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Danko 30th July 2005 04:01 PM

UCC208 current mode IC usage help NEEDED
 
2 Attachment(s)
Hello Everyone!

I have designed (okay okay, copied :D) a current mode PSU with UCC2808 IC, from Texas. I read all the available documents (how works the current mode, current mode VS. volate mode, etcetcetc).
On the UCC2808 IC's datasheet, there's a DC/DC converter for 5V@10Amps.
I modified a little bit that circuit, to produce 2x50V @ 3A. I did this by changing the resistors neard the TL431, for higher voltage. And for the higher current limit, I changed the voltage divider near the current sense resistor. I reduced 2.8k.

But... When there's no load (just 2 1k soldered under the ouptut capacitors) the output voltage is OK. The feedback is taken from only 50V, not from 2x50V. Oh, and the "50V" is not really 50V, it's 54V, becouse I don't have 1% resistors, and I wouldn't like to use trim.pot.
When I load the regulated 54V with 100ohm, then the voltage is fine, about 54V. But when I change the load from 100ohm to 16ohm, the voltage drops to 40V. Why?

And another thing.
Sometimes the PSU is oscillating. I can hear it from the transformer (ETD49). Maybe, should I solder some bigger capacitor near the TL431, to slow down the feedback?

I attached the original schematic.

Danko 30th July 2005 04:13 PM

2 Attachment(s)
Here is my schematic. You can see how is regulated the output. The difference between the regulated & unreg. output is only a few volts.

Unfortunatelly, this kind of secondary-arrangement, and rectifying is not good :( I realised it, while I was winding the transformer.
The transformer's window area isn't optimally utilized, becouse of that 4 secondary windings.

mzzj 30th July 2005 04:43 PM

Re: UCC208 current mode IC usage help NEEDED
 
Quote:

Originally posted by Danko
Hello Everyone!


When I load the regulated 54V with 100ohm, then the voltage is fine, about 54V. But when I change the load from 100ohm to 16ohm, the voltage drops to 40V. Why?
.

Noise coupling to current sense or feedback? check controller ic all pins for abnormal waveforms or spikes, especially current sense.

How is your board layout look like, it can have bigger importance than actual schematic and component values.

N-Channel 30th July 2005 04:54 PM

Voltage Drop
 
Danko-

Schematically, your circuit looks good. However, how are things we can not see? Like the transformer core size? The wire size? and the MOSFET capacity? What is your input voltage? Is it still the 36-72V as called for in the original Texas Instruments schematic, or is it +12V?

Also, try changing the current sense resistor to a MUCH lower value, like 0.01Ohm. This is because at a 30A draw from the primary side (at full power output), your 0.22 Ohm resistors will develop something like (30A x 0.22 Ohm) = 6.6V at the UCC2808's current sense pin. As I recall, the current-limit threshhold at this pin is 1.0V. A 0.01 Ohm resistor will develop only 300mV across this resistor, and with the current spikes from each cycle, this will just about approach (but not exceed) the '2808's current-sense threshhold. TI should have a good tutorial for calculating the value of the current-sense resistor.

Also, I realize the IRF540s are very robust MOSFETs, but you might try paralleling them, 2 per side of the primary. This will let them run much cooler :cool: at full load with considerably less stress and provide a backup if one MOSFET should fail.

Or better yet, you could try a MOSFET with much lower Rds(on). like the MTP75N06E from OnSemi. This is a 60V, 75A unit with an Rds(on) of only 0.010 Ohms (ten milli-ohms), while the IRF540 is 0.077 Ohms. This is a seven-fold reduction on Power Dissipation :hot: for conduction losses. :D

Hope these tips help. Let me know how these tips work out.

Regards,

Steve

Danko 30th July 2005 05:17 PM

2 Attachment(s)
I've attached the board layout.
The Q1 Q2 are the FETs. They are IRF540N. RDS_on=44mOhm, max. Current is 33Amper. They are mounted on the component side of the PCB, and that big place is used az a heatsink. The another 2 TO22 package are the two diodes. FEP16JT from Vishay. They are mounted the same way, as the FETs. U$5 is the output choke. The transformer is an ETD49. The primary winding is 2x6 turns. The secondary is 4x15 windings. I paralelled 0.4mm diameter CuZ copper wires. 12 paralelled wires in the primer, and /AFAIK/ 6 paralelled wires in the secondary.

This "stuff" is only a testing-circuit, to learn how current mode works, what are the basics, etcetcetc. I'm "in switched mode" for about 1 year. Under this time, I read a loooot of docs, made a few buck regulators, boost-converters.
In this circuit I can play /almost/ without risking my life, and my wallet, with burning expensive 500V/20A MOSFETs :)


Thank you for you patience! :-)

Danko 30th July 2005 05:29 PM

oh, I forgot one thing!

The circuit is designet for 50V input voltage.
I use my powersupply (stabilized, with a current limit) for this.
The UCC2808 has a separated supply, from a small transfomer+7815+Ampermeter.

Eva 30th July 2005 08:48 PM

Could you tell us transformer turn ratios, output inductor and capacitor values, and operation frequency?

Concerning the transformer, a single center-tapped secondary winding would have done the job for symmetrical output.

Danko 31st July 2005 08:04 AM

Quote:

Originally Posted by Danko
The transformer is an ETD49. The primary winding is 2x6 turns. The secondary is 4x15 windings. I paralelled 0.4mm diameter CuZ copper wires. 12 paralelled wires in the primer, and /AFAIK/ 6 paralelled wires in the secondary.

:D

The frequency is about 140kHz.

The ouput choke: 2x40-45 turns on an Al=200 iron powder core.
The maximum induction of that iron powder core is about 780AmperTurns. The output capacitors are 2x1000uF paralelled with 1-1uF WIMA foil capacitors. Unfortunatelly they /the 1000uF/ are not LOW-ESR type, but the foil-capacitors have very low ESR.

I know, that a single center-tapped secondary winding would have done the job for symmetrical output, but haven't got any common anode double-diode at home :(
Using 4 diodes would infrease the PCB area.
In the next version PCB i'm going to figure out somethin else, becouse this arrangement of secondary winding wastes a lot of space on the transformer.

mzzj 31st July 2005 08:14 AM

Quote:

Originally posted by Danko
Quote:

Originally Posted by Danko
The transformer is an ETD49. The primary winding is 2x6 turns. The secondary is 4x15 windings. I paralelled 0.4mm diameter CuZ copper wires. 12 paralelled wires in the primer, and /AFAIK/ 6 paralelled wires in the secondary.

:D

The frequency is about 140kHz.

The ouput choke: 2x40-45 turns on an Al=200 iron powder core.
The maximum induction of that iron powder core is about 780AmperTurns. The output capacitors are 2x1000uF paralelled with 1-1uF WIMA foil capacitors. Unfortunatelly they /the 1000uF/ are not LOW-ESR type, but the foil-capacitors have very low ESR.

I know, that a single center-tapped secondary winding would have done the job for symmetrical output, but haven't got any common anode double-diode at home :(
Using 4 diodes would infrease the PCB area.
In the next version PCB i'm going to figure out somethin else, becouse this arrangement of secondary winding wastes a lot of space on the transformer.


Have you checked waveforms? (and use short ground-lead for scope probe)
CS pin might reveal something on this problem

Eva 31st July 2005 02:51 PM

Well, with : 50V in, 54V out and 6:15 trafo, the converter is going to reflect 125V to each secondary winding and to operate at approx 44% steady state duty cycle when in continuous conduction (200V output diodes required).

With 280khz oscillator, 125V reflected to the secondary, 54V output and 44% duty cycle, we would get 0.34A p-p current ripple in each 320uH buck inductor. Should the inductance value decrease with increased current, ripple would also increase proportionally.

Loading one of the outputs with 16 ohms, output current is going to be 3.38A, adding half of the inductor ripple we would get 3.54A peak, that would translate into 8.86A reflected to the primary side.

8.86A is going to produce 2V voltage drop across your 0.22 ohm current sense resistor. Taking into consideration the 2k/2.8k attenuator, and neglecting the effect of slope compensation, 1.17V peak would be applied to the CS pin of UCC2808.

However, CS input signal is specified to be between 0 and 500mV for normal operation, and overcurrent protection is triggered somewhere between 700 and 800mV.

Furthermore, if we take 40V output and 16 ohm load and calculate the peak voltage applied to CS pin in these circumstances, we get 0.84V. It's just above the current limiting threshold, so current limiting is taking place.

You should redesign the circuit to use a current limit consistent with your maximum expected load.


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