EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: marios on March 27, 2011, 05:02:41 pm
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hello everybody. i have as a project to find design a dc converter, i have done everything so far but i can figure out how to find the efficiency i saw the episode where David did something similar but i didnt understand how he did it.
is it just divide output over input and multiply by a hundred??
please help
thank you in advance
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the efficiency is typically on the datasheet but you can find it by mearuring voltage and current in and out and calculating the percentage (Wo/Wi)*100
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yep, no more magic there. Of course the efficiency differs with load and over parameters like input voltage so if you have an operating circuit check input and output power in at least few different cases of output current (load) and input voltage (actually it's about changing difference between input and output).
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Make sure you are using the real power in Simon's formula.
Measure the true RMS voltage and current at the input and calculate Wi. Then repeat for output with a fixed dummy load to calculate Wo. You can change the input voltage or output load and do a sweep for an effciency curve.
If you have overspecified input bypass and output filter capacitors then a simple multimeter can be used as all AC components would have been filtered out, otherwise you should be using an oscilloscope or true-RMS multimeter with high bandwidth.
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I don't know what the common practice is, but I'd measure RMS input and DC output (at least for voltage).
You usually don't want to take count in the reactive power (which will be with RMS output measurement), otherwise the converter will gain efficiency as the output ripple increases.
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I have built the circuit in a PCB without load it delivers 14.96v 0.15A but when i connect a 100ohm 2W Load it has as an output 10.96volt and 0.15amps, this is logic right? so this has as efficiency around 73%
using ((Vout*Iout)/(Vin*Iin)) *100
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I have built the circuit in a PCB without load it delivers 14.96v 0.15A but when i connect a 100ohm 2W Load it has as an output 10.96volt and 0.15amps, this is logic right? so this has as efficiency around 73%
using ((Vout*Iout)/(Vin*Iin)) *100
No. You need to measure the voltage and current going into the circuit and the voltage and current going out of the circuit.
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can you explain this a little bit more? is that you are saying what i did? or you mean to measure the input voltage and current and output voltage and current while i have the load connected?
thanks for replying
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I have built the circuit in a PCB without load it delivers 14.96v 0.15A but when i connect a 100ohm 2W Load it has as an output 10.96volt and 0.15amps, this is logic right? so this has as efficiency around 73%
using ((Vout*Iout)/(Vin*Iin)) *100
No. You need to measure the voltage and current going into the circuit and the voltage and current going out of the circuit.
can you explain this a little bit more? is that you are saying what i did? or you mean to measure the input voltage and current and output voltage and current while i have the load connected?
thanks for replying
Of course, when unloaded, the efficiency must be null, since no power exits the converter. Your measurement is very strange, as you say that without load you had the same current as with load...
Connect the load and measure both input V,I and output (load) V,I.
If the converter is regulated, voltage shouldn't vary so much between unloaded and loaded condition.
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do exactly as scrat said. When you have no load automatically you have no current due to open circuit. Connect the load and then measure both input and output currents and voltages. The measurement itself will bring a small error but it's not important for now.
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i am using the MC34063A for control the switching time(i attached the Datasheet). The new measurments i took came out like this Vout 9.61Volt and current 0.10 A the input voltage is 5 and the current 0.211A, (all this when i have the load connected). hence the efficiency is about 91.09%(its seems good if what i am doing is correct )
but my question is that i desing my DC converter to have an output of 15V when i connect the load the output is 9.61V which is way bellow of what i need? that is not correct right? is my load too big? how can i measure how many watts load/resistor should i used?
thanks
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Without more information it is hard to say - The first thing to do is check that the current sense resistor is correct - if you have set it too high the switcher will turn off before you get the correct current so the output voltage will be low. Remember that the current sense resistor is for sensing the PEAK current through the switching FET, not the output current.
If it is correct then you may have problems with noise getting into the feedback path and triggering the device too early. Try adding a small capacitor to ground to stop this. It might be worth adding one to the current sense as well, switching currents can sometimes have spikes on them which will cause the overcurrent detect to trigger early.
Neil
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you're using Step-Up topology like on page 5 ? For test you can connect pin 7 directly to 6 without Rsc
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yes i am using a step up topology. i will try that tomorrow and i will let you know guys thanks for the help. the fact is that i am using an Rsc resistor of 1ohm although in the calculations the value for the Rsc resistor came out as 0.5.
Neilm where should a connect the capacitor to the ground, from which point?
i will attach some waveforms i got from the diode the first one is with the load and the second is without and load connected
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try adding cap to pin 5. If the calculated value of Rsc is 0.5 and yours is 1 ohm then just connect other one in parallel. Although I would get rid of that at least for now (short it).
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i did connect a 10microfarad cap on pin 5 and change the Rsc resistor to 0?33J 5W resistor(it was the only only i could find in the lab), the Vout was incresad to 14.67V but the load(100ohm 2W resistor) is getting hot.
any suggestions? shall i replace the Rsc resistor with only a wire? the calculated value was 0.5ohms
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The load resistor must get hot, it is over its rated power (15^2/100 = 2.24W)!
Rsc is for switch safety, it doesn't have to interfere with control under nominal conditions (no shorted output). Increasing the Rsc value means lowering the maximum current on the switch.
If you post the values of the components in your circuit it could be easier to find if there's something wrong.
In particular, it could be useful measuring the current waveform on the switch (or the voltage on the known value Rsc), to know if the current sense is the problem. The loaded waveform you posted is very strange, it looks like the switch is off almost all the time.
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hello scart i used the formulas from the datasheet of MC34063 to calculate my values so
ton/toff time was 2.7
using as frequency 100kHZ then the ton+toff 10?s
making toff=3.7?s and ton 7.3?s
Ct=292pF i used a 330 because it was the only available
Ipk=0.74A so diode shoud be a least 1A (i am using Schottky diode 1N5817)
Rsc=0.3/0.74=0.4 Ohms
Lmin=39?H
Cout=66?H
and feedback resistors 110K with 10K
I tried to remove the Rsc and just replace it with a wire but this result to burn the MC34063 I order new ones and hope i get them soon
since i burned the ic i cant get any measurements today, hope i have a new by tomorrow or before the end of the week.
Scart my email is mariosmen@hotmail.co.uk if you have any more info or tips for DC converter and the MC34063 we can talk from there
thanks a lot
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There are at least two other threads on the MC34063 here.
The best way for communication is inside this thread, since so anyone other needing help will find info without asking.
Please post the V,I specs (VinMIN, VinMAX, Vout, IoutMIN, IoutMAX), and the configuration you are using (you can just tell which figure of the datasheet or of the app note you are following).
The Ton/Toff ratio you posted seems strange, if (as I guess, but don't know for sure) you are trying to convert from 5V to 15V.
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yes that is what i am trying to do i want to go from 5V dc to 15V dc and i am using the step up configuration
for the ton/toff i used the formula
the Vin min and max is +-10%
(Vout+VF-Vin.min)/Vin.min-Vsat
(15+0.4-4.5)/(4.5-0.5)=2.725
Vf is 0.4 from the VI curve for the diode- forward voltage of the diode
and Vsat is the output switch from the data sheet which is actually 0.45 but it does make any difference using 0.5
i want to draw current up to 100mA and the voltage ripple to be close to 100mV(which i think it could be an acceptable value)
I think i am on a good way but today i shorted circuit something so the IC got burnt.
after i change the Rsc the Vout was increase to an acceptable level of 14.76V which is not that bad which if i use the power formula is V^2/R is 2.17W
BUT unfortunately i cant do any further work until i got new MC34063
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when you replace IC try to use potentiometer instead of one feedback resistors (or in series with a smaller value) and check if you can trim the output voltage to 15V. Also if it's possible try to change input voltage or at least change the load and then check the output voltage. If it drops or increases significantly than something is seriously wrong. And could you give us exact model of inductor you've used ? Maybe there is some problem with saturation.
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try adding cap to pin 5. If the calculated value of Rsc is 0.5 and yours is 1 ohm then just connect other one in parallel. Although I would get rid of that at least for now (short it).
neoone can you explain that a little bit more for me what does this capacitor will do, is it just to smooth the output a little bit?
in the Datasheet the MC34063 is saying something about an optional filter,i guess is for reducing the ripple, but is it the same like the capacitor in pin5.
i haven't receive the IC yet i will just do a little more research now :)
thanks
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well like Neilm said there is possibility that there is some noise on feedback so the cap would filter it and also you can consider it like a small delay between that what is happening on output and IC's reaction so it's operation won't be so "violent" (if that is indeed where the problem lies).
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Hello guys i did it and its working very good. now i am trying to measure my output ripple. i connet the ocilloscope the ground leg on my source ground and the other leg on my output but i get a line in the screen its not like the usual ones you should get
i think i am doing it wrong. any suggestions how should i connect the ocilloscope, my ciruit is still in breadboard
thanks
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post a photo of the oscilloscope screen
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VOLTAGE RIPPLE
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hm I'm no expert but that looks more like actual switching spikes not output ripple
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yes I would say so to. It looks like there's almost no ripple and low pass filter should get rid of that spikes. They are of quite high frequency so it shouldn't be a problem. Do already have any filter on the output ?
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no i haven't put any filter in the output yet, but because i am doing this for a small project, i will make just a comment of that it can be use a low pass filter as a further development,
so do you have nay suggestions for further development that may improve the converter?
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try adding some inductor for example 1uH in series and capacitor 100nF to ground after that and optionally 100uF as well. If you have more inductors with the same value as used in DC-DC you can use that for now.
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I built an dc converter and first i used a 47?? inductor but then i change it and used a higher value 170??
is it possible to have less input power when i am using a higher value of inductor? but the output power of the converter at the same loading condition to be the same(1.5 W) with both inductors.
thanks