Generate +12V from -48V, Non-Isolated
Hi PI-Spock,
I have read a recent thread (6 weeks ago) posted on this forum for a request for a DC-DC converter with single source of Voltage (http://www.powerint.com/en/forum/power-supply-design/dc-dc-converter-single-source-voltage). In your reply, you have suggested a simple Buck-Boost topology using the DPA425.
I have expanded on this concept of yours and prepared a schematic that I would like you to take a look at.
My requirement is to generate a +12V DC output from a -48V DC input. The Output and Input share a common ground (non-isolated). I need to generate 2.5A output current (30W max).
I have enclosed the proposed schematic. I would like your help in dimensioning the Inductor and the Rectifier Diode. Please also advise me on any other design changes/corrections required.
I look forward to hearing from you in this Regard.
Comments
Hi PI_Crusher,
Thank you for the feedback. This is my first stab at building such a Circuit. I would appreciate your help in selecting an appropriate Inductor to help me achieve an output current of 2.5A.
Regards,
Smokey
Hello smokey,
Here is what you need to do:
1) Write down the value of input voltage (Vin), output voltage (Vout), maximum load current (Iload), switching frequency (fs) and choose the ratio between ripple current trough the inductor and peak output current coefficient (kp)
2) a good first kp choice: kp = 0.3
3) For DPA425 with F pin connected to Source you selected: fs = 400kHz
4) Now calculate the inductor value: L=(Vout*(Vin-Vout))/(Vin*fs*Iload*kp)
The inductor will be adjusted later to optimize cost, size, and efficiency; just increase or decrease the value and measure the efficiency. If you hit the current limit you will see the output voltage going down. Also, the inductor must be design for the current limit without saturation and with copper wire big enough to withstand your RMS current. You have to Google around to see how this is done or buy from Digikey one inductor with higher current rating than your peak current. Pay extra attention to heat dissipation and design your application to deal with this problem.
Cheers,
PI_Crusher
Hi PI_Crusher,
I did the math and I am getting a value of 32uH for the Inductance value using your formula. This value seems very low. I do have a 100uH inductor lying with me in my lab. Can I try a value of 100uH instead ?
Regards,
Raj
smokey, it is possible to be the right value. Just start with whatever inductor you have and work your way from there. You will see if your output power is limited or not, this is what you have to check. Fine tuning allows you to select best price/size and to optimize the efficiency.
Hello smokey,
Your application is almost functional, with no major problems. For start, a 200uH inductor is good enough to provide around 0.7A output current (just check the maximum current of your inductor). Later you can check the current variation trough the inductor (dI/dt=V/L) and the energy transferred trough the inductor (E=0.5*L*I^2) in order to select the right inductance value for your output power.
There is one single problem you must pay attention to: in your example the control pin is decoupled to -48V rail, not to Source of DPS425. You have around 5.8V between C pin and S pins, and that leaves you with more than 40V applied to C37 and C38. Also, it is possible to experiment instabilities and then you will have to move the decoupling components to S pin (the analog reference for DPA425).
Cheers,
PI_Crusher