Timeline for DC DC Boost Converter - Why that high Ampere?
Current License: CC BY-SA 3.0
13 events
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| Sep 1, 2017 at 9:03 | comment | added | Michael Karas | (continued from above) Feedback comes into play where you want two key things. First off is the automatic adjustment of the output voltage when the load of the boost circuit changes over a wide range. Secondly the feedback circuit is using some type of reference against which the output is being compared. The reference is designed to be fairly immune to changes in operating voltage and temperature variations. Thus with feedback the converter can have an output that adjusts to cancel out other variations in the circuit due to component value variation, voltage and temperature. (end) | |
| Sep 1, 2017 at 8:57 | comment | added | Michael Karas | (continued from above) optimum in terms of component count and performance over circuit variations. A simple booster circuit made with a fixed duty cycle 555 design (even a trimpot adjusted duty cycle) will operate open ended and end up producing a certain boosted output voltage. As long as the load on the booster output is constant the output voltage will stay pretty much the same. This may be just perfect for the high voltage low current application suggested in the Adafruit article where a display would work just fine with a slight variation of the high voltage. (continued) | |
| Sep 1, 2017 at 8:49 | comment | added | Michael Karas | The XL6003 IC has the FB pin. FB is for feedback. This pin samples the output voltage (which may be scaled via a voltage divider) and adjusts the PWM duty cycle in an attempt to keep the output voltage constant. If the voltage is too high it will lower the duty cycle to lower the output voltage and if it is too low will increase the duty cycle to raise the output voltage. This feedback type of control is not present in a typical 555 chip PWM generator. It takes additional components around a 555 to achieve an electronically controlled 555 duty cycle and it will be less than (continued) | |
| Sep 1, 2017 at 8:01 | comment | added | dessi | learn.adafruit.com/diy-boost-calc/overview the last sentence on that article "These sorts of designs can be easily made with a 555, once you have the PWM output, connect it up to Q2!" - I look at the datasheet of the XL6003 IC - and it looks like an 555 o.O What is the huge difference that you recommand that instead of the 555? Just for my understanding :) | |
| Aug 31, 2017 at 11:57 | comment | added | Bimpelrekkie | Even using a microcontroller as the basis for a DCDC converter could be a better choice (although personally I would never do that, what if the software crashes and the low-side switch is stuck on, the inductor will saturate and the switching transistor will be destroyed). | |
| Aug 31, 2017 at 11:57 | comment | added | Bimpelrekkie | and even on the Adafruit site it's done with an NE555 Show me the link. That it's on Adafruit doesn't mean it is a good design/idea. Learning is excellent ! Then let me suggest you get an IC that is dedicated for the task like an XL6009. You will have plenty of trouble getting that to work. Choosing the right inductor, keeping the high-current loops small etc. | |
| Aug 31, 2017 at 11:41 | comment | added | dessi | @Bimpelrekkie But if I don't try I won't learn it or get better ô.O And even on the Adafruit site it's done with an NE555. And it's not answering my question... I already have a ready-to-go-module. But I want to do it by myself to learn and improve my skills; that's why I ask here ;) | |
| Aug 31, 2017 at 11:35 | comment | added | Bimpelrekkie | Also: if you're that inexperienced (not knowing how to make a certain DuCy with the 555) then I strongly advise you to buy a ready-made module (find one on Ebay) to do this and not to build your own. There is a very high chance that your circuit will simply not work due to your lack of experience. I have 30 years of building circuits experience but I also just buy a ready made module because it will be cheaper and it will just work. | |
| Aug 31, 2017 at 11:30 | comment | added | Bimpelrekkie | Don't use an NE555 for this. Just don't. Sure it can generate a PWM signal but you need also a feedback loop to control the PWM signal so that the output voltage will be 12 V and not 100 V ! Yes that is possible with boost converters ! So find a dedicated DCDC boost converter IC and forget about using a NE555 for this. It is simply unsuitable for the job. | |
| Aug 31, 2017 at 11:24 | comment | added | dessi | I now changed the MinVin to 2V and now the Ampere dropped down to 3.6A. But now I have the question: How do I know what exact duty cycle I need to calulate the resistors for the NE555 chip? :S Or is it just the Max. duty cycle when I have minVOut and maxVOut the same? Or is it depending on the VIn? | |
| Aug 31, 2017 at 10:24 | comment | added | Barleyman | @Bimpelrekkie As I mentioned in my answer, the diode is the most abused part of a boost converter. The actual conduction time may not be that short either, depending on the output load and duty cycle. But the real problem is that you have non-trivial amounts of power dissipation due to the voltage drop. | |
| Aug 31, 2017 at 10:22 | comment | added | Bimpelrekkie | I was also thinking along these lines but then realized that in a boost converter the diode is in series with the output. However, when the input voltage is low, the inductor needs to be charged more (longer) so the discharge cycle must be short. During the (short) discharge cycle the inductor must quickly discharge through the diode and this results in a high (pulsed) current. | |
| Aug 31, 2017 at 9:57 | history | answered | Michael Karas | CC BY-SA 3.0 |