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LM5000: Negative in, Positive out buck-boost

Part Number: LM5000

My question is about transient response or  compensation component selection for stability on this buck-boost converter. I have made a prototype with -12V input ( range of -9V  ~ -15V) , and 3.3V output converter with LM5000. Con output current range is  25mA ~ 250mA. With Rc=1k, Cc= 0.47uF, this converter is stable, but load  response slow. its output voltage  is within desired range with a fix load.  It is not good on load step response. 

With Vin=-12V, and load changes from 28mA to 138mA, output voltage has about 0.5V drop ( about 0.5 ms width). 

With Vin =-12V,  and load changes from 28mA to 244mA, output voltage drops about 1V with about 0.8ms width.

I understand that I have set a pole at 0.4Hz  with Rc=1k, Cc = 0.47uF, which is not within  the range of 10hz to 100Hz that  is suggested in LM5000 datasheet ( page 13 of 17 ). I have tried to make it within 10 to 100 HZ range, say Rc=102k, Cc=4.7nF ( and other Rc and Cc values)  , the circuits are not stable. it gets oscillation with heavier load ( say 150mA or more ), it also has ringing at start up with 28mA load.  My calculation shows that the pole formed by load ( 132 Ohm with 20uF cap ) fp is around 60Hz, and the right half plane zero is 7.3khz. Since we are using ceramic output caps, the output capacitor ESR zero is higher than RHP, which is 7.3Khz. The phase lay of Gp(s) at fzRHP will approach -270 degree and the crossover frequency fc has to be selected below the 60 Hz ( fp). I think this is why I have to set a pole at 0.4HZ with Cc and Rc to make this converter stable.

My questions are:

1. Is there anything that we can do to improve load step response for the converter?

2. is there anything that I have done wrong to make it can not meet 10 ~ 100Hz range requirement in the datasheet while choosing Rc and Cc.

Please see the schematics of this power supply. Rcc is R104 and Cc is C177 in this schematics.

startup waveforms with Rc= 1k and Cc= 0.47uF with 244mA load:

output voltage load step response ( output current changes 28mA to 138mA ), Rc=1k, Cc= 0.47uF:

output voltage load step response ( output current changes 28mA to 244mA ), Rc=1k, Cc= 0.47uF:

start up waveforms with Rc =102k and Cc = 4.7nF:

  • Hi Jake,

    Thanks for the detail description. Give me some time to look at this issue.
    Regards,
    Zack
  • Hi Jake,

    Can you use current probe to look at the inductor current at same time when you do the load transient test?
  • Here is the scope screen shot on inductor  current response when I do the load transient test.

    Yellow channel is connected to a the current probe, 10mV/1mA.  Red channel is monitoring output voltage transient with AC coupling. 

    Image below shows that  inductor current changes from 40.6mA to 251mA when load changes from 28mA to 188mA.

    Image below shows that  inductor current changes from 40.6mA to 326mA when load changes from 28mA to 244mA.

  • I have looked at the inductor waveforms with different loads. Some waveforms look good. They are similar to what I see on application note SLVA059A. But some of them are slightly different. Please see scope screen shots. A 0.47 Ohm shunt resistor is used to get the current waveform.

     Inductor current waveforms with  28mA load, scope screen shots:

     Inductor current waveforms with  244mA load, scope screen shots ( AC coupling ):

  • Hi Jake,

    Please refer to page 12 about the recommendation range of Rc and Cc. Rc range is 5k-20kohm. If using a 102kohm resistor, it's highly that the system will be unstable. Cc range is 680pF-4.7nF. If using 0.47uF, the compensation capacitor is too big, the system will too slow.
    Please double check the inductor waveform with current probe if possible.
    Regards,
    Zack
  • It is hard to make it stable while maintain Rc and Cc  within the ranges of 5k-20k and 680pF ~ 4.7nF. I had tried it with different values before I posted it at TI E2E and had no luck. I have tried it again today and have similar results ( no luck).   They all have  over shoot at start up with 28mA load and gets oscillation when I apply  more load, say 138mA load.  I have captured some waveforms, but for some reasons,  I can not upload those waveforms and share with you today.

    Could you please confirm if the  my circuits have  right  half plane zero  of  7.301khz? The maximum duty cycle is 27.8% and switch frequency is 700khz. Maximum current load is 250mA.

    based on the formula on the datasheet, My calculation shows that the RHPzeron  is 7.301khz. If it is true, we may need to make the  pole and zero(  formed by Cc, Rc, and output impedance of the error amplifier (850k) ) much lower than RHPzero ( 7.301khz). With the suggested ranges of Rc and Cc, It is hard to make the zero formed by Cc and Rc much lower than 7.3khz, say with Rc = 20k and Cc = 4.7nF, zero will be at 1.69Khz, It will be hard to get enough phase margin for stability since it adds 90 degree lag at RHPzero ( 7.301khz ).  Please correct  me if I was wrong.

  •  I put those waveforms in Microsoft Word and  past it here. it works!  Please see the waveforms that I captured yesterday.

    They are startup waveforms with different Rc and Cc within suggested ranges. All the waveforms are with 28mA load current, Vin = -12V.

      1. Startup waveforms with Cc=4.7nF and Rc = 20k

      1. Startup waveforms with Cc= 4.7nF and Rc =10k

      1. Startup waveforms with Cc= 4.7nF and Rc =5k

      1. Startup waveforms with Cc= 2.2nF and Rc =10k

     

     

  • It looks like that those waveforms are still not shown after I posted it .
  • Here are Jake's waveforms:

    1. Cc=4.7nF and Rc = 20k



    2. Cc=4.7nF and Rc = 10k



    3. Cc=4.7nF and Rc = 5k



    4. Cc=2.2nF and Rc = 10k


  • Hi Jake,

    The inductor in the schematic is a little bit high. Vin is 12V and Vo is 3.3V. The duty cycle can be calculated to be 25%. If using a 150uH inductor, the ripple is only 8.6%. And also the RHP frequency will be 31kHz. It's hard to set the cross frequency high and the load response will be slow.
    Please use a smaller value inductor like around 40uH. The ripple will be 30%. The RHPZ is around 107kH. The whole system will be more easy to compensate.
    And also, add the output capacitor will be helpful for the load drop.
    Tell me if my calculation has some mistakes.

    Regards,
    Zack
  • Hi Zack,

    Thanks for the directions on the improvements. Since my calculation result seems to be different from what you have on the RHPZ ( 7.1khz VS 31KHz ) with my current design, Could you please review my design and calculations? Especially on the calculations of RHPZ and minimum inductance.

    Power supply requirements:

    Vin :  -12V with range of -9V  ~ -15V

    Vout:  3.3V with the range of  3.135V ~ 3.465V  

    Iout: 150mA with the range of 25mA ~ 250mA. 

    I have chosen:

    1.  ΔILmax = 0.05A

    2. Io ( critical) =  ΔILmax/2= 0.025A , Iout need to maintain >=25mA to ensure in continuous mode.

    3. switch frequency:  700Khz, which means that Ts = 1.42857E-06 second.

    4. Continuous Conduction mode with minimum 25mA output current. 

    My calculations are based on application note SLVA059A and LM5000 datasheet:

    Calculations on duty cycles:

    Based on the equation ( page 6 of 31 on SLVA059A ) : Vout = -Vin * D/(1 – D)

          My calculation results:  Dmin = 17.29%, Dnor= 21.57%, Dmax = 27.8%

    Calculations on Minimum inductance :

    Based on the equation ( Page 11 of 31 on SLVA059A ):

    Lmin >= (-Vout * Ts* Vin(max) / ( 2 * Io(critical) * ( Vout – Vin(max) )

    With: Vin(max) = -15V, Vout =3.465V, Io(critical) = 0.025A, Ts = 1.42857E-06 second

          My calculation result: Lmin >=80.4uH

    I have chosen 150uH in my current design.

    Based on equation ( page 13 of 17 ) on LM5000 datasheet: RHPZ = Vout*(1-D)^2/( 2 * Ioutput* L)

    With Vout= 3.3V, D = 21.57%, I output = 0.25A, and L = 150 uH

    My calculation result: RHPZ = 8.615 kHz.

     

    If the equations and calculations above are right, the minimum L that I can choose is 81uH, which gives us RHPZ of 15.954khz.   It looks like there is not much room for me to reduce the inductance and push RFPZ higher.

    With LM5000, can you please confirm RHPZ of flyback is same as RHPZ OF Buck-boost as the equation I am using is from its datasheet which is for flyback comverter?

    If my calculation is right, to push RHPZ higher, it looks like I need to increase the minimum output current  ( Io ( critical) to reduce the inductance. It looks like there is not much room for that and will require our application to maintain higher minimum current level. What else can I do ?

    Thanks,

    Jake

  • Hi Jake,

    First, Do you really need to make sure the converter works at CCM when output current is 25mA? If not, we can have a smaller inductor.
    Second, the equation to calculate the RHPZ for inverting buck-boost converter and flyback converter is different from boost converter. RHPZ = Vout*(1-D)^2/( 2 * Ioutput* L*D). That's why my result is around 4 times higher than yours.
    If using a smaller inductor and with the right RHPZ equation, the real right half plane zero will be much higher, you'll have a lot of space.

    I'd like to introduce a tool for calculation: www.ti.com/.../powerstage-designer
    Hope this will help you.

    Regards,
    Zack
  • Hi  Zack,

    Thank you for the RHPZ calculation equation and the Power stage calculation tools that you introduce to me.

    I will try a smaller inductor, like  47uH. Based on the equation, RHPZ should be around 100kHZ.  This application should allow me to increase the minimum current draw  from 25mA to 50mA. I will let you  know how it goes once I have component and test it.

    Thanks,

    Jake

  • Hi Zack,

    The problem has been solved! Thank you for helping me to get through this issue.

    As you suggested, I have reduced the inductance of the inductor from 150uH to 56uH. It works fine with Rc=2k and Cc=0.1uF.  The output load step response is much better now. With 30mA to 140mA load step change, the output transient is measured to be  0.196V. With 30mA to 192mA load step  change, the output transient is measured to be 0.262V. Now the supply works well.

    thanks,

    Jake