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OPT8241-CDK-EVM: questions about opt8241-cdk-evm laser driver.

kyeongmin choi
Intellectual 590 points

Part Number: OPT8241-CDK-EVM

hello

I am in charge of hardware engineer and will design 3D Time of flight camera using OPT8241 and OPT9221.

I have some questions about illumination board.

1.Why are there two invertor for illum_en_g . Is it impossoble to connect from illum_en to illum_en_g directly?

2. what is puspose this circuit (mosdriver) . just for making high level 3.3v ?

   if we use trig_g which has 5v high level, this circuit  is  no need?

3. How to caculate value of inductor (L5) ?

    duty : 0.5

    delta v means sum of series laser forword voltage ?  ex) vf : 2v  . 4 string --> 8v

    f: modulation freq : 40Mhz

   I OUT means laser  peak current? ex) 2.5A 

   delta I : 0.25A ( asumption 10% of peak current (2.5A)

   We can get value(L) using below fomula. The value is 500nH.

   Is it correct caculation we did ?

4. Why does out put voltage  twice compare to inputvoltage   when we use shunt circuit .

thank you.

 

  • 0
    TI__Expert 3495 points

    1.Why are there two inverters for illum_en_g . Is it impossible to connect from illum_en to illum_en_g directly?

    >> First is an inverter (U17), the second is a "enable" switch (U16) needed as fault-tolerance for laser safety.

    2. what is purpose this circuit (mosdriver) . just for making high level 3.3v ?  If we use trig_g which has 5v high level, this circuit  is no need?

    >>  MOS driver(U13) is required to drive the MOSFET (U15) at high Vgs while delivering enough current to overcome the input capacitance. Inadequate MOS drive current will cause MOSFET (U15) not switching fast enough, or operate in linear region (heating) which can damage the device.

    3. How to calculate value of inductor (L5) ?

    >>  Ripple current for shunt topology is ∆ = ∆ ∙ /( ∙ ). Where is modulation period and ∆ is voltage across inductor, and is duty cycle (usually 0.5). Usually ∆ < 0.1*x.  With these inputs, compute .

    4. Why does out put voltage  twice compare to input voltage when we use shunt circuit .

    >>  When U15 is closed, current supply will build up energy in the inductor (L di/dt); when U15 is opened, inductor voltage polarity will instantly switch, causing the output side of the inductor to rise to nearly 2 * V_supply, as dumps built-up energy through the laser diodes.

    Please checkout this video, or this app note, which address your questions.

  • 0 in reply to Larry Li
    Intellectual 590 points

    i want to know more detail about your reply.

    1.Why are there two inverters for illum_en_g . Is it impossible to connect from illum_en to illum_en_g directly?

    >> First is an inverter (U17), the second is a "enable" switch (U16) needed as fault-tolerance for laser safety.

    I think  u20 also invertor. 

    This is enable progress.

    ILLUMEN_EN -->(U20) -->ILLUM_ENz-->(u17)-->illum_En_G      

    If ILLUMEN_EN is high , ILLUM_ENz is low  and illum_En_G is high.

    i think, Both  ILLUMEN_EN and  illum_En_G are same signal level. So why does circuit need two invertor (like u20,u17)?

    When illum_En_G  is high at U16 , laser and u15 can swich ?

    2. what is purpose this circuit (mosdriver) . just for making high level 3.3v ?  If we use trig_g which has 5v high level, this circuit  is no need?

    >>  MOS driver(U13) is required to drive the MOSFET (U15) at high Vgs while delivering enough current to overcome the input capacitance. Inadequate MOS drive current will cause MOSFET (U15) not switching fast enough, or operate in linear region (heating) which can damage the device.

    We try to supply 5v at vcca , vccb (U15).  That is short between vcca and vccb . i don't know , why supply voltage should be defferent  between vcca and vccb.

    another question ! If we can supply  high vgs(5v)  at TRIG_G , mos driver  no need?

    3. How to calculate value of inductor (L5) ?

    >>  Ripple current for shunt topology is ∆ = ∆ ∙ /( ∙ ). Where is modulation period and ∆ is voltage across inductor, and is duty cycle (usually 0.5). Usually ∆ < 0.1*x.  With these inputs, compute .

    4. Why does out put voltage  twice compare to input voltage when we use shunt circuit .

    >>  When U15 is closed, current supply will build up energy in the inductor (L di/dt); when U15 is opened, inductor voltage polarity will instantly switch, causing the output side of the inductor to rise to nearly 2 * V_supply, as dumps built-up energy through the laser diodes.

    That operate like boost convertor operating?

  • 0 in reply to kyeongmin choi
    TI__Expert 3495 points

    i want to know more detail about your reply.

    1.Why are there two inverters for illum_en_g . Is it impossible to connect from illum_en to illum_en_g directly?

    >> First is an inverter (U17), the second is a "enable" switch (U16) needed as fault-tolerance for laser safety.

    I think  u20 also invertor.   

    This is enable progress.

    ILLUMEN_EN -->(U20) -->ILLUM_ENz-->(u17)-->illum_En_G      

    If ILLUMEN_EN is high , ILLUM_ENz is low  and illum_En_G is high.

    i think, Both  ILLUMEN_EN and  illum_En_G are same signal level. So why does circuit need two invertor (like u20,u17)?

    When illum_En_G  is high at U16 , laser and u15 can swich ?

    >>>>  Some of the "extra" inverter are added also for purposed of logic level translation (3.3V to 1.8V), for instance.

    >>>>  OPT8241 will close U16 switch before it starts switching U15.

    2. what is purpose this circuit (mosdriver) . just for making high level 3.3v ?  If we use trig_g which has 5v high level, this circuit  is no need?

    >>  MOS driver(U13) is required to drive the MOSFET (U15) at high Vgs while delivering enough current to overcome the input capacitance. Inadequate MOS drive current will cause MOSFET (U15) not switching fast enough, or operate in linear region (heating) which can damage the device.

    We try to supply 5v at vcca , vccb (U15).  That is short between vcca and vccb . i don't know , why supply voltage should be defferent  between vcca and vccb.

    >>>>  I'm not sure what you meant by 'vcca' and 'vccb'.  U15 is a MOSFET.   If you mean U13, SN74LVC2T45 is a dual voltage rail driver.

    another question ! If we can supply  high vgs(5v)  at TRIG_G , mos driver  no need?

    >>> > For MOSFET like CSD16301Q2, higher Vgs,  lower Rds_on.  That means higher Vgs reduce power loss to heat.

    >>>>  It is not just Vgs, but current capability is also important.  So whatever you drive TRIG_G with, it must be capable of delivering high frequency (>50MHz) current to keep Vgs high (5V is good).

    3. How to calculate value of inductor (L5) ?

    >>  Ripple current for shunt topology is ∆ = ∆ ∙ /( ∙ ). Where is modulation period and ∆ is voltage across inductor, and is duty cycle (usually 0.5). Usually ∆ < 0.1*x.  With these inputs, compute .

    4. Why does out put voltage  twice compare to input voltage when we use shunt circuit .

    >>  When U15 is closed, current supply will build up energy in the inductor (L di/dt); when U15 is opened, inductor voltage polarity will instantly switch, causing the output side of the inductor to rise to nearly 2 * V_supply, as dumps built-up energy through the laser diodes.

    That operate like boost convertor operating?

    >>>>> Basically, yes.