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As I am using LTspice to develop different types of dc-dc boost converter topologies and wanted to define the best inductor to use when the voltage source only produces a current of 30 μA's, I was reading these very useful equations in answer one from this link:

How to design a boost converter? And how to specify the inductor and capacitor values?

Everywhere I look, all formulas assume the inductor will accept nJoules/μSeconds based on what the load needs. Of course this is not true. If the source can't provide energy, then the duty cycle or frequency must slow down or the expectation has to be lowered.

I'm (attempting) writing a calculator to take this into account. For my calculator I have energy in (joules) per cycle and energy required per cycle. The latter is discussed in formulas, and the first I could do through guess work, but is there an actual formula to present it?

Edit. Here's an example of the formulas used:

Energy in. Vin = 1.6V, Iin = 30μA Freq = 10Khz Thus E = 4.8E-09 J E = (V * I) / Freq

Energy Out. Vout = 5, Iout = 10mA, Thus E = 5E-6 J

Proposed Inductor: L = ?? (6.8 μH), Current in Inductor = 0.383482494 A I = sqrt((2 * Jout)/ L)

Duty Cycle: V = L*di/dt thus Amps/S (0.3834 A/ .0001)=5 V/6.8 μH = 0.5%

Which of course is incorrect as it doesn't take into account that the 'energy in' can't provide 383 mA in 0.5% duty cycle of 10KhZ unless I'm totally not understanding this. But, my simulations pretty much prove this. At 25% duty cycle on 20KhZ I can load a 6.8 μH Inductor to 604.35 μA's with the Voltage Source shown above.

Edit for Jonk: Remember the voltage source for this? The blocking oscillator works well, but I think I can improve on it in a dc-dc boost topology.

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    \$\begingroup\$ I use Vin, Vout, Iout, and at least Bmax (and a few other details) as my starting point. Taking into account switch details and diode details I can work out the actual Von and Voff across the inductor and the duty cycle. Is the 30 uA you mentioned your output need or input limitation? \$\endgroup\$
    – jonk
    Mar 31, 2022 at 18:36
  • \$\begingroup\$ @jonk Hey Jonk. I'm barely back in action... The 30uA is input limitation. \$\endgroup\$
    – RobMcN
    Mar 31, 2022 at 18:40
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    \$\begingroup\$ If so, I think that's your Ipeak that has to be equal to 2*Iout/(1-duty)? (Where the duty cycle is Voff/(Von+Voff).) \$\endgroup\$
    – jonk
    Mar 31, 2022 at 18:43
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    \$\begingroup\$ Vin = 1.6V, Iin = 30μA means your input is 48μW, which means if your 100% efficient, you'd get 48μW out. The mA output seems way off for this. \$\endgroup\$ Mar 31, 2022 at 20:14
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    \$\begingroup\$ @RobMcN What exactly are your behavioral sources in that newly added diagram supposed to actually be? I don't recall ever seeing anything like what I see inside the dashed line boxes. And it looks seriously wrong to me. \$\endgroup\$
    – jonk
    Mar 31, 2022 at 20:19

1 Answer 1


To Start the Discussion

I'll write up a few things from memory. Since your input is limited to \$30\:\mu\text{A}\$ and since I never start with that limitation but instead work out what I need to supply as an output, not as an input to the design, what I have to say will have to be considered with that in mind.

Also, I'm keeping in mind that this is a boost application. So it must be the case that \$V_{_\text{IN}}\lt V_{_\text{OUT}}\$.

Let's say that my semiconductor switching device, when on, drops \$V_{_{\text{Q}_\text{ DROP}}}=100\:\text{mV}\$ and my semiconductor flyback diode drops \$V_{_{\text{D}_\text{ FLY}}}=300\:\text{mV}\$. (I'm just picking this out of thin air, for now, to simply establish that there are some realities that have to be accepted and dealt with.) Then the applied voltage across the inductor during the on period is \$V_{_{\text{ON}}}=V_{_\text{IN}}-V_{_{\text{Q}_\text{ DROP}}}\$ and the inductor's flyback voltage during the off period is \$V_{_{\text{OFF}}}=V_{_\text{OUT}}+V_{_{\text{D}_\text{ FLY}}}-V_{_\text{IN}}\$. (I'm assuming a little something about the topology there. Hopefully, this comports with your case.) The duty cycle is then \$D=\frac{V_{_{\text{OFF}}}}{V_{_{\text{ON}}}+V_{_{\text{OFF}}}}\$ and the peak inductor current needs to be \$I_{_{\text{PEAK}}}=2 I_{_{\text{OUT}}}\frac1{1-D}\$.

Since you are specifying \$30\:\mu\text{A}\$ as an input limitation, I take this to mean that \$I_{_{\text{PEAK}}}=30\:\mu\text{A}\$ as an input to the design. So this means that \$I_{_{\text{OUT}}}=\frac12\left(1-D\right)I_{_{\text{PEAK}}}\$. That's going to be your first moment of validation as to whether or not you can achieve what you want to achieve, at all. If the output current doesn't meet your requirement, you are done. You cannot do what you want to do with what you have. Box the idea up and put it on a shelf. Game over.

Note that frequency and Joules haven't even entered into the above. The frequency will be proportional to the inductance, \$L\$, the peak inductor current, \$I_{_{\text{PEAK}}}\$, and inversely proportional to a concept that kind of looks like taking \$V_{_{\text{ON}}}\$ and \$V_{_{\text{OFF}}}\$ in parallel, \$\frac{V_{_{\text{ON}}}\cdot V_{_{\text{OFF}}}}{V_{_{\text{ON}}}+V_{_{\text{OFF}}}}\$. Obviously, of the above, you are missing \$L\$. So we can't speak of frequency, just yet. The inductor needs to be designed. That's the entire point, here. And now, without frequency, there's no point in talking about Joules, either.

You need to fill out the details, now. If you do, I may expand this into a real answer. For now, it's here to get you to improve the question. It was more than I could consider writing as comments.

Playing With Your Added Details

  • \$V_{_\text{IN}}=1.6\:\text{V}\$
  • \$V_{_\text{OUT}}=5\:\text{V}\$
  • \$I_{_\text{IN}}=30\:\mu\text{A}\$

I've no idea what your practical switch and diode are, but I'm going to stick with the numbers I started with:

  • \$V_{_{\text{Q}_\text{ DROP}}}=100\:\text{mV}\$
  • \$V_{_{\text{D}_\text{ FLY}}}=300\:\text{mV}\$

From here I find:

  • \$V_{_{\text{ON}}}=1.6\:\text{V}-100\:\text{mV}=1.5\:\text{V}\$
  • \$V_{_{\text{OFF}}}=5\:\text{V}+300\:\text{mV}-1.6\:\text{V}=3.7\:\text{V}\$
  • \$D=\frac{3.7\:\text{V}}{1.5\:\text{V}+3.7\:\text{V}}\approx 0.712\$

Now, let's add in another specification you gave:

  • \$I_{_\text{OUT}}=10\:\text{mA}\$

From this, I find:

  • \$I_{_\text{PEAK}}=2\cdot 10\:\text{mA}\cdot\frac1{1-0.712}\approx 70\:\text{mA}\$

Note that this is quite a bit more than \$30\:\mu\text{A}\$.

Now, if you plan to use this output supply only for very short periods of time and can spend a lot of time, beforehand, charging up an output capacitor, then perhaps there's more to discuss. In this case, we can say:

  • \$I_{_\text{OUT}}=\frac12\cdot 30\:\mu\text{A}\cdot\left(1-0.712\right)\approx 4.3\:\mu\text{A}\$

And you may use that to help charge up a capacitor for occasional use.

Unfortunately, some of the earlier assumptions made in developing those earlier relationships I wrote above (an equilibrium state, which is no longer true) also no longer apply in the same way as before. Which means developing new equations for this specific purpose, if so.

I'll leave it there, for now.

  • \$\begingroup\$ Some details added. I'll add my boost design - which is simple for testing purposes only. \$\endgroup\$
    – RobMcN
    Mar 31, 2022 at 19:55
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    \$\begingroup\$ @RobMcN Need some details about the switches and diodes. Those really can't be ignored. Not at these microwatt levels. And I think you are putting a cart before the horse in what you added. But maybe I'm wrong. I'm heading into another meeting in one minute -- an hour long away -- so I may not be able to write much. Oh... meeting canceled. Cool. \$\endgroup\$
    – jonk
    Mar 31, 2022 at 19:58

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