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Designing a boost step up converter

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engr_joni_ee

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Hi, I am having 24 V as voltage source and I need to design a boost converter having output voltage from 24 V to 32 V with step size of 50 mV. I am sit sure if such device exist to convert the DC and can deliver the output current 500 - 600 mA ?
 

There's plenty of 24V DC-DC modules capable of the current and
voltage. The fine resolution variable output probably wants you
to pick a "bring your own feedback resistors" type module, and
make one of your resistors a tapped ladder through an analog
mux tree. But 8V / 50mV is 160 steps, 8 bits. That's a lot of taps
on a linear ladder. Maybe use the to-GND leg and add binary
weighted shunts, if you can stand a somewhat nonlinear VOUT-
vs-code characteristic. Or a binary RDAC on the top leg, which
will be more linear and only need 8 low-Z-shunted taps.
 

Even if you can get the RESOLUTION, I don't think your accuracy is going to be that good. I'm sure this can be done, but it sounds like a $10000 lab instrument.
 

I was also wondering before to get an DC-DC converter with output vary between 24 V to 32 V with control pin which can be driven by FPGA but not able to find such module. Do we have such module ?
 

Not sure if there are any voltage controlled converters, but you can add some circuitry to do this. There is generally a voltage divider that sets the output voltage. The bottom of this divider is usually connected to ground. If you instead connect the bottom of the divider to an adjustable voltage, you can effectively change your output voltage. You could connect your FPGA to a DAC to generate this control voltage.
 

Yes, that is a nice idea to connect the bottom of the voltage divider to an DAC that is connected to FPGA on digital side but how about the power rating of the resisters in voltage divider because there will be then 600 mA @ 30 V which is 18 Watt, too big resisters.
 

Yes, that is a nice idea to connect the bottom of the voltage divider to an DAC that is connected to FPGA on digital side but how about the power rating of the resisters in voltage divider because there will be then 600 mA @ 30 V which is 18 Watt, too big resisters.
Where are you getting 18 watts????
600 mA is your output current, not the current through the divider.
 

Hi,

Again: THE 600 mA IS NOT GOING THROUGH YOUR DIVIDER!

below you can see the typical application of an (arbitrary chosen) boost converter [1]. As you can see, there is a feedback network, where the voltage drop across R2 is a portion of Vout and used to set/regulate the output voltage. What you can clearly see, there is no high current flowing across this resistors (R1 & R2) as they are in the kOhm range (12 V / 19400 Ohm = 619 µA).

LM2577_Typical_Application.PNG


[1] https://www.ti.com/lit/ds/symlink/lm2577.pdf

BR
 


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