diy boost converter question
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diy boost converter question
Same thing as the adafruit boost
http://www.ladyada.net/library/diyboostcalc.html
This may seem like a terribly dumb question, and i know it's listed at the bottom of the adafruit site. A 30% duty means that the switch will be on 70% of the time, so eight bit pwm will need to be set to 179. I always thought duty was the on time?
I'm not getting anything near what these equations state. The equation assumes 0 voltage drop for the switch, so it also assumes 0 resistance for the inductor? I'm getting about 50v, but I should be getting 150 volts. Everything in the system can do 200V, and I can get 200v with feedback control.
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Re: diy boost converter question
Q2 conducts when the gate is pulled low, so the duty-cycle logic is inverted.I always thought duty was the on time?
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Re: diy boost converter question
oh found it. yes the equation above is for an ideal system. The real equation takes all resistances into account.
http://focus.ti.com/lit/an/slva061/slva061.pdf
pg9
http://focus.ti.com/lit/an/slva061/slva061.pdf
pg9
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Re: diy boost converter question
sorry to raise the dead, but I have a bone to pick at.
i peak = D*Vin /(L*f)
If Duty is the OFF time, then I really don't understand the logic.
Below considers duty as the ON time - when the inductor is charging.
When the circuit first turns on, the inductor is going to have close to zero current running through it. The inductor will slowly start accepting more current as time elapses. If the duty is low (0 < duty < .1), the inductor won't have enough time to allow as much current. If the duty was high (.9 < duty < 1) then the inductor is will have more time to accept more current.
Since it tries to maintain current, a high duty (inductor charging) means that there will be more current running through the inductor before the switch is off (inductor discharging). This is the current that the diode will be exposed to.
i peak = D*Vin /(L*f)
If Duty is the OFF time, then I really don't understand the logic.
Below considers duty as the ON time - when the inductor is charging.
When the circuit first turns on, the inductor is going to have close to zero current running through it. The inductor will slowly start accepting more current as time elapses. If the duty is low (0 < duty < .1), the inductor won't have enough time to allow as much current. If the duty was high (.9 < duty < 1) then the inductor is will have more time to accept more current.
Since it tries to maintain current, a high duty (inductor charging) means that there will be more current running through the inductor before the switch is off (inductor discharging). This is the current that the diode will be exposed to.
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- Posts: 352
- Joined: Tue Aug 19, 2008 4:36 pm
Re: diy boost converter question
brainfart never mind...adafruit_support wrote:Q2 conducts when the gate is pulled low, so the duty-cycle logic is inverted.I always thought duty was the on time?
Please be positive and constructive with your questions and comments.