help with two quadrant bidirectional dc-dc converter

[--jason--]

hi,
i am designing a two quadrant bidirectional dc-dc converter whereby at the input side, it is an input sources 48v (can be an energy storage unit:ultracapacitor) and at the output side it can connect to a dc motor drive with a regulated dc output voltage 100v.
i am using an inductor with 5.7mH (choke inductor) and an output capacitor with 0.01F. The inductance and capacitance values that i get are based on the current ripple and voltage ripple formula from the boost converter.
At the first, i tried at an open loop condition by input a fix duty cycle without load and after that i loaded it with a resistor of 36ohm.
without the load, the current ripple is small and when i'm loaded with 36ohm resistor, the current ripple was high and not the same as what i am calculated from the formula of boost converter.
somebody can help me? i will very appreciate your help, thanks.

 

[rvionics]

The output capacitor value should be selected for the required ripple voltage at the required load current, you can use C x V= i x t equation where t is the power delivery time of capacitor or(no power delivery from inductor in Boost)
i.e. capacitor should supply your load current during no power delivery phase, also make sure the ripple current rating of the output capacitor is
well within derating (gen 80%)

 

[--jason--]

Thanks rvionics,

btw, let the V output be 100 and the capacitor that i'm using is 0.01F, and i am applied a load resistor of 36ohm, so the output current i= 100/36=2.7778A and the switching frequency is 25kHz. Since the input voltage (48v)need to boost up to 100v hence the required duty cycle is about 0.52. So the t (power delivery time of capacitor)=0.52 *(1/25K)= 2.08e-5 sec.
from what i'm calculated the C*V is much larger that i*t.
But the current ripple is still high and work in DCM mode.
can someone please provide me some note regarding to designing of bidirectional dc dc converter. Thanks.

 

[rvionics]

Thanks rvionics,

btw, let the V output be 100 and the capacitor that i'm using is 0.01F, and i am applied a load resistor of 36ohm, so the output current i= 100/36=2.7778A and the switching frequency is 25kHz. Since the input voltage (48v)need to boost up to 100v hence the required duty cycle is about 0.52. So the t (power delivery time of capacitor)=0.52 *(1/25K)= 2.08e-5 sec.
from what i'm calculated the C*V is much larger that i*t.
But the current ripple is still high and work in DCM mode.
can someone please provide me some note regarding to designing of bidirectional dc dc converter. Thanks.

The CV product is 1u, 0.01uFx100 = 1u, which is less than 111u (2.777A x 40uS)so CV is smaller by factor of 110 by it means you will need a big capacitor

---------- Post added at 08:41 ---------- Previous post was at 08:38 ----------

Here we have not considered the operating duty cycle which will have big impact again on the Capacitor value. (If we consider 50% then i x t product will be 55u

---------- Post added at 08:44 ---------- Previous post was at 08:41 ----------

so the capacitor requirement goes down, still CV product is 55 times lesser than i x t

 

[--jason--]

hello rvionics, thanks for the response,

the capacitor that i am using is 10,000uF, so is 0.01F.
how about the inductor? i am using a common mode choke with 5.7mH. will the problem come from the choke?Thanks.

 

[rvionics]

3.11Amps will be the rough RMS Ripple current through your output capacitor

---------- Post added at 09:11 ---------- Previous post was at 09:07 ----------

Sorry I missed u in the capacitor value,
If your choke saturates that can creat problem as there will be no energy to transfer to the load, you can varify by putting current probe in the inductor and monitor the current it should be linear ramp. Even the peak current ripple will go up big way

 

[FvM]

i am using a common mode choke with 5.7mH. will the problem come from the choke?

Very surely. A common mode inductor will saturate at a small fraction of the specified current when using it single ended. You have to select an inductor with sufficient rated current.