TL7705B: Questions about RT and Asserting graph of TL7705B.

Hi Kazuya-san,

Please see my response in red below:

Q1. RT is explained in "10 Power Supply Recommendations" section on page 14 of the device datasheet (Ver.SLVS037O).

        But I can't understand why RT is needed. Could you please tell me the reason?

See attached picture below:

 Q2. What is happened when the RT resistor value is increased or decreased?

        Also what is happened when RT is not used?

See above attached 

 Q3. What do Fiure 6-3 (Assertion Time vs Load Resistance), Figure 6-4(Deassertion Time vs Load Resistance),

        Figure 6-5(Assertion Time vs Load Capacitance), Figure 6-6(Deassertion Time vs Load Capacitance) mean?

        Are these same meaning of  TPHL, TPLH and td in Figure 6-1 on page 7?

Yes, you are correct.  The timing diagram in Fig 6-1 is represented in figures 6-3 to 6-6.

Ben

 Hello Ben,

 Thank you very much for your reply.

 Could I ask you a few additional questions?

 Q1, What is the source current from CT to the external capacitor? Is it about 70uA?

 Q2. When RT is inserted, does the delay time become shorter than CT equation (td ≈ 2.6 × 10^4 × CT)

        in 8.3.2 Adjustable Pulse Duration section?

 Q3. Do you have any equation for td in case of RT inserted?

 Thank you again and best regards,

 Kazuya.

Hi Kazuya-san,

Please see my answers in red below:

Q1, What is the source current from CT to the external capacitor? Is it about 70uA?

I cannot find the information in the datasheet.  Is the 70uA what the customer measured from the part?

 Q2. When RT is inserted, does the delay time become shorter than CT equation (td ≈ 2.6 × 10^4 × CT)

        in 8.3.2 Adjustable Pulse Duration section?

Base on the datasheet, the simple answer is no, RT does not change the CT equation.  See below:

 Q3. Do you have any equation for td in case of RT inserted?

See above.  The equation is complicated and is dependent on the duration of the fault condition.

Ben

Hi Kazuya-san,

Thanks for pointing out that the CT voltage is 1.8V.  I did not see that voltage in the morning.  Now, I know how you were able to solve for 70uA.  Thanks!

Ben