For the power flow across an inductor, since you have taken V1 and V2 to be the amplitudes, there will be a factor of 2 in the denominator of the average power calculation. Pavg= (V1*V2*sin(delta))/(2*w*L) at 11:25
hi Tim, thanks for the video. However, i think there is a minor mistake in your AC power transfer calculations. I remember that the normal representation in power systems usually takes V
First of all thanks for the detailed information. Sorry for next silly question, when you make reference to Iout you mean output current from the secondary winding, not the load current. thanks 😁
1. Transistors are very different at right and left (high voltage vs high current). Different timings, a complex controller (if exists) etc. 2. A classical sync rect with a dual secondary do the same job without additional heat losses and money 3. Why people still teach "a dual active bridge" if no one uses such architecture in a real life?
Thanks for the detailed explanation, it helped me to understand about dual active bridge in greater detail. So for real life application, there will be a dead time between Q1,Q4 --> Q2,Q3? Also, for ZVS, Q1 and Q2 will be on before transitioning to Q2,Q3?
Dead time is always needed in hard switching applications. Q1 and Q2 on at the same time would result in shoot through. To achieve ZVS we simply allow body diode conduction before turning the switch on. In other words, the switches would be off and we allow natural commutation to determine what conducts.
Would the final power equation be different if the input and output were chargeable batteries? (when one of them functions as a load, the other acts as a power source, allowing bidirectional power flow)
I apologize for the delayed response. I didn't get a notification for this comment. The equations do not change! In fact, a colleague of mine has done this to achieve battery balancing between cells (and also power a load) for a paper. ieeexplore.ieee.org/document/7845680, patents.google.com/patent/US8779700B1/en (this has some pictures) Another paper does exactly what you're describing: mplab.ee.columbia.edu/sites/default/files/content/Publications/BB_APM_APEC2018.pdf In this beginning of this discussion I (implicitly) assumed small ripple approximation on the output voltage, effectively treating it as a constant voltage source. Sounds like a battery to me :) Typically this transfer of power between batteries (or battery cells) would be to ensure they have a similar state of charge so that they can be better utilized over the full power cycle of the battery pack.
The leakage inductance of the transformer is a representation of flux that doesn't link both the windings, it is that part of flux that completes its path through the air, then how can we say this leakage inductance is responsible for power transfer?
Very good point to bring up. It seems we're talking about two separate phenomena here. The transformer couples the primary side to the secondary side and as a result transfers power from the input to the output. However, as demonstrated in the motivating example, there doesn't need to be a transformer at all and only an element which defines some current between the input and output is needed. Continuing with your reasoning, we can simply think of the leakage inductance as an air core inductor we have placed in series with the transformer. It doesn't couple the primary side to the secondary side, but it does relate voltages V1, V2 and the current through the governing equation of an inductor. The inductance, frequency, and phase shift are what define the peak current in the system, which defines how much power is transferred. So yes, the transformer does transfer power from the primary to secondary, but this flux linkage does not regulate current. If there was no series element (in this case leakage inductance) to define the current, then power transfer would not be regulated in a controllable way as it is here.
Please make a video on ZVS operation between the snubber capacitance across the switches and leakage inductance in transformer of Dual Active Bridge. I was looking for design criteria for calculating the output filter capacitor, but couldn't found much as DAB does not have an inductive filter. How to choose its value.
For some reason I did not receive a notification for this comment. I apologize for the late reply. I'm currently uploading a video which briefly discusses ZVS operation in the DAB. It's not equation heavy, but I have provided a reference which does look more deeply into the equations. As for the output capacitor, the process for selecting capacitance is the same as any other converter. First find the difference between the DC load current and the current output from the secondary bridge. This is the "AC" current absorbed by the output capacitor. We can then calculate the positive charge (the positive area under the curve) from this current. This charge takes the output voltage from its minimum to its maximum, so it represents a change of 2deltaVo. Then C = deltaQ/2deltaVo. For this DAB, this is a little bit more complicated and the equation for the general case is nasty. After a little bit of arithmetic, it seems Io < I1/n. This means that calculating the positive charge we send to the output requires summing three different areas (which can all be found geometrically) . deltaQ = (I1/n-Io)(Ts/2)(1-D) + (1/2)(Ts/2)(1-D)(I2-I1)/2 + (1/2)(tx)(I2/n-Io) I1 and I2 are as we've defined them and D = phi/(Ts/2). tx is the time needed for the reflected inductor current to fall from I2/n to Io. I highly recommend you try this out for yourself. What I've calculated (and not checked) is: deltaQ = (1-D)(Vg(1-2 D' D) + Vo/n)/(8 f^2 L n) + (1-D)^2(Vg - Vo/n)/(8 f^2 L n) + (D' D)(Vg(1 - 2 D' D) - Vo/n(1 - 2D)) /(16 f^2 L n)* (Vg)/(n (Vg + Vo/n)) Hopefully you can actually interpret this, I didn't want to use too many *. Finally, C is this deltaQ/2deltaVo. Just a reminder that this has not been check over too deeply. I may go over this in a tutorial video. Good luck.
I have not tried this method, but it seems like a valid method. I imagine you have an infinite sum containing terms with sin(n*w+phi). Considering the direction calculation presented in the video is well established, I'm not sure if you could get further insight into the operation of the DAB using Fourier analysis.
@@timmcrae3831 Thank you very much for your reply! I can understand the classical method of power calculation, but the first part of your video inspired me, By doing Fourier analysis, I got an infinite sum: (8*V1*V2)/((i*pi)^2*omega*L) * Sin(i*phi), where i is odd number. But the plotted figure is different from the classical calculation derived in the video. I'm really curious where is the mistake.
Thank you professor for the lecture. I have a query regarding circulating currents. What exactly are the circulating currents in single-phase shift modulation and eliminating in dual-phase shift modulations? Could you please clarify this? Thank you.
High circulating currents refers to the fact that in some situations, the peak current in the leakage inductance may be very high, but the load (DC) current may be low. Dual phase modulation effectively allows the peak current to be lower for the same load current, reducing circulating current and thus conduction losses. If I'm not mistaken, dual phase modulation (for a DAB) can be thought of a single phase shift modulation, but you also modulate the duty ratio of the active bridges. So instead of a a 50% duty ratio voltage applied to the inductance, you may apply +V, 0 and -V.
HELLO PROFESSOR. I took The Dual Active Bridge as a design homework. Is there a book, article or source where I can find even the most detailed information about this subject, thank you.
As far as textbooks go, I'm not familiar with any that focus on DABs. Most information I've read has been from theses and peer reviewed publications. A colleague from my lab looked at DABs for their thesis and it may serve as a useful reference to you or at least direct you to other references if needed. tspace.library.utoronto.ca/handle/1807/70201
At time 6.55; you have mentioned V2(t)=V2cos(wt - theta); & below that (V2/wL) sin(wt - theta). Why this difference in values of V2? Why cos(wt - theta) & sin (wt - theta)??
The sine equation describes the current I. At approximately 5:40 I explain that because there is only a reactive element (an inductor), the current is phase shifted by 90 degrees with respect to the voltage. Mathematically we see this as a division by the imaginary unit j. I chose a cosine reference for the phasors, so the phase shift effectively turns the cosines into sines. The resulting equation for power then contains a factor with cosine and a factor with sine.
Not according to the equation! We have D(1-D) in the numerator, thus we have zero power transfer when D is zero. This D can be mapped to the phase shift.
Based on the way I defined my turns ratio (1:n) it seems to me that I should be dividing by n, which I correct as I finish writing out the equation. n1*I1 = n2*I2 n1 = 1, n2 = n I1 = I_L, I2 = I_o -> 1*I_L = n*I_o -> I_o = I_L/n So while this isn't a precise way of doing it, it shows that when reflecting current from the primary to the secondary, I should divide that value by n for this particular definition of turns ratio.
During practical operation, calculation of the required phase shift is done by the controller. For analysis, the phase shift can be calculated from the conversion ratio.
For the power flow across an inductor, since you have taken V1 and V2 to be the amplitudes, there will be a factor of 2 in the denominator of the average power calculation. Pavg= (V1*V2*sin(delta))/(2*w*L) at 11:25
Nicely spotted. Thank you.
from my calculation i think w should also be squared in the result. maybe i am wrong, but please check it.
the best video on DAB I have seen on youtube, Thank you!
This is the best video on DAB I have seen on youtube🙌
Thank you!
TRULY
Brilliant explaination about DAB converter which is hottest topic in power electronic now😍
Thanks for the Lecture. Looking forward for more Lectures on DAB and Resonant Isolated converters
Thanks a lot for the amazing lecture !! Requesting lots of more interesting lectures from you, Sir.
Hey Tim! Thank you for the video and the theses reference in one of the comments!
Full of information in DAB, great lecture sir.
Thank You for the lecture. Hope for more and more lectures.
can we use DAB for two identical (same voltage) system
Thanks for the great lecture. I hope you can make a design example of the DAB in the future.
Thanks professor for the very organized explanation 🤞🌺🙏
THANK YOU
Thanks for explaining DAB in great detail. Could you also please make a video on the analysis of Three Phase DAB ?
hi Tim, thanks for the video. However, i think there is a minor mistake in your AC power transfer calculations. I remember that the normal representation in power systems usually takes V
Thank you Tim
Thank you so much for this explanation...can you please do the same explanation on dual active bridge converter with coupled inductor
thank you soo much , thats perfect
Thank you for the video...could you please tell me how to control the shift angle? to control the power flow from one side to another?
First of all thanks for the detailed information. Sorry for next silly question, when you make reference to Iout you mean output current from the secondary winding, not the load current. thanks
😁
very nice explanation
1. Transistors are very different at right and left (high voltage vs high current). Different timings, a complex controller (if exists) etc.
2. A classical sync rect with a dual secondary do the same job without additional heat losses and money
3. Why people still teach "a dual active bridge" if no one uses such architecture in a real life?
Thankyou for a such great lecture .
Sir , please share reference paper for this lecture.
@timmcrae3831 At 6:12 why you took minus theta, according to me it should be plus because V2 leads V1 by theta??
Thanks for the detailed explanation, it helped me to understand about dual active bridge in greater detail. So for real life application, there will be a dead time between Q1,Q4 --> Q2,Q3? Also, for ZVS, Q1 and Q2 will be on before transitioning to Q2,Q3?
Dead time is always needed in hard switching applications. Q1 and Q2 on at the same time would result in shoot through. To achieve ZVS we simply allow body diode conduction before turning the switch on. In other words, the switches would be off and we allow natural commutation to determine what conducts.
Great lecture...please post proper sizing lectures.
Would the final power equation be different if the input and output were chargeable batteries? (when one of them functions as a load, the other acts as a power source, allowing bidirectional power flow)
I apologize for the delayed response. I didn't get a notification for this comment.
The equations do not change! In fact, a colleague of mine has done this to achieve battery balancing between cells (and also power a load) for a paper. ieeexplore.ieee.org/document/7845680, patents.google.com/patent/US8779700B1/en (this has some pictures)
Another paper does exactly what you're describing:
mplab.ee.columbia.edu/sites/default/files/content/Publications/BB_APM_APEC2018.pdf
In this beginning of this discussion I (implicitly) assumed small ripple approximation on the output voltage, effectively treating it as a constant voltage source. Sounds like a battery to me :) Typically this transfer of power between batteries (or battery cells) would be to ensure they have a similar state of charge so that they can be better utilized over the full power cycle of the battery pack.
The leakage inductance of the transformer is a representation of flux that doesn't link both the windings, it is that part of flux that completes its path through the air, then how can we say this leakage inductance is responsible for power transfer?
Very good point to bring up. It seems we're talking about two separate phenomena here. The transformer couples the primary side to the secondary side and as a result transfers power from the input to the output. However, as demonstrated in the motivating example, there doesn't need to be a transformer at all and only an element which defines some current between the input and output is needed.
Continuing with your reasoning, we can simply think of the leakage inductance as an air core inductor we have placed in series with the transformer. It doesn't couple the primary side to the secondary side, but it does relate voltages V1, V2 and the current through the governing equation of an inductor. The inductance, frequency, and phase shift are what define the peak current in the system, which defines how much power is transferred.
So yes, the transformer does transfer power from the primary to secondary, but this flux linkage does not regulate current. If there was no series element (in this case leakage inductance) to define the current, then power transfer would not be regulated in a controllable way as it is here.
Please make a video on ZVS operation between the snubber capacitance across the switches and leakage inductance in transformer of Dual Active Bridge. I was looking for design criteria for calculating the output filter capacitor, but couldn't found much as DAB does not have an inductive filter. How to choose its value.
For some reason I did not receive a notification for this comment. I apologize for the late reply. I'm currently uploading a video which briefly discusses ZVS operation in the DAB. It's not equation heavy, but I have provided a reference which does look more deeply into the equations.
As for the output capacitor, the process for selecting capacitance is the same as any other converter. First find the difference between the DC load current and the current output from the secondary bridge. This is the "AC" current absorbed by the output capacitor. We can then calculate the positive charge (the positive area under the curve) from this current. This charge takes the output voltage from its minimum to its maximum, so it represents a change of 2deltaVo. Then C = deltaQ/2deltaVo.
For this DAB, this is a little bit more complicated and the equation for the general case is nasty. After a little bit of arithmetic, it seems Io < I1/n. This means that calculating the positive charge we send to the output requires summing three different areas (which can all be found geometrically) .
deltaQ = (I1/n-Io)(Ts/2)(1-D) + (1/2)(Ts/2)(1-D)(I2-I1)/2 + (1/2)(tx)(I2/n-Io)
I1 and I2 are as we've defined them and D = phi/(Ts/2). tx is the time needed for the reflected inductor current to fall from I2/n to Io. I highly recommend you try this out for yourself. What I've calculated (and not checked) is:
deltaQ = (1-D)(Vg(1-2 D' D) + Vo/n)/(8 f^2 L n) + (1-D)^2(Vg - Vo/n)/(8 f^2 L n) + (D' D)(Vg(1 - 2 D' D) - Vo/n(1 - 2D)) /(16 f^2 L n)* (Vg)/(n (Vg + Vo/n))
Hopefully you can actually interpret this, I didn't want to use too many *.
Finally, C is this deltaQ/2deltaVo. Just a reminder that this has not been check over too deeply. I may go over this in a tutorial video.
Good luck.
@@timmcrae3831 Thank you very much for the response.
hi, Tim. I want to use Fourier Series to derive the power flow expression of DAB, but the result is different. Have you tried this method?
I have not tried this method, but it seems like a valid method. I imagine you have an infinite sum containing terms with sin(n*w+phi). Considering the direction calculation presented in the video is well established, I'm not sure if you could get further insight into the operation of the DAB using Fourier analysis.
@@timmcrae3831 Thank you very much for your reply! I can understand the classical method of power calculation, but the first part of your video inspired me, By doing Fourier analysis, I got an infinite sum: (8*V1*V2)/((i*pi)^2*omega*L) * Sin(i*phi), where i is odd number. But the plotted figure is different from the classical calculation derived in the video. I'm really curious where is the mistake.
Thank you professor for the lecture. I have a query regarding circulating currents. What exactly are the circulating currents in single-phase shift modulation and eliminating in dual-phase shift modulations? Could you please clarify this? Thank you.
High circulating currents refers to the fact that in some situations, the peak current in the leakage inductance may be very high, but the load (DC) current may be low. Dual phase modulation effectively allows the peak current to be lower for the same load current, reducing circulating current and thus conduction losses. If I'm not mistaken, dual phase modulation (for a DAB) can be thought of a single phase shift modulation, but you also modulate the duty ratio of the active bridges. So instead of a a 50% duty ratio voltage applied to the inductance, you may apply +V, 0 and -V.
@@timmcrae3831 Thank you, professor. How and why the circulating currents will takes place in this situation? could you please explain? Thank you.
Could you please tell how to calculate maximum leakage inductance im a dual active bridge converter
HELLO PROFESSOR. I took The Dual Active Bridge as a design homework. Is there a book, article or source where I can find even the most detailed information about this subject, thank you.
As far as textbooks go, I'm not familiar with any that focus on DABs. Most information I've read has been from theses and peer reviewed publications. A colleague from my lab looked at DABs for their thesis and it may serve as a useful reference to you or at least direct you to other references if needed.
tspace.library.utoronto.ca/handle/1807/70201
Thank you.
How to know the maximum leakage inductance??
At time 6.55; you have mentioned V2(t)=V2cos(wt - theta); & below that (V2/wL) sin(wt - theta). Why this difference in values of V2? Why cos(wt - theta) & sin (wt - theta)??
The sine equation describes the current I. At approximately 5:40 I explain that because there is only a reactive element (an inductor), the current is phase shifted by 90 degrees with respect to the voltage. Mathematically we see this as a division by the imaginary unit j. I chose a cosine reference for the phasors, so the phase shift effectively turns the cosines into sines. The resulting equation for power then contains a factor with cosine and a factor with sine.
I guess the phase of V2 should be -theta for sin(wt-theta) so it's back 2:18
sir i have one doubt please try to clear it. If phase shift is zero whether power transfer is possible when voltage gradient is present?
Not according to the equation! We have D(1-D) in the numerator, thus we have zero power transfer when D is zero. This D can be mapped to the phase shift.
When finding the output current, I think that you should have multiplied by n. Am I incorrect?
Based on the way I defined my turns ratio (1:n) it seems to me that I should be dividing by n, which I correct as I finish writing out the equation.
n1*I1 = n2*I2
n1 = 1, n2 = n
I1 = I_L, I2 = I_o
-> 1*I_L = n*I_o
-> I_o = I_L/n
So while this isn't a precise way of doing it, it shows that when reflecting current from the primary to the secondary, I should divide that value by n for this particular definition of turns ratio.
@@timmcrae3831 thank you
@@timmcrae3831 thank you
How do you calculate the phase shift though
During practical operation, calculation of the required phase shift is done by the controller. For analysis, the phase shift can be calculated from the conversion ratio.
@@timmcrae3831 thank you for the response, furthermore could you refer some material which deal with calculating the same
tell draw back of DAB Converter
i think when Qa Qd are on, Iout=I_L/n
You're correct, there is a factor of n which I forgot in the breakdown.
@@timmcrae3831😀