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Help with my simple SMPS design

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Electric kid

New Member
Hello,

I have a little problem.
I chose a project for school which I've managed to get working...
All the circuitry before and after the SMPS is working properly,
my only concern here is, that the SMPS is only working at an efficiency of around 54%.

I've added my schematics for the SMPS part... feel free to use it even though its not good :)
http://obrazki.elektroda.pl/6095477100_1493012792.jpg

I would apriciate any help or suggestions to bump the efficiency higher!

Ps. I can't feel any heat beeing generated... so where am I loosing all the power???
PPs. Sorry for my bad english, it's a bit rusty even though I'm 19...
 

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dr pepper

Well-Known Member
Most Helpful Member
Welcome here.

Looks like a reasonable circuit, though you dont show ground connected on the feedback rectifier.
The efficiency depends on the output power, the circuit itself will use some energy, so if your only pulling a watt or so it will be inefficient.
U4 needs a decoupling cap, in fact all ic's would benefit from one.
Also your switching the fets hard a resistor inline with the gate to control turn on time might improve things, 10 - 22 ohm is common.
I'm not sure your method of feedback is that great, using one of the op amps on the '494 might be a better idea, or possibly pin 4 the dead time input on the '494.
Still not a bad try, with a little tweaking it will improve.
 

tomizett

Active Member
Also, do make sure that your efficiency measurements are correct. As you have mentioned, if the efficiency is low then you should feel parts of the circuit getting warm - if that is not happening, perhaps your meter is making incorrect readings due to ripple on the input or output?

No need to apologise for your English - there are plenty of 19 year olds here in England who don't write as well as you...
 

schmitt trigger

Well-Known Member
Also, do make sure that your efficiency measurements are correct. As you have mentioned, if the efficiency is low then you should feel parts of the circuit getting warm - if that is not happening, perhaps your meter is making incorrect readings due to ripple on the input or output?

.
Those are my same thoughts.
You don't mention the total output power, but 54% is a very low efficiency value, and if the measurement is correct then the heat must be going somewhere.

Most likely some measuring instrument is getting fooled by the ripple current.

How are you measuring voltage and current? Show ALL the details.
 

alec_t

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Most Helpful Member
Do you have a link to the MOSFET driver IC? That IC seems unusual in not having the bootstrap cap connected to the high side FET source.
 

kubeek

Well-Known Member
Most Helpful Member
It is L6384 and you are right, there should be a connection between Vout and the middle of the transistors.
 

Electric kid

New Member
Thanks for the replies...

dr pepper, i cant use the op amp in the tl494 otherwise the the output will be low when the opamp triggers, that will set the output of the tl494 to low thus one of the mosfets high during that time... yea the feedback is wierd but thats what i came up with lol. there is actually a 8.2 ohm resistor between the gates and reverse diodes for fast discharges...

its kinda weird the more i push through the circuit the higher are the losses?!?!?

tomizett, I've used 4 flukes for the measurements current voltage on each side... ive tested them at school the are prettey darn accurate.
and i get really ugly spikes at the output ill post pics this afternoon

well thanks but i've still a lot to learn :)
 

Electric kid

New Member
alec_t and kubeek im soooo sorry you are right, there is actually a connection on my breadboard I totally forgot to draw that in there...
 

Cicero

Active Member
Those are my same thoughts.
You don't mention the total output power, but 54% is a very low efficiency value, and if the measurement is correct then the heat must be going somewhere.

Most likely some measuring instrument is getting fooled by the ripple current.

How are you measuring voltage and current? Show ALL the details.
This. Please show all your measurement details.
 

Electric kid

New Member
all at 5v out

7w input 4,5w out

20w input 10w out

40w input 17w out

it is kinda weird, the more i push through the circuit the less effective it gets! shouldn't it be the other way araound?!?!

there is also a big voltage drop! when i leave the smps free (without) a feedback i get a voltage drop from 17 v to 5v when i connect a 1A load to it (input voltage 30v)
 

Cicero

Active Member
Wait, 40W through a breadboard!?

But seriously, if your power measurements are correct, losing 10's of Watts of power is gonna make a lot of heat. It just has to.
 

kubeek

Well-Known Member
Most Helpful Member
tomizett, I've used 4 flukes for the measurements current voltage on each side... ive tested them at school the are prettey darn accurate.
They may be accurate, and they even may be measuring RMS values, but multiplying Irms x Vrms still does not yield real power, there still is the power factor that needs to be considered. You either need to get better filtering on the input and output so that both voltage and current are very close to DC, or you need a meter that calculates the power on the fly.
 

Pommie

Well-Known Member
Most Helpful Member
You do realise that your three 5V outputs will vary dependant on load? Is this what you intended?

Mike.
 

dr pepper

Well-Known Member
Most Helpful Member
If nothing is getting warm and your sure efficiency is pants, then maybe the transformer is a place to start, stray inductance/capacitance and coupling factor will all affect efficiency, however if your dumping lots of energy something will get warm.
 

schmitt trigger

Well-Known Member
all at 5v out

7w input 4,5w out

20w input 10w out

40w input 17w out
If you are truly dissipating 23 watts, something must be getting so hot that you should be able to smell the burning plastic, feel the searing heat and see the smoke curls. And eventually hear a loud pop! and crack!
 

dr pepper

Well-Known Member
Most Helpful Member
Trigs got a point, even the tranny would be hot if it were dumping that energy.
 

Pommie

Well-Known Member
Most Helpful Member
As I pointed out earlier, the outputs aren't regulated to 5V. If he's assuming 5V then his calculations could be miles out.

Mike.
 

ChrisP58

Well-Known Member
Most Helpful Member
How quiet are your input and output voltages and currents at the measurement points? If they're noisy, you're not likely to get valid measurements.
Look at all of those points with an o'scope to ensure that they're nice, flat DC lines under all operating conditions.
 

dr pepper

Well-Known Member
Most Helpful Member
Yes thats true pomm's, you'd have to measure and add effective power on all o/p's to measure efficiency.
 
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