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UCC28084: CTRL and CS roles in the current mode feedback

Part Number: UCC28084
Other Parts Discussed in Thread: LM25019, TL431

Hello,

This is my first time fully designing an isolated power supply. I am struggling with managing all of the design constraints: magnetics, power, slope compensation, isolated feedback, etc.
I am having some difficulty understanding the full chain of events that controls the OUTA and OUTB duty cycling on the UCC28084. I will do my best to explain what I think is happening and any guidance or correction would be much appreciated.

The design I am trying to achieve is:
UCC28084
Power: 0.5 - 5.0 W (1.0 W nominal)
Vin: 12-48 V (24 V nominal)
Vout: Isolated 5 V (push-pull)
VDD: 10 V auxiliary supply provided from LM25019 (this is the voltage supplied to R.p that sets the CTRL voltage through the current from the output of the PS2801-1 opto-coupler)
f.sw = 600 kHz (I am hoping to adjust this for slope compensation / efficiency once my understanding is stronger)
Isolated type II feedback with TL431 and PS2808-1 opto-coupler (1.75 < CTR < 2.85 for 9 mA diode current, 10 mA into cathode of TL431 with 1 mA from a bias resistor)
Coilcraft HPH1-1400L with 4 windings on primary and 2 windings on secondary for turns ratio of 2:1
FDT86246 power switching MOSFETs


I used SLUA584A to set a reasonable value for R.f = 499 Ω (R.ramp) and R.CS = 1.13 Ω
I am trying to use many resources (TI SMPS Comp Made Easy, Ridley Designer Series XV: Designing with TL431, and many others)
To start my slope compensation I need to define the power stage first. I am having difficulty defining the gain A.vc of the power stage.
The way I understand it is the upward current ramp of the output buck inductor (referred to the primary) travels through the MOSFET and R.CS creating a voltage ramp
Additional 'downslope' voltage ramp is added from 5*I.set*R.f (in my case 150 uA * 499 Ω = 0.075 V)
This voltage ramp should peak below 0.47 V as the maximum input signal to the CS pin
This voltage ramp, +0.3 V (from the functional block diagram), is compared against the CTRL voltage * (60/(60+80) = 0.429*V.CTRL
One of the things I do not fully understand is that 0.47 V + 0.3 V = 0.77 V is the maximum CS input into the PWM comparator.
But that means my control voltage can't be larger than 0.77 / 0.429 = 1.80 V (is this true? If so my auxiliary supply of 10 V might be too large for the opto-coupler to bring the CTRL voltage low enough?)
Is the Current Sense Gain from the datasheet 1.9 < A < 2.5 V/V the equivalent of inverting the voltage divider on the CTRL signal? 1/0.429 = 2.33 So this is just accounting for 1% tolerance of the resistors?
What is the meaning of the CTRL to CS offset voltage in the datasheet? 0.70 V nominal (It would be silly to add this as an additional offset to the 0.3 Vref in the block diagram?)

I tried to organise this information as best I can, but with some of my holes in understanding I can read that I've rambled a bit. I hope this sheds some light on what level of understanding I have and someone can help me get the rest of the way there.

Thank you in advance for your time and comments. Kind regards,


  • Hi Kyle,

    1. V_CTRL will be set by the optocoupler current to regulate output voltage (through the resistor between VDD and CTRL). 10V for your auxiliary is a good value.

    2. Current sense gain is nothing but VCTRL/VCS. You can also estimate it with the voltage divider and +0.3V. 

    3. CTRL to CS offset is nothing but VCTRL when VCS=0V.

    Thank you

  • Hello Manuel,

    Thank you very much for your reply. If I may I just want to ask a couple clarifying questions:

    This makes sense and agrees with my understanding. One issue I am having is knowing what steady state voltage on the CTRL pin sets my nominal output of 5.0 V. I tried running the UCC28084 in open-loop, with no opto-coupler feedback from the output, using a potentiometer between VCC and CTRL so see if I could find a CTRL voltage that brings my output value to 5.0 V. Is this the normal way to do this? I'm worried I might damage some of the components in open-loop, for example the push-pull configuration can have flux imbalance which current mode control is supposed to help prevent.

  • Hi Kyle,

    If you want to test the IC in open loop, you can use a potentiometer from VCC. Make sure VDD is powered with a voltage higher than UVLO. Set you switching frequency fsw with RT resistor (check Datasheet). You can leave Iset floating and in the case of VCS pin you can connect a resistor from VCC (check typical value VCS in Datasheet). 

    What you will expect is PWM pulses at the output A and B with 1V high state (see below) with a fsw set by RT.

    Thank you.