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UCC3808x - Datasheet clarification/error?

Other Parts Discussed in Thread: UCC38086, TL431

I'm working on a consumer product which looks like a perfect fit for a UCC38086.  Summing a sloping compensation current to the current signal filter looks like an excellent idea.  I'm eager to try it but I'm having trouble understanding some of the details in the datasheet:

The Current Sense filter cap, CF in fig 1 is 220uF, application schematic (page 8) is 220pF, figure 7 is 220kOhms.  A value of 220pF makes sense.  Is it 220pF?

Figure 1 labels the signal from the current sense resistor (RS) connected to the top of RF as “VCS”.  This looks like it’s supposed to be VRS?  If it is then figure 1 would match figure 3.

I can’t make out the meaning of the term “RAMP VOLTAGE HEIGHT”, which is in equation 2 on page 6.  If I’m not mistaken this is the crux of understanding the part.  Here where I’m getting confused:

Referring to the internal block diagram on page 5, the resistor connected from Iset to GND (pin2) forms a current sink, whose output is Vct/Rset.  The block diagram labels the + input of the amp as “CT”, but that really should be Vct, which is the voltage developed across CT?  More on this in a moment.

Vct should always range from 0.2V to 1.5V, shouldn’t it (neglecting temperature and delay effects)?  Ct is charged by a current set by Rt and it’s discharged by the fet shown next to Ct?  The window comparator for the osc uses 1.5V and 0.2V as limits.  Is this correct?

Back to Islope: The current sink controlled by Rset is turned into a current source and multiplied by 5 to form Islope, therefore Iset = Vct/Rset and Islope = 5(Vct/Rset)?  Let’s consider the quiescent case of 0 output current (Vrs=0) and Rset=50K.  If Rset=50 then Islope would start at 20uA when Vct=0.2 and rise to a peak of 150uA @ Vct=1.5V?  If the recommended value of Rf=1K is used, then the voltage at the CS pin would rise (after charging CF) from 20mV to 150mV?  In this case would the value of equation 2 for  “RAMP VOLTAGE HEIGHT” be 130mV or 150mV?

Let’s take the case of the output voltage at 0, and the current in the sense resistor causes Vrs to ramp from 0 to 150mV.  The voltage at the CS pin (Vcs) would then ramp from 20mV to 300mV, using the compansating ramp of 150uA described above?

Referring to the “PWM Comparator/Latch” section on page 5, it appears that if the voltage at the CS pin exceeds 0.5V then the PWM latch would be triggered directly through an OR gate and the PWM comparator is not involved.  This is the confusing part for me.  The - (middle -) input to the PWM comparator must then be < 0.5V for the part to be useful?  If the voltage at the CTRL pin exceeds 1.17V then the output of the PWM comparator would never trip the PWM latch??  I thought this part was designed to run when the CTRL pin is pulled low?  How is that supposed to work if the middle - input to the PWM comparator is at 0V??  Wouldn’t the comparator always be high in that case, because of the 300mV offset summed internally to the + input of the PWM comparator?

I would sincerely appreciate your clarification.

Steve

Edit as of 7:18PM:

Correction to that last paragraph, the application schematic shows that the device will run when the CTRL pin is NOT held low, got that backwards.  My basic question still remains, I'm still not clear about the gain of the PWM comparator section.   The + input of the PWM is the terminal where is the ramp is present.  When the + input is greater than the middle - terminal then the PWM latch is set.  If the max input of the CS pin is 0.5V (when the latch is set anyway), then is the full swing at the + input (when duty cycle = 0) when the + input is at 0.8V?  If so then the middle - input should not exceed 0.8?  And the swing from 50% to 0% occur when the occurs when the middle - input of the PWM comparator (which is the slow moving voltage control input) moves from 300mV to 800mV?

Edit as of 7:53PM:

I just found SLUA584A which answered most of my questions regarding the current injection into the filter on the CS pin.  I'm still not clear about the voltage input , specifically the recommended values of the pullup resistor between CTRL and VDD, and how that  value is computed. Should this value be chosen so that - input of the PWM comparator is at in the middle of the range, for example 0.5V?  Figure 1 in the datasheets suggests a value of 10K, but no voltage is given for the VDD pin.  A 10K value would set a resistor divider ratio of 0.4 from the CTRL pin to the - input of the PWM comparator.  When VDD=5V that would mean 2V on the - input of the PWM comparator.  That doesn't seem right.  Is a divider ratio closer to 0.04 closer to the values presented? That would be a 1.36Mohm resistor?

 

  • Hi Steve,

    I will ask one of the engineers here to take a look at this.

    Regards

    Peter
  • Hello

    A 220pF filter cap is correct. Figure 7 is wrong and 220micro farads is definitely wrong. - Please note that the purpose of the filter is to reduce the amplitude of the 'leading edge' spike which occurs when the switching devices are turned on. Layout has a big impact on the amplitude of this spike. In general, the less filtering you can use the better and reducing the amplitude of the unfiltered spike is the first step here.

    VCS in Figure 1 is indeed the voltage developed across the current sensing resistor -

    There is a lot of information about slope compensation available, for example in the TI Power Supply Design Seiminar article - slup113.pdf In this context the term RAMP VOLTAGE HEIGHT refers to the amplitude of the ramp as a function of RT and these are plotted in Fig 13, 14 and 15. Essentially, is is a way to set the amplitude of the slope compensation ramp that is added to the current sense signal ramp.

    CT in Figure 1 is the voltage developed across the timing capacitor. The amplifier forces the ISET pin to follow the voltage on the internal capacitor.

    "Vct should always range from 0.2V to 1.5V, shouldn’t it (neglecting temperature and delay effects)? Ct is charged by a current set by Rt and it’s discharged by the fet shown next to Ct? The window comparator for the osc uses 1.5V and 0.2V as limits. Is this correct?" Yes - you are correct

    I would expect that the FET would discharge the Capacitor down to 0V rather than 200mV. the RAMP VOLTAGE Height is given in Fig 13, 14 and 15

    The CTRL input is the output from the voltage control loop - an external error amplifier in this case. The CTRL input sets the demand current at which the PWM comparator will trip. A higher voltage at the CTRL input will cause the PWM comparator to trip at a higher current - (a higher voltage at the CS input)
    The CS circuitry block is used only for over current protection.
    The Input to the CS pin is a ramp, the comparator will trip when the CS signal rises to the demand level set by the CTRL input. The third (upper) input to the comparator is used to force a soft start ramp. During the Soft Start, this third input is ramped from 0V. The error amplifier output (at CTRL) will terminate the PWM cycle only once the soft start ramp has increased past the level set by the error amplifier. Note, the CS ramp is a sawtooth at the switching frequency, the soft start ramp is much, much slower and is effectively unchanged during a given switching cycle.

    The reference schematic shows a 5.1k resistor from CTRL to VCC AND a connection from CTRL to the collector of the optocoupler U3. U4 is the error amplifier and its output controls the current in the diode of U3. The resistor is chosen such that the error amplifier has enough control range to pull CTRL from 5V (no current in the optocoupler output) down to about 700mV. (700mV = 300mV at the junction of the 80k/60k potential divider). The calculations should allow some margin for various offsets and for optocoupler CTR variations from part to part and for CTR degradation over time. In this case, with 5k vrom CTRL to VCC, a starting point would allow the TL431 to pull 0 to 1mA in the opto phototransistor. Assuming that the optocoupler CTR could vary from 100% to 200%, this would require a 1mA current in the Optocoupler diode. Given a 50% lifetime degradation in CTR (this depends on the specific optocoupler and is a function of LED current and temperature and what your expected product lifetime is) you would want the TL431 to be able to drive 2mA in the LED.


    Hope this clarifies things.
    Regards
    Colin
  • "There is a lot of information about slope compensation available, for example in the TI Power Supply Design Seiminar article"

    As I mentioned further down my post, I found SLUA584A, which details the calculation of Rs, Rf, and Rset.  

    "The CTRL input is the output from the voltage control loop - an external error amplifier in this case"

    Yes, it's clear that an external error loop is controlling the CTRL pin in the reference schematic.  My question was regarding how the CTRL input to the PWM comparator works.  Referring to the application schematic on page 8 of the datasheet, please let me know: When the opto output transistor is sinking 0 current (meaning full load demand), how should the value of the pullup resistor between CTRL and VDD be chosen to allow for maximum range of Ton?  If the CS pin reaches 0.5V then the overcurrent comparator will tripped anyway, regardless of the voltage at the CTRL pin.  If, for example, I want to trip the end of cycle at a CS pin voltage of 0.45V, then the voltage at the + input of the PWM comparator would be 0.75?  So the target value at the middle - input in this case would be 0.75V?  If the VDD pin is at 12V, then the resistor value to achieve this would have to be 820K!  

    The reference schematic shows a 5.1k resistor from CTRL to VCC

    Really?  Where?  The reference application on page 8 of the datasheet shows no value for this resistor and a Vdd of 12V.  The schematic of Figure 1 on page 6 shows a value of 10K but no VDD value??  Where do you see a 5.1K value?  Why use a 5.1K resistor for any value of VDD?  With a 5.1K resistor and VDD=4.1V (VDD-min) then the PWM comparator - input would be at 2.1V and it would never be tripped?

     

  • Hello Steve

    The reference schematic shows a 5.1k resistor from CTRL to VCC -- this is R4 in the schematic in Fig 5 on page 11 -- the resistor is shown without a value in the schematic on Page 8

    The linear operating range of the CS pin is 0 to 400mV. If the CS signal exceeds 500mV (actually 470mV to 570mV) then the PWM Comparator/Latch will trip and terminate the present cycle. This is effectively a current limit function. If for some reason the CS signal gets to 750mV the Soft Start and Fault Latch will trip.

    Now, assuming that the CS signal is somewhere in the 0 to 400mV range -
    The CTRL input sets the current demand level for the current control loop. This demand is the output of the voltage control loop - effectively the output of the error amplifier. The CTRL input has to be able to set a demand anywhere from 0% to 100% of current. This corresponds to a voltage range at the PWM comparator between (0mV +300mV) and (400mV +300mV). Effectively a control range from 300mV to 700mV. This corresponds to a control range at CTRL of 526mV to 1.23V. The CTRL input is multiplied by 0.57 (80/(80+60)) of course.

    So the expected linear operating range for the CTRL input is from 526mV to 1.23V.
    The resistor from CTRL to VDD provides a pull up to the collector of the Opto Coupler. The Opto Coupler operating conditions should be chosen so that it can pull its collector over this range - with some margin at both ends of course. With a 5.1k pullup and Vc = 1V then Ic would be about 2.1mA - which is reasonable.

    You will get a current limit point by the action of the 500mV comparator in the CS circuitry block. When the CS signal gets to 500mV then irrespective of what the CTRL input is doing - the present cycle will terminate and so limit the current.

    The CTRL input is the main control input to the chip - it sets the demand current. Its authority is limited to a max CS signal level of 500mV. If the error amplifier sees Vout too high, it will set a lower current demand at the CTRL pin, the lower current will then cause Vout to reduce. Similarly if the error amplifier sees Vout too low, it will set a higher current demand at the CTRL pin, the increased current will drive Vout up - back to its setpoint - providing of course that the error amplifier is not demanding a current greater than the 500mV limit in the CS Circuitry block allows.

    Regards
    Colin

    There is some explanatory material at www.ti.com/.../slup075.pdf