# DC DC Boost Converter - Why that high Ampere?

I have a question for the electronic experts of you! I want to create an DC-DC boost converter circuit for a project to get from 5V to 12V.

Calculated with help of this site: https://learn.adafruit.com/diy-boost-calc/the-calculator

There is this result: My question: Why there is an Ampere size of 5.76A!! oO I don't get it why this number explodes that much... Can you explain this to me (or if I can choose a lower Ampere tolerant Schottky diode?

Thank you! :)

• That is the peak current, not the average. It arrives in pulses Aug 31, 2017 at 9:52

If you assume a converter efficiency of 100% (not possible in reality) the output power equals the input power.

Output power = 12V * 1.5A = 18W

Input power = Vin Min * I In = 18W

Input Current = 18W / 4V = 4.5A

Under the less than ideal conditions the input current will be higher. The calculator must be assuming some level of efficiency less than 100% which is not directly visible in the information presented in your question.

• 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 10:22
• @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 10:24
• 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 11:24
• 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 11:30
• 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 11:35

Boost circuits definitely do stress the components a lot. It's a fact of life. That form omits some important parameters such as average current and inductor ripple. Average current creates heat and this is separate from max current capability.

Do note that the output diode is the hottest component of a boost converter. Also the diode specs need to be considered carefully. 5A diode will usually NOT survive 5A current as it will overheat and die. On that kind of current you need to start paying attention to heat dissipation solutions, using PCB surface copper is a popular solution as it requires no extra hardware. For high power loads it may not be sufficient.

You may want to try texas webench which will give you thorough analysis of the circuit you need.

When the boost circuit component stresses get out of control, it would be time to start considering a flyback which will do the voltage stepping "for free". It's a more complex beast of course with that coupled inductor and all.

Texas SLVA372C gives you basic equations for determining boost converter values. http://www.ti.com/lit/an/slva372c/slva372c.pdf

Diode current rating equaling the output max current is indeed sufficient in theory but in reality will overhead without a cooling solution. Check datasheet for thermal parameters, they usually give you temperature rise values for "minimum" PCB footprint and with a blob of surface copper to dissipate the heat.

• Just a small comment, if I may: "average current creates heat", needs to be reformulated. In a diode, the total conduction loss is expressed by $P_d=V_{T0}I_{d,avg}+r_dI^2_{d,rms}\approx V_fI_{d,avg}$ in which $V_{T0}$ is technology-dependent and $V_f$ is the drop at the considered average current. For the MOSFET, the conduction loss is $P_d=r_{DS(on)}I^2_{d,rms}$ in which $r_{DS(on)}$ is taken at a 100-°C junction temp. It is only for the input source where the power is defined using the average current: $P_{in}=V_{in}I_{in,avg}$. Aug 31, 2017 at 12:12
• @VerbalKint In most cases you can get away with $P_d = V_f * I_frms$ where Vf is taken at RMS current. Vf does vary by the current and temperature but the delta isn't really that big in most cases. In fact $V_f$ goes down with higher temperature so it self-compensates to a degree the fact that $V_f$ goes up with current. The major consideration must be to evaluate the diode power dissipation versus package versus PCB mounting and any other considerations like forced convection or heatsinks. MOSFETs are especially obnoxious with their "55A" specs where it'll cook at 10A without cooling. Aug 31, 2017 at 12:31
• the conduction loss in the diode - in its simplified form - is $V_fI_{d,avg}$. Average and rms are close to each other if the ac ripple is very low as the rms current definition is $I_{rms}=\sqrt{I^2_{ac}+I^2_{dc}}$. However, I agree that you select a diode not truly based on its current capability - a 3-A diode for a current of 2 A for instance - but not more on its thermal capability heavily linked to the die size and package. Aug 31, 2017 at 13:37
• @VerbalKint Not so fast. Average and RMS voltage of even a fully rectified signal are not the same. Real heating power is given by $I_{rms} * V_{rms}$ or $V_{rms}^2/R$ See this answer for example: electronics.stackexchange.com/questions/40341/…. Aug 31, 2017 at 14:14
• It is a bit more complex actually, the average power is $I_{rms}V_{rms}$ only if you deal with a resistive load where the power factor is 1. As long as a) you have distorted signals or b) a phase displacement, this formula no longer returns W but V.As. For a dc source, power involves the average current as V is constant and for the diode, you have more calculation details in this AN from ST: st.com/content/ccc/resource/technical/document/application_note/… Aug 31, 2017 at 14:25