just going to finish that that lecture series up with maybe an

example you know going through the motions doing all the algebra and stuff or designing a dab

uh and maybe i will there have been some requests to do other things if you guys have

specific requests uh the more specific the better that i can you know answer your questions better in video

format feel free to ask i'll try i'll do my best to answer those questions

and yeah keep asking if if i have the power to answer i will cool so today resonant converters well

what are resonant converters basically resonant converters employ resonance so in a sense like dabs they are ac power

transfer converters or they somehow convert a dc

voltage to an ac voltage they do some kind of resonance stuff and then they produce

a dc output typically you can do other stuff of course but yeah right now we're thinking of dc

to dc converters that's what we've been doing this whole time so why would you want to use a resonant

converter well typically it's it's a topological solution to the switching problem basically or

like hard switching people want to avoid hard switching for certain applications you know maybe

high voltage stuff higher voltage stuff higher power stuff whatever whatever the application may be

but where in situations where switching loss is significant

you can use resonant converters to employ soft switching if you want to use a transformer and a

dc dc converter resident converters can do that as well so

there are benefits and there are trade-offs right so maybe in resonant converters

there's a you might have higher circulating currents in the system which might increase conduction loss a little

bit so really there's you have to pick and choose where you want to use these but again resonant

converters are very common and we're going to kind of look at uh how to construct them so right now we're

really thinking about the fundamentals so how kind of develop this idea of resonant converters

so if we think back to [Music] the dab we had this idea of converting a

dc voltage into an ac voltage putting it through that transformer right

with some leakage inductance and then rectifying the output and that's pretty much what we do here

right so we have some input source which is connected to let's say an inverter right and they're

different kinds of inverters you can use just like with the dab and then that is connected to

a resonant network right and this is where the the resonance comes from in the name

and really you can think of this as a filter we analyze it in pretty much the same

way and then finally we want to convert it back to dc right because we're using this as a dc to dc

converter so we we pass it to a rectifier right just like the dab really the only

difference is this middle bit and then finally we have the load which i'll just put as a resistor rl so we

have vg over here so we kind of have like a dc world over here we have a dc world over

here and then in the middle we have these you know ac voltages and currents and maybe maybe

just in this whole world that's what we're trying to do and that's the basic structure

so the way i want to approach this at first so that you know you understand what's going on or

easy to understand what's going on i want to we have to analyze each of these

different sections so we have the inverter we have this resonant network and then

we have the rectifier so the reason i want to look at them separately is because we can choose

different blocks to put in here and choosing different blocks produces different circuits or produces different

topologies you could say specifically with the resonant network so you could put in one kind of filter

and you would classify it as one kind of topology there would be specific voltages and currents you'd

have to worry about you could put in a different filter and get you know different results and it really comes

down to what you're trying to do typically what's involved is something

called the first harmonic approximation and this is important i

mean it is an approximation and we'll see kind of where it breaks down it's going

to be a few uh this might take a few videos to get through all the nuances of these

converters but we'll see kind of where this first harmonic approximation breaks down

but basically what it does is we only consider the first harmonic of the

waveform we generate right so usually inverters generate square waves

and you can make different inverters obviously with uh different which produce different kinds

of waveforms you use multi-level converters which produce waveforms with multiple steps

but in the end there there are always these sharp corners which you try to filter out with some

kind of filter typically right so the first first harmonic approximation is just saying let's

ignore let's ignore uh all the upper harmonics and only consider

the first harmonic a filter operates or acts differently on different frequencies right so if we

were to consider each different harmonic of the waveform we'd have to calculate you know the

the uh the filter response at each of those frequencies and then add them together it would get

pretty complicated so really this is like a tool we use to understand how this

uh resonant converter works and i'll be clear when we're using the first harmonic approximation

all right so first first right so the inverter we want to generate

a model to help us understand what's going on the inverter i'm going to consider

is a full bridge con inverter so we have our input vg and we have our output voltage

which we'll just call v1 we also have an input current and we have

our output current which i guess i'll call the resonant current ir so we kind of already know what the what

kind of waveforms we can make with this inverter there's actually a few different kinds but we're just going to

consider 50 duty ratio and what we get when we do a 50 duty

ratio with this inverter is a square wave

right so we have ts tso42 you've all seen this before right so we can we can think about v1

the real v1 of te like this square wave where it goes from vg to minus vg

so if you recall the fourier series of a square wave right we end up summing

something like 4 over pi times the amplitude which is here vg times sine

of there's an n over here times sine n omega t right where n is odd right we have all the odd integers and

we get this this fourier series if we only consider the first harmonic

what that means is we consider a sine wave whose amplitude is four over pi

times this amplitude vg and whose frequency is the same as the fundamental frequency

of the square wave right so our fundamental harmonic of this square wave

is just this sine wave right where this amplitude is four over pi

v g right so it's a little bit higher than pg so this is kind of like half of the

model right we we know when we apply this uh first harmonic approximation to this

inverter on one side we have a dc voltage source and it produces on the other side

an ac voltage source with an amplitude of 4 over pi vg the other side is the current

so what does the current look like well again we're considering the first harmonic

so if there's some kind of filter over here then if we apply some sinusoidal voltage

right this 4 over pi v g sine omega t then we're going to get some

scaled version of that that's phase shifted right so that an example current might be something like

i don't know let's just phase shift it some amount with some amplitude variation right

maybe the current looks something like this so that there's some phase shift between

v and i and the amplitudes are slightly different so this is amplitude ir right so that's what the output current

is going to look like but we want to know what the input current is how does the

input current relate to this sinusoidal current so what we're going to do

is simply look at the average of this uh ir well not the average actually

the average of the rectified current right relative to uh the input

right so what do we know about ac power transfer well one

when current and voltage are in phase that means that we're transferring real

power right and when they're out of phase it's all reactive power so this kind of

makes sense in the ac world right and really the thing that encapsulates this

is power factor right and the power factor for single harmonic single frequency systems

is simply the cosine of the angle between the current and voltage somehow we want this

term to appear in the model for our input current again the input current in is going to

be a rectified version of ir right let's draw one half period of the current

so again this current is going to depend on the phase shift so maybe when they're completely aligned

this input current looks something like this right and then maybe when there's some phase

shift between them maybe the current looks something like this right when they say the 99 degree

90 degrees phase shift between the voltage and the current on this side we can describe this phase

shift phase shifted current pretty easily just using

the sine equation right even over one half period right so i n of t the time varying input current

maybe i should write this as a little lowercase ion of t really this is just equal to the peak

which in this case happens to be related to the peak of the ir right

so i'll just put ir here times sine and i'm kind of using sine arbitrarily but it

works out sine omega t plus phi right and this accounts for the

phase shift between the voltage and the current basically this we're describing this phase shift

so knowing this this equation we can pretty easily find the average input current right that's

what we're trying to do we're trying to find the average input current so i n of t the capital i n of t is really

just the average of the lowercase ion of t over in this case one half of the

switching cycle right if we can calculate this we know what the average input current is

so let's do that all right this is a pretty easy equation so we're trying to find

average oh yeah right for our model so what does it look like

well i n is going to be equal to in this case 2 over ts we're averaging over half a switching period

the integral from 0 to ts over 2 of this thing right the time varying version right so ir

sine in this case omega is 2 pi over ts times t plus the phase shift phi

dt okay so i mean the real way that you would want to do this integration

is you know you do some substitution find the real answer i'm just going to give you the real answer because this

isn't a calculus course and what you end up getting is 2 over pi times ir

right the amplitude of the resonant current times cos five right so this is like the power

factor like what i was saying in the end this comes out in the wash you end up getting this power factor in there so

when the current and voltage are completely out of phase 90 degrees out of phase

then the average input current is zero and you can kind of see that here right so when the current

is 90 degrees out of phase in one half of a switching cycle is going to be equally positive as it is

negative which means the average current is zero and then the only time you get maximum

current is when no part of this current goes below zero right and that is when voltage and

current are completely aligned so this is what our average input current is so we kind of have all the

pieces of information we need to construct the model for the in the uh the inverter right our

model for resonant converter so what does the model look like well on the one side we know what our

input is it's just a dc voltage source and we know what the the the input current draw is

the input current draw is exactly this right so we have i'll just write it explicitly we have vg

we have some input current i n we know what this is equal to it's equal to this value

we represent this with a current source right we suck this current out of the input

whose value is equal to 2 over pi i r cos phi great so that's the input side right on

the or the dc side right on the ac side what do we have well

the inverter effectively creates a fundamental frequency or fundamental harmonic

with amplitude 4 over pi vg right so what we get on the ac side is an ac voltage source right

whose amplitude is 4 over pi vg right and now this is this hooks up to the remaining ac circuit

and we can say this is v1 so this is for a full bridge right so for four controlled switches

you could do this for a half bridge and basically the difference would be this would be half as big right

right that that is basically the difference the current would i believe still be the same so

this is our model for the inverter right so this isn't this is ac over here so now

let's look at the rectifier there are the to the two flavors we're going to look at

are current fed and voltage fed or current driven and voltage driven voltage drift and these have

actually two different models slightly different models we use these different rectifiers

so that we can match up a something that is providing a voltage with something that draws a current

and something that provides a current with something that provides a voltage right so it's just

kind of to match up what the resonant filter is doing so what do these two things look like

well i'm going to include a transformer just for completeness they don't have to have transformers in the rectifier part

but it just will be convenient to include it here so on the current driven rectifier

what we have is a transformer and let's say it has turns ratio of 1 to n

and this goes to a full diode bridge rectifier again you could you could have a full wave

rectifier that's totally cool you could probably even do a single rectifier if you really

wanted to but this is connected to just a cap

right and this rectifier again produces a dc output voltage

right but if you think about it this is a constant voltage so if you're connecting if there's a voltage over

here this is driven by a voltage you're kind of what you end up doing is you connect a voltage source

right say this is v2 a voltage source to another voltage source which produces you know current spikes and

stuff typically not what you want so right so this is something that we want to

use with currents right if we're we want to drive this with currents because we have a voltage source over

here so we call this a current driven rectifier the voltage-driven rectifier

was maybe you can guess what it would look like it's pretty much the same

right we still have this transformer again with turns ratio 1 to n let's say

and we also have a full bridge diode rectifier and again this could be

a different architecture but the principle is similar and instead of having just a capacitor

we kind of have a an lc filter right and the difference here is that this inductor is going to act

as a something like a constant current source right so we can see maybe this is i out so again

if we were to drive just a current through this well we're kind of connecting a current

source something that's providing a current to another current source typically what we don't want it would

cause cause voltage spikes or something similar right so we want to drive this rectifier

with a voltage right this one we want this to be a voltage driven rectifier

cool and again we have a v out over here to develop this ac to dc model of this inverter what we have to do is look at

the waveforms of the voltages and currents what we're trying to do is relate

the ac currents and voltages to the dc currents and voltages so how do we do this well we're going to

inspect i out of t right this time varying current that occurs just after the

rectifier right we know that the average of i out of t is going to be equal to

the average output current right the current that's flowing through the load the load resistance right so you could

say you could say that i out is actually v out over r load

right and we also know that i out of t is simply the rectified version of i r of t so we kind of already have a

relation between these two currents so let's just draw that out so it's clear

right and don't forget we have this transformer here so i'm just going to do a full waveform

here right full ts so tst is over 2 and on one side we have i r of t right which

for here looks just like a sine wave with zero phase why because on the dc side we only have a resistor

our load is is we can say resistive what i mean by that is that

it absorbs real power it only absorbs real power there's no imaginary power that is that it's taking

i mean you could for some applications but really most loads do real work right what that means

is that on the ac side because only real power is being provided

the current and voltage are going to be aligned right and that's kind of true even with

just this rectifier so obviously non-ideologies with the rectifier are going to change things but for the most

part ir and v2 are going to be aligned because we're providing real power to

the output so because of that and because we're only looking at one

section of the converter i'm going to choose to align ir with zero right i can kind of do that

freely just because we're looking at one small section right this just makes it easier to do integration

eventually so we have i r of t which looks like a sine wave right

with peak let's say ir i out of t is the rectified version of this right

so when i r is positive i out is positive when i r is negative i out is still positive

note that we have this transformer here right so the peak of i out

is going to be scaled compared to ir right it's going to be scaled by n right so

whereas here we have ir i out is going to have peak ir divided by n

all right because one times i r is equal to n times i out basically

so i out the peak must be ir over n cool so that's the current and then as we were saying with the voltage

because we're providing real power to the output the current and voltage are going to be aligned here

right so let's just consider what happens then so again this is a current driven rectifier which means the

current decides the state of the of the system basically

right so when the current is positive when ir is positive it's going to forward bias

this diode and this diode right which means that we're going to have

positive v out over here right when say 1 and 4 are on we're going to have a positive yield

over here which is going to then get reflected to the to the primary side right

which means v2 for this first half of the switching cycle is going to be positive

it's going to be related to v out it's not going to be exactly v out it is actually going to be

v out over n right because of the transformer and then when the current is negative

well as you would expect the voltage also goes negative right when the current is

negative we can say that diodes two and three two and three are turned on

which means we're going to have negative v out over here which is going to get reflected to

the primary side right and again it's not going to be v out it's going to be v out over n

minus v over n cool so this is a square wave we have not yet applied

for uh first harmonic approximation when we do what we get is a

voltage a sine with peak related to the to the peak of the square wave

but scaled by a factor of four over pi right so the peak of this sign is four over n pi

v out cool so with these two waveforms we have related

the ac voltages and currents to the dc voltages and currents basically right we know how they relate

now we have this ac waveform for i out or this time varying waveform for i

we want to know what the dc of io is right so i out the average is equal to

[Music] the average of i out of t over one half of a switching cycle right

because this first half is the same as this second half so let's do this integration let's do

this averaging right now so what do we have this is equal to again over half a switching cycle two

over t s we integrate from zero to t s over two and then we have a sine wave basically

right in this first half with amplitude ir over n sine omega t again this is

2 pi over t s times dt right so we we can do the the into the integration two over t

s i r over n and then we we get some scaling factor we're gonna

divide by omega again this is related to ts and then we're gonna get cos omega t

evaluated from 0 to ts over 2. what this ends up being is 2 over n pi

times ir right so again note here that there is no phase shift involved

why is that it's because basically this is uncontrolled if we're using diodes we don't we can't control the phase

relationship between the voltage and current when the current is positive the voltage

is positive right so basically they're always going to be aligned so the average i out is equal to 2 over

n pi times ir so this kind of gives us a piece of our model what is the other piece well

the other piece is kind of right here right so we know that v2 the peak of v2

amplitude of v2 is simply equal to 4 over n pi v out

cool cool we're we are you know close to what we want now what does this say it tells us

what the the current and voltage actually look like right now we can write an equation for i r of t

i r of t we know what the amplitude is it's n pi over 2 times i out the average i o times sine omega t

right and this also tells us what the waveform right what this waveform of v2 looks like

so v2 of t is equal to this v2 which is 4 over n pi vo this is the amplitude times sine of omega t

so we're really close to completing our model basically right now i want you to note something

here these are always aligned right the voltage and current on the ac side are

always aligned which means that effectively they're feeding something that looks like a resistor

what is that resistance well it's some kind of equivalent resistance what does that look like

right where is this resistance well if the voltage and current are aligned then it means that looking in here

it looks like a resistor right when we look into this primary side of the transformer it

basically looks like we're looking at a resistor what is this resistance what is our equivalent

well we can find it pretty easily right we can find the equivalent resistance by dividing voltage by current right so

our equivalent is simply equal to v2 of t over ir of t and we have equations for both of these things right

v2 of t is 4 over n pi v out sine omega t right ir is equal to n

pi over 2 i out sine omega t and really because these are completely aligned we can just

cancel these out right which means that the equivalent resistance if we do some stuff

do some simple algebra we get 8 over n squared pi squared times v out over i and remember

v out over i out is our load resistance right which means this is equal to

eight over n pi squared n squared pi squared times our load so on the ac side of this model this

inverter or this rectifier sorry what we end up with is a scaled version of our load resistance

right scaled by this factor 8 over n squared pi squared right so what does this mean our model

is saying that the ac side looks like a resistor and the dc side just looks like a

current source connected to our cap and load right so what does our model look like

basically we have some equivalent resistance right over here which is equal to again

8 over n squared pi squared times our load all right we have v2 we have

ir of t you can say this is v2 of t and then on the dc side we rectify this ac current right this is

a current driven rectifier so we expect some ac some dc current on the dc side which

feeds basically our cap which feeds our load and you can arguably eliminate this right so this

current is simply equal to two over n pi i r right and this is v out all right we have r load

so this is this is our model right this it encapsulates the resistance and the effect of the rectifier

so to be even more clear this bit is from the rectifier all right the 8 over pi squared if

there's no if there's no transformer you'd still get this 8 over pi squared factor

this bit accounts for our transformer now what does the model look like for

the voltage driven rectifier well the analysis is very very similar right so what we're going to do is

pretty much the same thing again we have a we have a resistance as our load which

means that the voltage and current are going to be aligned but in this case instead of it being

driven with the current it's driven with a voltage right so what we're going to do is we're

going to look at voltage and current waveforms to relate

the ac side to the dc side so let's do this let's see what the ac current and voltage looks looks like so

again this is a voltage driven rectifier and you can tell that because we have this

current at the output right so maybe what i want to draw here is that we have some v out of t

over here right so the voltage is going to decide if this is positive or negative

and then this filter is going to implies that the average of v out is capital v out right so in other

words v out is actually d the average revered over t over

one half of the switching cycle cool so back to the waveforms i'll draw again the full switching cycle so you see that

we're actually thinking about a rectifier so we have ts we have ts over two

right so again v2 of t all right i guess it's an orange v2 of t we're assuming is an ac waveform

right and i'm going to choose that this aligned with zero just so the integration is easier to do

so this voltage driven rectifier sees some input voltage that is v2 some peak v2

right and then so this is v2 of t v out of t is a rectified version of

this right so v out of t is simply whenever it's positive it's positive and whenever it's negative it's

positive right and it's going to be scaled by this transformer so we scale

this input voltage and we make it always positive and the peak of this output voltage v

out of t is actually n times v2 of t cool so this is the voltage and we can

do something similar with the current so again you can imagine that this output

forces there to be a constant current we have a constant current here so when v out is positive

one and four will be turned on basically and that means that i out is going to flow in this direction

right again i'll just i'll make make sure the dots are here

so if it's flowing out of this dots flowing into this dot whenever i out is positive a constant dc

current ir is going to be is going to match that right and then similarly when vote is negative

diodes 2 and three will be on and then the current will reverse so we're gonna have a square wave current right

that whose polarity is determined by the voltage and again because our load is resistive

we're it's all real power there is no reactive power which means that the voltage and current

are going to be aligned so let's draw that so again when the current is positive ir

of t is going to be equal to i out positive io

when v2 is negative it's going to be equal to negative i out right so this is i out

this is minus iot and obviously there is going to be uh a factor of n right so there's n

and io correct because of the transformer now we have not yet applied first harmonic approximation if we do

that again the sine the first harmonic is real is related to the amplitude of

the square wave by a factor of four over pi right so the peak of this thing

is going to be n times 4 over pi i out so that is the peak of ir of t which means we can now

we have an equation for this right we can figure out what the time dependent current is

with this amplitude cool however first of all what i want to do is look at the average right we already

had an equation stating what the average was right so the average v out

is actually equal to the time average of v out of t over one half of the switching cycle right

and we can do this integration right so we have two over t s from zero to ts over 2 times the peak

n times v2 times sine omega t so it's very similar to the previous example right

and again if we if we do this if we do this integration what we end up getting is n times 2 over pi

times v2 that is what the average output voltage is and again we can figure out using this

we know what the expression for v2 of t is right this is the amplitude of v2 of t

which means which means that we can write the time varying expression for v2 of t

in the following way n over 2 pi over 2 n right v out times sine omega t

right and again no phases here and again similarly over here i r of t we know what the amplitude is right it's

n four over pi i out so we have n four over pi i out and again sinusoid with no phase

difference sine of omega t and again what does this mean looking into

this input it looks resistive what is that resistance what is that resistance our equivalent

well it's equal to uh the ratio of v2 to i to ir right that is the definition of

resistance right the ratio of current to voltage basically we have expressions so we have pi over

2n times v out sine omega t over n four over pi

i out sine of omega t simplifying this cancels with this they're aligned right because it's a

it's a uncontrolled rectifier and what you end up getting is pi squared over eight

n squared right so again what do we have also v over i out and we can simplify this even further pi

squared over eight n squared our load is equal to the equivalent resistance

right so these two these two rectifiers do two different things right

so now because with this voltage driven rectifier we actually have a factor of pi squared over eight as opposed to 8

over pi squared and then we still have this factor of n squared which is related to the terms

ratio that isn't going to change right because we expect the turns ratio 1 to n to

divide the load resistance by n squared all right so just looking at the difference between these two things so

so for the current driven rectifier we have the equivalent resistance is 8 over n squared pi squared times our

load and for the voltage driven rectifier we have the equivalent resistance is pi

squared over n squared i'll say eight our load right so just this flipping of pi

squared over eight is the difference between the voltage and current driven rectifier

so i believe this has already taken quite a bit of time this is where i wanted to get and

actually we can we can you know finish this off the one final step is to make the model

for the voltage driven rectifier right so on the one side the ac side actually sees an equivalent

resistance right it sees this thing pi squared over n squared eight our load

this is v2 we have ir going in here and then on the dc side because this is a voltage driven rectifier we expect a

voltage over here a dc voltage right with an amplitude of n2 over pi v2 which we know is actually v

out and this you know feeds to our dc voltage right over here cool so that this is basically what the

the model is and even though i haven't completed it as an example maybe what i can do let's

just zoom out a bit yeah so the they look pretty similar the difference is for the current driven

rectifier on the dc side we see a current for the voltage driven we see a voltage what we've done

is we've been able to insert or now we can insert a a block or an a model into these blocks right we

still haven't done this middle block yet but for these two outer ones the inverter rectifier we can we can

stick stuff in depending on what we have so we have our load right so for the for the inverter

basically what we have is you know it looks like a current source followed by an ac

voltage source then we have the resonant tank the resonant filter whatever you want to

say right and then in here we're going to stick something that looks like an equivalent resistance and then either

a voltage or current source depending on what type of rectifier you're using

over here right and this is basically what the model is going to look like so we're going to have

some kind of voltage source over here some kind of resonant tank and their equivalent resistance and with that

when we figure out what the model is for the resonant tank we can solve the conversion ratio for a

resonant converter so that's it for this because this video has gone on for

quite a while i believe and the next video we're going to investigate this middle

this middle section this is what we're going to look at in the next video how do we

model this and funnily enough we already know right we know from

other courses right it's this it's the same as any other kind of filter we're going to

use bode plots and yeah that's going to be what determines our game awesome

i'll talk to you guys later keep asking questions i'll do my best to answer them and i'll see you later