Lm2679-ADJ, will it work ?

Thread Starter

edmundopt

Joined May 4, 2011
60
Hello

first of all , thks for all the help, this is a very good forum.

I have build my switching power circuit, so I will post it here, and I accept any sugestions, if I have any mistakes please report.
If this is very bad, please try to understand that it's my first switching design :)

Here is the BOM details:

qtd - label - details
2 - J1 + j2 - connector 2.5mm metric 5mm
1 - R0 - resistor zener D1
1 - D1 - zener for MOSFET
1 - Q1 - Mosfet P-channel FQPF47P06
3 - C3 , C2 , C1 - capacitor elect CIN 15uF 50V
1 - C4 - capacitor CINx .47uF 50V
1 - D2 - diode schotky SB560(sugested is 80SQ045, but i has 2mA of leak current
1 - R1 - resistor 1kOhm 1/4w
1 - R2 - pot 10k (5 volts 3.132ohm) package PRE-SQ2
1 - U1 - IC DC LM2679-ADJ
1 - R3 - resistor 7425ohm
1 - C6 - capacitor boost 0,01uF
1 - C5 - capacitor soft start 3.3nF
1 - L1 - indutor toroidal 22uh 5.22A
1 - C7 - capacitor COUT ceram 0.47uf
2 - C8, c9 - capacitor COUT elec 180uF 50V
1 - R4 - resistor led 1Kohm
1 - D3 - diode LED, 10mA

Designed for use with input voltage of regulated DC 12volts or regulated Dc of 24volts
Output voltage, fine adjust between 5 and 5.5volts( best case scenario of 2.5amps, worst case scenario of 3.2amps)
 

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mcgyvr

Joined Oct 15, 2009
5,394
Did you calculate worst case dissipation to see if a heatsink will be needed? You don't have much copper area for dissipation there.. (might be close to 40deg C rise over ambient)

Have you seen the "webench designer" on the TI.com site.. You simply enter Vin/Vout/Iout and it virtually builds the schematic for you.. (its slightly different from what you have)
 

RichardO

Joined May 4, 2013
2,270
A word of warning... When a pot used as a variable resistor, you should always connect the wiper to one end of the pot. This is important. Without this connection, if the wiper ever goes high resistance then the resistance of the pot goes way higher than the value of the pot. In your case it would cause a very high output voltage.
 

Thread Starter

edmundopt

Joined May 4, 2011
60
the package I choose for the IC is very similiar to TO220, and at the back there's some space for the heatsink. RDS(on) of the IC is 120mOhm, so the worst case scenario would be 3.5Amps P = 3.5^2 * 0.120; P = 1,47W,but I do not know the formula were the 260khz switching goes with the math:/ (some lack of knowledge here...)

I've started with the webbench, then I have a look at their SNVA013D, wich features a version of whith 5volts 5amps, and then, I had a extensive look at the datasheet wich is so good that I made all the calculations myself, I oversized the inductor, choose a diode with less leakage current.
I've just notice that the trough whole capacitators value is wrong.. will update, thks
 

Thread Starter

edmundopt

Joined May 4, 2011
60
A word of warning... When a pot used as a variable resistor, you should always connect the wiper to one end of the pot. This is important. Without this connection, if the wiper ever goes high resistance then the resistance of the pot goes way higher than the value of the pot. In your case it would cause a very high output voltage.
If I read you correct I shall conect the middle pin of the pot to the unused pin ? I've just tried that with another pot and the resistance value is the same, were can I read more about it ?
 

Thread Starter

edmundopt

Joined May 4, 2011
60
I run into a major problem, I do not know to calculate the IRMS of a capacitor based on the datasheet impedance, I've tryed IRMS = VRMS / Z, using the UPM1V222MHD1AA from nichicon, that as a working voltage of 35volts and a impedance of 0.020ohm, the value is of 1750amps, so what is the formula ? the datasheet form LM2679 rates this capactior at 2.68Amps.

Edit:
I have found some info, I will study it and then try to calculate the capacitors value properly. thks
 
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#12

Joined Nov 30, 2010
18,224
The problem with the pot is in case of failure some time from now. The wiper gets dirty and does not connect. You should connect the wiper to the "unused" pin so it will not be unused.
 

RichardO

Joined May 4, 2013
2,270
If I read you correct I shall conect the middle pin of the pot to the unused pin ?
Yes, you understood right. I know it looks odd. :)


I've just tried that with another pot and the resistance value is the same, were can I read more about it ?
As you saw the action of the pot under normal circumstances in completely unchanged. But -- The wiper can get noisy if the pot is adjusted a lot or when it just gets old. With the extra connection, even when the wiper goes open circuit the worst that can happen is the resistance becomes the value of the pot.

I am not sure that I have ever read this anywhere. If I have, it was so long ago I have forgotten where. One possible place I may have read about it is in the book Troubleshooting Analog Circuits by Robert Pease.
 

ronv

Joined Nov 12, 2008
3,770
The cap needs to supply 1/2 the output current, so it is okay. The layout for the IC doesn't look correct. Having said that if you can get about a square inch of copper for the heatsink it will be about 50C per watt. It will be toasty, but within spec.
 

Thread Starter

edmundopt

Joined May 4, 2011
60
Thank for all your replyes, I have mistake the slot were the card goes, and I had to redraw the whole board, it is bolted at the top part, the part visible is the bottom one, were the conectors and led diode are.

The pot pins are corrected, the capacitors are still a work in study(progress), I have two drawings, one has a ground plane, the has a ground path, please advise the best one, or a third one from your sugestions :)
 

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Thread Starter

edmundopt

Joined May 4, 2011
60
The value for the input capacitors is for Cin are 3 * Panasonic 120uF 50volts with 1,14Arms, and for Cout 2 * Panasonic 50volts 220uf with 1.65Arms
 

bountyhunter

Joined Sep 7, 2009
2,512
Thank for all your replyes, I have mistake the slot were the card goes, and I had to redraw the whole board, it is bolted at the top part, the part visible is the bottom one, were the conectors and led diode are.

The pot pins are corrected, the capacitors are still a work in study(progress), I have two drawings, one has a ground plane, the has a ground path, please advise the best one, or a third one from your sugestions :)
The ground pin of the IC is far away from the ground points of the input and output connectors, which is the worst possible approach.

Some basic rules for doing switcher layouts:

1) Identify the paths of high switching currents, such as between the input/output caps, catch diode, switch devices, etc. maker SURE these paths are minimum length and make sure they don't go along ground paths which will make circuitry bounce.

2) Keep key ground nodes together for a "single point ground". The ground of the IC, and the input/output ground must be close to gether.

What you have now is a ground loop running around the board between the power connectors and the IC. Remember, the IC regulates with respect to it's ground node. If that node is bouncing with respect to the output ground node, your output will be noisy, bouncing, and vary with loading.


The best approach is a two layer board where switching currents are serviced on one layer and the other layer is the clean ground plane. IMHO, getting a single layer layout to work (and be clean and stable) will be very difficult, but may be possible.

Attached is a reference showing the main current paths in the buck converter. In your case remember that the source of "VIN" current will be from the three input capacitors.
 

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Thread Starter

edmundopt

Joined May 4, 2011
60
Thks, will work based on your sugestions, I can make it 2 layer, but it will take two weeks to make the pcb, and I can only make 1 layer pcb.
So I'll try a better design with one layer only.

Can you take a look at the LM2679S-5_EVA, the pcb is specified at the SNVA013D, I will attach to this post
 

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bountyhunter

Joined Sep 7, 2009
2,512
Can you take a look at the LM2679S-5_EVA, the pcb is specified at the SNVA013D, I will attach to this post
That is a better looking layout. The ground for both the input/output caps are close together and the IC ground goes directly to the cap ground via short path. That board was designed at National when I was there.

NOTE: you can see it's laid out for surface mount tantalum caps for CIN and COUT. make SURE the caps you use are good enough for high frequency switching use, generic electrolytics will not work. The data sheet for the parts always have recommended capacitors to use.

You really need to read the data sheet carefully:

http://www.ti.com/lit/ds/symlink/lm2679.pdf

Make sure to read about how the caps/inductor affect loop stability.
 
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RichardO

Joined May 4, 2013
2,270
Thks, will work based on your sugestions, I can make it 2 layer, but it will take two weeks to make the pcb, and I can only make 1 layer pcb.
So I'll try a better design with one layer only.
One way to make your own double sided PCB is to use a double sided unetched panel and only etch one side. The other side becomes your ground plane acessed by putting bare wires through holes in the board.
 

Thread Starter

edmundopt

Joined May 4, 2011
60
These are the requirements, pcb size 100x60 or 100x100, connectors and led at the 100mm side, maximum height 30mm.

so this is what I have to start with(attached requ board)

ok, so the most ugly pcb ever is done, please sugest changes
respecting the lm2679 specifications, the resistors R1 and R2 are not near the pin as specified, I can move the pot, but that space is reserved for the heatsink

capactiors are panasonic FC series, input have a total of 3.2Arms, and output have a total of 5Arms, there's a post here http://forum.allaboutcircuits.com/showthread.php?t=89492

updated BOM:
qtd - label - details
2 - J1 + j2 - connector 2.5mm metric 5mm
1 - R0 - resistor zener D1
1 - D1 - zener for MOSFET
1 - Q1 - Mosfet P-channel FQPF47P06
3 - C2 , C1 - capacitor elect (no value defined)
1 - C3 - capacitor CINx .47uF 50V
1 - D2 - diode schotky SB560(sugested is 80SQ045, but i has 2mA of leak current
1 - R1 - resistor 1kOhm 1/4w
1 - R2 - pot 10k (5 volts 3.132ohm) package PRE-SQ2
1 - U1 - IC DC LM2679-ADJ
1 - R3 - resistor 7425ohm (Imax = 5amps)
1 - C4 - capacitor soft start 3.3nF
1 - C5 - capacitor boost 0,01uF
1 - L1 - inductor toroidal 22uh 5.22A
1 - C6 - capacitor COUT ceram 0.47uf
2 - C7, C8 - capacitor COUT elec (no value defined)
1 - R4 - < 420ohm resistor led
1 - D3 - diodo LED, 30mA max current


 

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bountyhunter

Joined Sep 7, 2009
2,512
Layout isn't terrible. I would move the physical placement of the round filter caps (C1, C2, C7, C8) into the area between the power connectors to reduce the ground path length between them. Main current pulses flow from the input to output caps.

The path distance to the inductor is not so critical but caps are.
 
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