How can I find the magnetising inductance of DAB transformer?For example, finding the magnetising current of DAB converter and use V/I =1/jwL to find out L, assumed that leakage inductor is neglected…
I would like to express my great thanks to you for your excellent explanation and analysis. I just have a remark at time 42:35, I think you missed a D, it's D*Vg.
I really appreciate the time you spent for clarifying the DAB design. thank you M. Kim. However, I think that the equation of "tx" at time 43 min it is not correct. Would you please check it or explain how you did you get it. I tried with it and I found: tx= ( I2 - n * Io) * L/ (Vo/n + Vg ). knowing that I2 + I1= [2* D/(4*L*fs) ] * [ Vo/n + Vg ].
thank you for this great work. i have one question may you tell me name of good reference for DAB and SAB topology, circuit analysis and control methods
Thank you for this great video. It is indeed very useful with reasonable dept of explanation. I have a question though. How could you relate D to phi/(Ts/2). Isn't phase shift expressed in terms of angle. We know that duty cycle (D) in relation with Ts is a form of time. Could you clarify that please because I want to know what I am missing there. Thank you. Regards,
Dear TIM, Could you help me with magnetic inductance of transformer for DAB design? Here is no influence of value this inductance in real transformer for reactive power of DAB converter? I think here will be some differences in primary and secodary current waveform or not? If not I can not expain myself why not. I think bigger value is better than lower value of inductance, because she has bigger impedance for PWM (switching frequency) and lower additional reactive curent through this induktance. Thank you so much for help
You're correct, the magnetizing inductance will result in increased "reactive power" in this converter, or increased circulating current. Larger magnetizing inductance is better to reduce the resulting magnetizing current. Modelling this inductance on the primary side we expect to see +Vo/n and -Vo/n applied, resulting in a triangular current waveform. Depending on the application, this extra current could be significant. In addition to the magnetizing inductance, we can also see that the turns ratio and output voltage have a direct impact.
We can consider control of power flow to be load dependent voltage gain. Unfortunately due to the phase shift control in general, there is no clean load independent voltage gain. Perhaps there is a different modulation scheme, but I am not aware of it. Probably something worth researching.
Thank you very much for this amazing video. At 4:37, should the turn ratio be on the numerator instead of denominator? I have seen the same equation in TI application with n is secondary to primary turns ration at the numerator.
Thank you so much Dr. McRae for the lecture. It's really useful. I would like to know if I want to study the converter behavior under all modulations you mentioned(SPS, EPS, DPS and TPS), which modulation do I have to use to design the DAB? I understand that, depending on the modulation you use, the equation of average power differs.
Typically different modes of operation are used for a single converter to cover different ranges of input and voltage and load current. Design the DAB separately for each operating mode. Compare the resulting L, C, transformer and FET choices and see if there is overlap across the operating range of your converter. Selection of specific component values will determine which mode you can operate in for different voltages and currents. As these external factors vary, your controller will have to switch modes seamlessly by varying the phase shift and switching frequency.
Thank you professor for the lecture. Basically, when we design, ZVS range of the converter is essential right? could you give us some idea that how to consider the range and all. Thank you sir.
Hi, sir. Just a little bit i know. In Zero voltage switching, the inductor current should be sufficient to discharge and charge output capacitances of mosfets (1/2)LI^2=(1/2)CV^2. It must be done during dead time. Mr.Tim MCRae sir knows much better. If I'm wrong please correct me.
This is the correct idea. Ensuring you achieve ZVS over a wide range is basically ensuring that the current at the moment of transition is sufficient to change the voltage of the output capacitances of the MOSFETs. From the energy relation given above, you can see that it is related to the inductor current (at the moment of transition), and also the input voltage.
I"m confused with the rectified current. How is it ever negative (doesn't it always flow "to the right as drawn" producing positive Vo across RL)? I would expect Iout to always be >0, thus, in the first quadrant at time 21:07, io(t) in orange , i'd expect it to start as drawn when IL is negative, but then follow the IL once IL goes positive. I don't think the orange line should ever go below 0....could you please clarify this? Thank you. Love your lectures, by the way. They have the right amount of detail for me.
Remember there is an output capacitor, which stabilizes the output voltage. This means that for short portions of time, current can be drawn out of this capacitor and instead of the voltage immediately going negative, it simply decreases. As long as the average current into and out of this capacitor remains zero, this average voltage will remain the same. Also recall that the secondary side bridge is active. When the switches are on current can flow both ways. This means that if the inductor current goes negative, the reflected and rectified current will also be negative at that time. We can still send a positive average current despite this.
D' (said D prime) is analogous to the D' used in standard converters: 1-D. The only difference here is that D is defined over the half-cycle and as a result we want D(Ts/2) + D'(Ts/2) = Ts/2. Relating it back to phase or the phi notation: D' = (Ts/2-phi)/(Ts/2)
Can anyone help me out please, I want to make a 20kW bidirectional charger that will take input 600V and output 48V, please help me out with that will DAB is suitable for this kind of design.
There is not always a clear answer to this. Different applications will have different requirements. For high input voltage applications you want to choose a scheme which reduces switching loss on the primary bridge. For high current applications you might want to choose a modulation which reduces the RMS current in the inductor.
That would depend a great deal on how the converter is designed; you absolutely could build one that outputs 10kV-ish if you use really high voltage semiconductor switches (or switches comprised of many switches in series), and a very high-ratio transformer.
If you tune the turns ratio appropriately, then the primary side current will look similar regardless of output voltage on the secondary side for the same power. Consider the slope in the second portion of the current: (Vg-nVo)/L. No matter what Vo is, you can always choose an n such that this slope is 0 (or near 0). The slope in the first portion of the current will always be 2Vg/L (or near it). The peak current will vary depending on the desired output power but the shape will be similar. The actually implementation will look different because the transformer design will vary depending on the turns ratio, resulting in different losses.
@@chai5632 what exactly are the circulating currents? how do we understand these from the waveforms? and why they are only before zero crossing in the waveform?
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This is what the internet is really for. Thank you so much for your work!
Thanks for a very detailed tutorial on design of DAB.
Thank you for the video, sir. Kindly say the reference book or journals for designing DAB converters.
How can I find the magnetising inductance of DAB transformer?For example, finding the magnetising current of DAB converter and use V/I =1/jwL to find out L, assumed that leakage inductor is neglected…
Thank you for great lecture...at 23.58 , it should be Ip/n instead of Ip/2 in delta q2 expression ..thank you once again
I would like to express my great thanks to you for your excellent explanation and analysis. I just have a remark at time 42:35, I think you missed a D, it's D*Vg.
I really appreciate the time you spent for clarifying the DAB design. thank you M. Kim. However, I think that the equation of "tx" at time 43 min it is not correct. Would you please check it or explain how you did you get it.
I tried with it and I found: tx= ( I2 - n * Io) * L/ (Vo/n + Vg ).
knowing that I2 + I1= [2* D/(4*L*fs) ] * [ Vo/n + Vg ].
thank you for this great work. i have one question may you tell me name of good reference for DAB and SAB topology, circuit analysis and control methods
Would this design and its derivations stay the same if the dab is a step up so if VgVg
Thank you for this great video. It is indeed very useful with reasonable dept of explanation. I have a question though. How could you relate D to phi/(Ts/2). Isn't phase shift expressed in terms of angle. We know that duty cycle (D) in relation with Ts is a form of time. Could you clarify that please because I want to know what I am missing there. Thank you. Regards,
Dear TIM,
Could you help me with magnetic inductance of transformer for DAB design? Here is no influence of value this inductance in real transformer for reactive power of DAB converter? I think here will be some differences in primary and secodary current waveform or not? If not I can not expain myself why not.
I think bigger value is better than lower value of inductance, because she has bigger impedance for PWM (switching frequency) and lower additional reactive curent through this induktance.
Thank you so much for help
You're correct, the magnetizing inductance will result in increased "reactive power" in this converter, or increased circulating current. Larger magnetizing inductance is better to reduce the resulting magnetizing current. Modelling this inductance on the primary side we expect to see +Vo/n and -Vo/n applied, resulting in a triangular current waveform. Depending on the application, this extra current could be significant. In addition to the magnetizing inductance, we can also see that the turns ratio and output voltage have a direct impact.
Great Video Dr. McRae , Thank you. How can we control the Voltage gain of the converter rather than looking at power flow ?
We can consider control of power flow to be load dependent voltage gain. Unfortunately due to the phase shift control in general, there is no clean load independent voltage gain. Perhaps there is a different modulation scheme, but I am not aware of it. Probably something worth researching.
Thank you very much for this amazing video. At 4:37, should the turn ratio be on the numerator instead of denominator? I have seen the same equation in TI application with n is secondary to primary turns ration at the numerator.
Thank you so much Dr. McRae for the lecture. It's really useful. I would like to know if I want to study the converter behavior under all modulations you mentioned(SPS, EPS, DPS and TPS), which modulation do I have to use to design the DAB? I understand that, depending on the modulation you use, the equation of average power differs.
Typically different modes of operation are used for a single converter to cover different ranges of input and voltage and load current. Design the DAB separately for each operating mode. Compare the resulting L, C, transformer and FET choices and see if there is overlap across the operating range of your converter. Selection of specific component values will determine which mode you can operate in for different voltages and currents. As these external factors vary, your controller will have to switch modes seamlessly by varying the phase shift and switching frequency.
what ks the relation between input and output voltage in terms of duty ratio?
can you teach the same for dual phase shift DAB converter?
Thank you professor for the lecture. Basically, when we design, ZVS range of the converter is essential right? could you give us some idea that how to consider the range and all. Thank you sir.
Hi, sir.
Just a little bit i know.
In Zero voltage switching, the inductor current should be sufficient to discharge and charge output capacitances of mosfets (1/2)LI^2=(1/2)CV^2.
It must be done during dead time.
Mr.Tim MCRae sir knows much better.
If I'm wrong please correct me.
This is the correct idea. Ensuring you achieve ZVS over a wide range is basically ensuring that the current at the moment of transition is sufficient to change the voltage of the output capacitances of the MOSFETs. From the energy relation given above, you can see that it is related to the inductor current (at the moment of transition), and also the input voltage.
Thank you Dr. McRae for making this tutorial. Could you please also make a tutorial on the analysis of Three Phase DAB ?
With Thanks,
nice video. is there also a tutorial on system modelling and PID control loop and stuff?
I haven't got to control yet. Soon though!
Pretty helpful. Could you make a video on Controller design of DAB?
make tutorial for high frequency transformer design
I"m confused with the rectified current. How is it ever negative (doesn't it always flow "to the right as drawn" producing positive Vo across RL)? I would expect Iout to always be >0, thus, in the first quadrant at time 21:07, io(t) in orange , i'd expect it to start as drawn when IL is negative, but then follow the IL once IL goes positive. I don't think the orange line should ever go below 0....could you please clarify this? Thank you. Love your lectures, by the way. They have the right amount of detail for me.
Remember there is an output capacitor, which stabilizes the output voltage. This means that for short portions of time, current can be drawn out of this capacitor and instead of the voltage immediately going negative, it simply decreases. As long as the average current into and out of this capacitor remains zero, this average voltage will remain the same.
Also recall that the secondary side bridge is active. When the switches are on current can flow both ways. This means that if the inductor current goes negative, the reflected and rectified current will also be negative at that time. We can still send a positive average current despite this.
if for example I have a unit transformer, what is the relation for output voltage if only input voltage is known?
very informative lecture. D=phase/(Ts/2). what does D' dash represent? whats the procedure for finding I1 and I2. Thank you for the tutorial.
D' (said D prime) is analogous to the D' used in standard converters: 1-D. The only difference here is that D is defined over the half-cycle and as a result we want D(Ts/2) + D'(Ts/2) = Ts/2.
Relating it back to phase or the phi notation: D' = (Ts/2-phi)/(Ts/2)
@@timmcrae3831 thanks for your response
Can anyone help me out please, I want to make a 20kW bidirectional charger that will take input 600V and output 48V, please help me out with that will DAB is suitable for this kind of design.
Thanks for the tutorial!!
Gained so much from your lecture sir.
May i know which is the best modulation strategy (sps,eps,dps,tps) for getting high efficiency?
There is not always a clear answer to this. Different applications will have different requirements.
For high input voltage applications you want to choose a scheme which reduces switching loss on the primary bridge. For high current applications you might want to choose a modulation which reduces the RMS current in the inductor.
@@timmcrae3831 thank you sir
What if your not given the current range? How do I find L?
can the DAB converter give a high voltage at the output of the order of 10kV
That would depend a great deal on how the converter is designed; you absolutely could build one that outputs 10kV-ish if you use really high voltage semiconductor switches (or switches comprised of many switches in series), and a very high-ratio transformer.
19:16
What is the value of n here?
please tell the calculation of LL
Can we use DAB for high gain applications, like vg/vo=10
I don't see why not. The transformer design is going to be a key aspect of this kind of application.
@@timmcrae3831 thank you sir
@@timmcrae3831 okay sir
But when voltage gain increases(or high)M=nVo/Vg ,where n=(N1/N2)=(V1/V2), will it effect the circulating current ?
If you tune the turns ratio appropriately, then the primary side current will look similar regardless of output voltage on the secondary side for the same power. Consider the slope in the second portion of the current: (Vg-nVo)/L. No matter what Vo is, you can always choose an n such that this slope is 0 (or near 0). The slope in the first portion of the current will always be 2Vg/L (or near it). The peak current will vary depending on the desired output power but the shape will be similar.
The actually implementation will look different because the transformer design will vary depending on the turns ratio, resulting in different losses.
@@chai5632 what exactly are the circulating currents?
how do we understand these from the waveforms? and
why they are only before zero crossing in the waveform?
Can DAB designed to be bidirectional Buck boost?
The DAB is typically used as a bidirectional converter. The gain is load dependent but the converter can be designed to buck or boost.
What is D prime?
How I can contact you