I am trying to make a boost converter using UC3843 with the following

  • the input voltage = 24 V
  • the input current = 3 A
  • the output I want variable = from 120-235 V
  • the current output = 0.3 A

I will use MOSFET IRF840 and diode UF4007.

Did did my design according to Texas Instrument PDF for boost converter: https://www.ti.com/lit/an/slva372d/slva372d.pdf

enter image description here

My calculations:

  • max duty cycle = 1- (V in min / V out max)* η
  • max duty cycle = 1- (24/ 235) *0.8 = 0.718 = 72%

I assumed η = 80%

  • I switch = I coil = I diode = ΔI/2 + I out max/ 1- D
  • I switch = I coil = I diode = 0.03 + 0.3/ 1-0.72 = 1.1A

(ΔI/2 = I load min = 10% of the I max = 0.03A)

  • inductor (L) = V in * (V out max - V in min) / ΔIFsV out)
  • inductor (L) = 24 * (235 - 24 ) / 0.06 * 100k * 235 = 3.5 mH

Fs = 100 kHz , ΔI = 2*I load min

I add R7 to the schematic just to make the circuit CCM boost converter (I am not sure about its value yet!).

What do you think about this boost converter will work?

  • 2
    \$\begingroup\$ Why boost over flyback? Flyback is much better here. \$\endgroup\$ Oct 12, 2023 at 11:57
  • 2
    \$\begingroup\$ For output adjustment the best way is to adjust the low-side divider resistor only. Because the low-side divider resistor doesn't appear in the transfer function, so the risk of affecting stability and overall performance can be minimised. To do this, short the wiper (centre) and one leg of the pot, and run this junction to FB input of the IC. You may need to re-calculate the divider resistors. \$\endgroup\$ Oct 12, 2023 at 12:12
  • 2
    \$\begingroup\$ What's the lowest current taken by the load where you feel you still need to remain in continuous conduction mode? Stating what the input current is does not solve anything (unless you mean it to be a maximum peak value). I suspect that your inductance might be 10x too high in value. You need to state what the minimum and maximum input voltages might be. You should put your calculated values against those formulas otherwise they don't contribute anything to your question. \$\endgroup\$
    – Andy aka
    Oct 12, 2023 at 12:41
  • 1
    \$\begingroup\$ R2 would generally be called the "high-side" position. Regarding CCM, it is unsuitable for peak current mode control; DCM is the simplest mode to implement here. Mild CCM (down to maybe 50 to 30% ripple fraction) is usable when slope compensation is included, which requires a modified circuit. \$\endgroup\$ Oct 12, 2023 at 12:46
  • 1
    \$\begingroup\$ @TimWilliams If i removed the R7 or add low resistor, the boost will be in DCM mode? \$\endgroup\$
    – Tito
    Oct 12, 2023 at 13:38

2 Answers 2


I would make several changes:

  • As commented, a transformer is still better. Even a 1:1 ratio would be helpful, e.g. a coupled inductor with the windings in series. Closer to 1:3 or 1:4 would be ideal, and these do exist, though with poor selection, and maybe not actually in this power level (50-100W). But 1:1s are pretty abundant.

    This is simple enough to implement, using the calculated inductance for the primary only:


simulate this circuit – Schematic created using CircuitLab

(A snubber (R+C) across the diode may be necessary, to deal with leakage inductance LL. This will have value \$C \geq 3 C_J\$ and \$R = \sqrt{\frac{L_L}{C_J}}\$, where CJ is the diode's capacitance.)

Anyway, proceeding assuming no transformer, just pure boost:

  • Inductance should be lower as mentioned, to move into DCM, or low CCM. 50µH is about right. You will most likely need slope compensation at low Vin and max output, or to reduce max output a bit to maintain DCM.

  • Most critically, Vmax is 20V, but Vin is 24V. See UCC28C43 Datasheet. It's not clear between your schematic and text which version of controller you intend to use. In any case, over 15V is quite high for gate drive purposes, and you will dissipate excess power developing such gate voltage swing.

    Simply put a 7812 regulator in front, to reduce supply to a more reasonable level. You'll need a 1uF or so bypass cap on its output / controller's supply. Two added components, easy fix.

  • IRF840 is rather small for this purpose. At nearly an ohm Rds(on), but needing 5 or 6A peak, the circuit will be very inefficient at full load. (With a transformer, peak voltage is reduced by the ratio, and something like IRF640 would suffice.) Consider something of lower Rds(on).

  • R6 is too large; for peak currents nearing 6A, 1V/6A = 167mΩ would be more typical. Less still for slope compensation. Which...

  • Slope compensation can be added per the datasheet page 20:

enter image description here

Notice the oft-quoted erroneous caution about 50% duty. There is absolutely nothing special about the 50% point in an ordinary boost or flyback converter. The reason this figure is quoted, is it's the typical design value for a flyback, i.e. you set the turns ratio so that, whatever your input and output voltages are, you get near 50% duty cycle under nominal/max conditions. Or when setting up a basic boost circuit for analysis, setting Vo = 2 Vin is a common starting point, to the same effect.

What they actually mean to say is what happens when crossing that level in a nominal-50%-max design. Current goes continuous. That's it. That's all they need to say, CCM. But everyone repeats the "50%" without giving it a single thought.

And in case it sounds like I'm being harsh on the authors, it is warranted; application notes, in general, are notoriously poor, and must always be read critically. They are, at best, starting points for further analysis and research. It's unfortunate that better information is not provided, but that is unfortunately the state of things. TI does provide a "Document Feedback" link and you are welcome to report anything you see in error. (But don't get your hopes up; I've yet to see anything I've reported, get updated.)

In any case, since you may enter CCM at max load, slope compensation may be a welcome addition. Typically Rcsf = 1k, Ccsf = 100pF or so (depends on output switching characteristics), and Rramp such that the CS pin is biased up to say 0.5-1V over the ramp waveform. Something in the 2-10k range. It's only two added components (emitter follower and dropper resistor) so it's well worth considering.

Note that slope comp superimposes upon the current sense waveform itself, thus reducing its range. Which means you need less voltage drop across R6, which is nice to improve efficiency as well. (The same biasing effect can be done without slope comp, by just wiring a resistor from VREF to CS; biasing up to about 700mV is reasonable, beyond which internal tolerances/errors take over, and if you need to push sense voltage lower than this, I would recommend a newer controller or regulator instead.)

  • Compensation components (R8, C5) can be improved. Most likely their value needs to be changed anyway, and you will want to do load step transient analysis to prove this out. Most beneficially, R8 should be in series with C5, not parallel. I have no idea why the appnotes all show them in parallel -- it is strictly worse for overall performance, it's just easier to compensate (because it kills loop gain). Chalk it up to the same appnote cynicism, I suppose.

As for tuning, typically start with a resistor around say 10 or 100k, and a too-large capacitor, like 100nF. First check that the output is regulating, stable, and it should be overdamped in step load response. Reduce C until it starts ringing, then adjust R to dampen out the ringing. Steps can be pretty chunky, like factor of 2 here or there; compensation isn't terribly precise, at least until you get close to optimal. Repeat until the transient is as fast as you can get, or are comfortable with, while being reasonably well damped (usually air a bit on the overdamped side, to account for possible load reactance). You may also need to change output filter capacitance in the process.

Speaking of output filter cap, you probably want some ceramic or film caps in parallel there; electrolytic ESR will be particularly relevant to output ripple voltage and efficiency (and thus its own lifetime due to self heating(!!)). (This is again due to the large step-up ratio.)

You can wire up a step load testing circuit by driving an IRF840 with a NE555 or whatever, and using that to switch a load resistor. Drive it at a low frequency, say 100Hz, slow enough to see the transient response of the supply. Put another load resistor in parallel for "idle" load, and size the switched resistor to bring it up closer to nominal-max load when on. A typical load step condition is 50% "idle", 100% during the step. Percentage of nominal max Iout, that is.

  • \$\begingroup\$ thanks Tim for your explanation, i will consider that, i just want to clarify iam using uc3843 not uuc28xx as i mentioned in the topic address (using uuc28xx in schematic because i did not found uc3843 spice model) anyway both are really quite similar but the uc8343 allow Vcc up to 30V but anyway iam going to use lm7812 as you mention, but i have a question, should i calculate the input power with 12v or 24v after i use lm7812? \$\endgroup\$
    – Tito
    Oct 13, 2023 at 12:23
  • \$\begingroup\$ Calculate what input power? The inductor shall still be supplied from 24V. Only input to the 3843 goes through the 7812. \$\endgroup\$ Oct 13, 2023 at 23:36

What do you think about this boost converter will work?

You appear to have used 0.06 amps as the ripple current (ΔI) in your equations and, of course, that will yield an inductor value of 3.5 mH but, the "normal" thing is to keep the converter operating in CCM at low load currents without going silly on how low that current might be. So, I asked you what's the lowest current taken by the load where you feel you still need to remain in continuous conduction mode and, you said this: -

the lowest current is assumed it as 10% of the max out current = 0.3 × 0.1 = 0.03

OK, so 30 mA would be reasonable as a low current load value but, that doesn't mean you can assume the ripple current should be 0.06 amps (well, that's what you appear to have done).

Assuming 60 mA ripple current leads you to an inductance value that is probably nearly 8x what it could be. Consider these two scenarios from my basic website boost calculator. The first scenario is the light load situation i.e. 235 volts output and a load of 7830 Ω (30 mA load current): -

enter image description here

I used a 470 μH inductor above and, as you can see, the circuit remains in CCM (desirable) at a load current of 30 mA. If you look at the peak and minimum inductor current values you will see that the ripple current (ΔI) is 458 mA.

Now if I drop the load resistance to 783 Ω we see this (300 mA load current): -

enter image description here

Ripple current (ΔI) = 3.168 amps minus 2.710 amps = 458 mA and, this is totally as expected when remaining in CCM.

The point I'm making is that you have mysteriously assumed an inductance value that is around 8 times higher than what you probably need. If I plugged a value of 3500 μH into the calculator, the peak current is 2.97 amps and the minimum current is 2.908 amps. This makes ΔI 62 mA but, it means you have a much more expensive inductor.

Try and get the inductance as low as possible so that it doesn't become overly expensive or, doesn't deliver the goods.

My basic website calculator is for an ideal converter but, it won't be more than 10% out on the inductor value calculation.

  • \$\begingroup\$ after researches i found you are right i was using large inductance i do not need, i also found that DCM mode will be better in this case as Tim Williams explained, i update the circuit schematic do you this the circuit is right now? \$\endgroup\$
    – Tito
    Oct 12, 2023 at 23:30
  • \$\begingroup\$ @Tito regarding the circuit, you must use a simulator and check that the circuit is functional as intended. It's easy for you to make a transcription error and, it's easy for me to miss some error you may have introduced so, double-check your circuit using a simulator. I use a simulator on every circuit (and sub-circuits) I design. It's a no-brainer for me; I just do it. The next step will be PCB layout and this requires a whole lot of other skills so, ask a question on that when you get there. \$\endgroup\$
    – Andy aka
    Oct 13, 2023 at 7:49
  • \$\begingroup\$ @Tito regarding DCM, it's for you to say and not somebody else. That is why I asked you the question: what's the lowest current taken by the load where you feel you still need to remain in continuous conduction mode and you gave me an answer of 30 mA. Maybe you didn't fully understand the question of course. Anyway, if you feel you can drop into DCM at higher than 30 mA, use the tool linked in my answer and play around with the inductor value until it hits the spot. One final note: DCM does produce a noisier regulated voltage so, if this is critical then don't do it. \$\endgroup\$
    – Andy aka
    Oct 13, 2023 at 7:51
  • \$\begingroup\$ @Tito I've also rolled back your question to where it was when I left my answer. Reason: the modifications made my answer look wrong or me incompetent. I'm sure you'll understand. If you want to evolve the question (after an answer is given) then raise a brand new question with the new circuit. \$\endgroup\$
    – Andy aka
    Oct 13, 2023 at 7:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.