low efficiency MC3403

[enridp]

Hi !
I'm trying to power a DC motor with batteries, the motor is 24V and a consumption from 100 to 300mA.
I was thinking in using 9V (6 AA batteries) and boosting to 25V with a MC3403.
But calculating the components with this page:
**broken link removed**
I found that from 9 to 25V I have a max Iout of 240mA (Imax supported is 1.5A for MC3403), so the efficiency is:
[(25*0.24)/(9*1.5)] = 44% !!

Am I doing something wrong? I thought that MC3403 was a very efficient DC-DC converter.

Thanks !!
Enrique.

 

[Genomerics]

The tool is applying the absolute maximum rating of the switch current, 1.5A, to define its limit and then appears to be making some strange choices as to component values. It also does not give you much design insight since the author is not telling you what is actually going on in the background. The final result is not indicative of achievable circuit efficiency and that might still not be overly impressive.

https://www.onsemi.com/pub_link/Collateral/MC34063A-D.PDF

I'd be inclined to avoid this IC like the plague...... I think it is some old legacy thing that got designed into stuff in the past and is refusing to die because masochists love to prove their worth by beating their heads over it only because it will satisfy their sadistic side because others will see it being used and think it might be a 'good idea'.

Still I suppose it works down to 3Volts and might be cheap.

ARGHHHHHH ARGHHHHHHH ARGHHHHHHH RUN AWAY RUN AWAY.

Sorry about that one, might be back later.

Genome.

 

[enridp]

I think the script of the webpage is only using the math equations from the datasheet.

But if MC34063 is not a good option, which is the good option then?

 

[Genomerics]

Sorry. I went a bit fruit bat.

The 'tool' is telling you that you have exceeded the maximum current rating of the switch inside the IC. Your boost converter will operate at a duty cycle of,

D = [VOUT - VIN]/VOUT

Or 0.64 in your case. With a maximum 1.5A available from the switch then the most you might hope to draw from the supply would be Ipk.D, an average, or 0.96A or about 8.5W and that assumes everything is ideal. By the time you start having to deal with 'worse case' and other things then you will not be able to realise what you are looking for.

I'll try harder later.... maybe. My suggestion may not be much better given it has its own 'quirks' and given things have moved on there is likely to be something more suited.

UCC3803 and an external Mosfet...

https://focus.ti.com/lit/ds/symlink/ucc3805.pdf

Genome.

 

[enridp]

Hi Genome!
I was searching more information about UCC3803, but it seems to be an IC for very low power (less than 2W).
I found this paper from TI:
Dual Output Boost Converter
application notes and circuits for Dual Output Boost Converter
direct link to pdf:
http://www.datasheetdir.com/go/-ml-jklm-ckcx.pdf

It's a boost converter from 5V to bipolar 12V (+12 -12).
If I can use -12 and +12 then I have 24V from 5V.
But the max. power supply is 2.5W, and efficiency is only 60% in that situation.

am I omitting something?

 

[FvM]

I agree with Genomerics, that 34063 isn't the best choice. But it's one of the cheapest switcher ICs on the market and still used in many products. I also choosed it for a new design some time ago - without any masochistic or sadistic intentions. It's serving it's purpose, but I won't suggest it for a DIY design.

1.5 A peak current of the chip isn't identical with the achievable average input current, it has to be reduced according to both duty cycle and designed current ripple. The expectable efficiency will be rather 70 to 80 %, I guess.

 

[Genomerics]

Yes, I have gone a bit over the top. For a given application the 34063 will do the job but it has its limitations.

The UCC380X IC itself is just a controller. What you place around it will determine the available power and efficiency and the example you have found is designed to achieve a certain function which in itself is low power.

Let's say you want to go from 9V to 24V and assume you expect the battery voltage at end of life to have dropped to 6V then assume you want a maximum output current of 0.5A If you ignore voltage drops and losses to begin with then you want 12W output power which will require an average current from your 6V source of 2A. Pick a switching frequency... 100KHz, higher frequency means more switching losses in the mosfet and diode due to parasitic capacitance and reverse recovery but it is a design trade off.

The duty cycle is D = [VOUT - VIN]/VOUT or 0.75 giving a switch on time of 7.5uS. A general starting point is to accept 20% ripple current in the inductor which would be 0.4A from that your inductor value would be L = VIN.TON/dI or 112uH call it 100uH and the ripple becomes 0.45A. The peak inductor/switch current will be 2.225A, average plus half ripple. For the switch treating it as a simple rectangular wave and ignoring the ripple then it's RMS will be 2*SQRT[D] or 1.73A.

Let's say you are happy to lose 0.5W in the mosfet and pretend half that will be switching losses with the other half being conduction losses. You would need a device with an RDSon of .25/1.73^2 or 0.084R or 85mR and be capable of being driven by a 5V gate drive signal . It would need a VDS rating in excess of 24V, plus one diode drop, so use a 30V device... IRLR2703 seems to fit the bill.

http://www.irf.com/product-info/datasheets/data/irlr2703pbf.pdf

You will need a fast diode and it will see peak currents of 2.25A, and some extra for possible transient overload. Its average current will be... simply the output current and its reverse voltage the output voltage. Something like a BYV27-50 will be more than up to the task,

**broken link removed**

You will need a current sense resistor in the source of your Mosfet. Now peak limit in the UCC380X is relatively high at 1V for low voltage supply operation. I might only assume that it has been made that way for compatibility with the older UC384X range. Pick a 0R1 sense resistor value and you will lose 300mW to generate a peak signal of 225mV.

The circuit will need slope compensation and in addition you can use the IC's reference to offset things further in order to match the 1V if required.

So far ignoring other switching losses we have 0.5W mosfet, 0.3W sense resistor. The diode will add about another 0.5W and I'll ignore the other contributions. Total is 1.3W for 12W out so efficiency is about 90%.

Jumping into LTSpice,

Linear Technology - Design Simulation and Device Models

This is not an exact representation since the suggested components are not available in the package.. it does show the basic set up though.



The LT1243 is 'similar' to the UCC3803.

Q1, D2 and R5 supply the slope compensation. During switch off time the current downslope is 18V/100u converted to a voltage of 18,000 in the sense resistor. Half that is 9K. In the LT1243 the ramp amplitude is 1.7V and with Fs at 100K we get 170K so we have to attenuate it by about 20 giving a value for R5 of 20K. R10 offsets the signal to match the internal 1V peak limit of the IC.

Apart from the compensation the rest is more or less as previously described. I'm afraid I've been evil and 'guessed' at values in the feedback network to get something that 'works'.

Start Up,



In regulation,



Output power is 11.93W. Input power is 12.73W. Losses are 0.8W, better than expected but the diode is a Schottky and the Mosfet is specced with a lower RDSon, 93% efficient. Depending on the models LTSpice might underestimate some of the switching losses.

Model attached.

Genome.

 

[enridp]

Wow! thanks for that reply Genome!
I will try to find those components (here in Argentina that's the most difficult part I guess).