LT8393 goes into infinite fault loop

[sadmak]

Hello!

I dont have to much expirience in power electronics, actualy it is my first serious project.
I have an issue with circuit, based on LT8393 buck-boost led driver IC. The design works fine (it reaches the designed 30V, 3A output (the input is 12V, ~8A)), but if a current rises up to fast, the IC jumps into an infinite fault protection loop and cant get back to normal working condition, besides the outpoot mosfet (D mosfet according to datasheet) starts to heet up and the input power of circuit is biger than the output (guess there is a short somewhere).
I suppose the problem is somewhere in the feed back loop.
I have tried a lot of configurations of external RC compensations on Vc pin, but could not reach the deserved output conditions. The best configuration is C= 10pF and R = 300K, the output is 30V, 2A stabil, but further decreese of capacitance gives no result (if some on have expeerience in, how the compensation works in this circuit, i would be glad, if you could explain it to me).
Decreese of Soft start pins capacitance also helps little bit (2,2A), but in that case, the IC ruins very quickly (now the SS pins capacitance is 1uF).
There is also a mistake in design according to datasheet, but i am not shure if it can cause this situation. The point is that the diming function is not used in this configuration, and according to datasheet in this case the PWM pin (11 pin) should be tied to INTVcc pin (pin 8) directly, instead it is tied to Vref pin (pin 12). The RP pin is tied to GND.

I would appreciate any help or addvice, especially in connection with RC conpensation.

 

[wwfeldman]

not familiar with LT8393
as you wrote:

There is also a mistake in design according to datasheet, but i am not shure if it can cause this situation. The point is that the diming function is not used in this configuration, and according to datasheet in this case the PWM pin (11 pin) should be tied to INTVcc pin (pin 8) directly, instead it is tied to Vref pin (pin 12). The RP pin is tied to GND.

i suggest you correct this mistake.
it is never clear from any data sheet what will happen if the circuit is built incorrectly.
there are too many ways things can be wired/built incorrectly that it is impossible to cover all of them.

starts to heet up and the input power of circuit is biger than the output (guess there is a short somewhere).

the input power is always bigger than the output power
there are always losses
do you mean that the efficiency goes down significantly?

 

[sadmak]

Helo, thank you for your help.

About the mistake. It would be the right thing to do, but to do so i have to cut the PCB, and it is more than two layers, so it is a little bit expensive. I would rather leave this at the end.

About power loses. Yes, i would rather say the efficiency lose. In normal working condition the circuit consumes 12V and ~8 - 9A, and the output is 30V and 3A, but if the current increases to fast (for example after connecting a normal 10 ohm resistor, instead of increasing it with electronic load) the output voltage starts to oscilate around 20V and the output current fals to ~100 mA, but the input still 12V and 4A.

 

[wwfeldman]

it may be expensive, but if the circuit is incorrect, there
is not reason to expect it to work correctly

kindly post what it is and what it should be

if you're using through hole parts, you may not have to cut the board
pull the problem components, bend leads of the incorrect circuits up, put the components
back and rewire the correct circuit by hand

if you increase the load slowly, it doesn't oscillate?
does it only oscillate when presented with a step change?

is the output oscillating, or is it going into and out of current limit?

 

[sadmak]

Yeah, may be you are right. I going to trie out to transport all the parts on the evaluation board and correct that mistake.

May be i shouldnt use the oscilate word, because according all signs the IC goes into fault and cant climb out of there(fault pin output, soft start input charges and discharges ...).
I guess there is some spike in feed back loop, but i cant filter it out. I think that a good combination of resistor and condensator on the RC compensation input would solve the problem, but i dont understand how it works.

Under the slow increasing i mean if you increasing current on electronic load in CC with mA -s. Than the out put can reach even 30V, 4-5 A.

 

[BradtheRad]

To slow jumps in current flow... Try installing an inductor in line with the power supply.
As to what Henry value, start with the same as your switching inductor.

 

[sadmak]

Wont it decreese the ammount of current you can gain when you short the inductor?
Because if you put another inductor in serial, with same characteristics you halve the current.

But it is easy to trie out, so i'll give it a shot

Thanks for the help!

 

[KlausST]

Hi

Because if you put another inductor in serial, with same characteristics you halve the current.

Don´t know how you come to this conclusion...Please expalain

Klaus

 

[wwfeldman]

Brad did not mean putting a second inductor in serial with the switching inductor
he meant to put it in series with the output - so that the output current goes
through this new inductor then to the outside world.

 

[sadmak]

Oh I thought on input

Than it is ok.

 

[KlausST]

Hi,

I also thought on input. Still the "half the current" statement makes no sense.

@BradtheRad: please clarify inductor postition.

Klaus

 

[BradtheRad]

By putting a choke on the 12v power supply, current rise is slowed to the IC despite sudden changes in load.
Input capacitor C1 is commonly installed.
My simulation has duty cycle 80 percent.

 

[sadmak]

Hello everyone!

First of all ishould clarifie the IC s operation. It is not buck-boost, rather buck + boost. It senses on FB pin what voltage we want from it and switches mosfets accordingly.

About inductor on input. Well, as i understand the conception of a boost curcuit, it shorts the inductor, and with this move it can acquire huge current on it (because the resistence of inductor is small). If you put another inductor in series, you double the resistance and halves the current (if the second inductor is the same).

The lates news about the circuit. I increased capacitance (150nF -> 4,7uF) in RC stabilizator (Vc pin) and decreesed inpedance (300K -> 20K). Now it can recover the normal operational codnditions, but it oscilates preety strongly if i drain to much current from it. (Now it fals almoust to zero )
Seems this RC circuit works as an average PI controller (Increesed P gives biger oscilation, but decreese offset error).

 

[KlausST]

Hi,

To me it seems the shown circuit does not match your textual description.
If so: this just confuses. Text and pitcure need to match for us to understand it´s operation.

About inductor on input. Well, as i understand the conception of a boost curcuit, it shorts the inductor, and with this move it can acquire huge current on it (because the resistence of inductor is small). If you put another inductor in series, you double the resistance and halves the current (if the second inductor is the same).

* "because of resistance": the resistance of an inductor has about no effect on current.
* the current in an inductor follows this rule: rate of current (in A/s): dI/dt = V / L. The average current depends on operation mode (CCM, DCM)

But ...nobody asked to put an inductor in series with another inductor.
See Brad´s picture above: L1 is not directly connected in series with L2.

About the stability problem: Can you show some photos of your circuit with wiring?

Klaus

 

[sadmak]

About IC -s operation. In my first post I attached the datasheet for controller IC, the whole process of votage conversion is written there precisely.
But in short, in boost mode A MOSFET is always open, B MOSFET is always closed, and D-C MOSFETS switching alternately. In buck mode A and B switches alternating D is always open and C is always close.

About coil current. I guess the maximum current is still defined by resistence of the coil.

ps: Sory for short answeres, I writing from workplace.

 

[KlausST]

Hi

About coil current. I guess the maximum current is still defined by resistence of the coil.

..still they no one said to connect them in series.

**
But:
If you have a LED rated for 20mA. .. now connect a second LED also rated with 20mA in series...what is the maximum current they may withstand now? 10mA?
If you have a 10 fuse and connect another 10 fuse in seires ... will the trip now at 5A?
If you have a coil rated for a maximum current of 10A ... and you put a second one in series....

What is your idea with the coils?

***
One thing I have to agree is: the max coil current is (mainly) defined by it´s resistance.
Still this does not mean the the max current changes if two are in series.

Klaus

 

[sadmak]

Hello every one, sorry for not answering for to long.
Well, my concerns is not about the totaly max current that the coil can withstand without burn out.
The IC watching the current ammount flowing throught the coil with measuring resistor, and switches ounly when it reaches itsmaximmum. So if you put anorher coil in the way of current, the resistence going to doubles, so the maximum current will be halved.

Anyway, I have tried out the output inductor, and it works beter.
My next goal is to make a couple circuits work in a cascad, so shall i put an inductor after each circuit, or rather put a bigger inductor at the end?