LM5022: LM5022

Hi Genie,

Thank you for using E2E forums, I am a bit confused could you elaborate what is your question? do you mean how to calculate the Vin at the start up?

BR,

Haroon

Hi Haroon,

Yes, I want to know how to calculate the voltage on U1 pin 1 at the start up. In other word, I can understand the schematic consist of R1, R2, R3, R4, R5, R6, R106, R121, REN, C121, D1, D2, Q100, Q101.

Best regards,

Genie

Hi Genie,

I suppose you can simulate this in Pspice and check from there, may I know why do you want to calculate this voltage? In the test report you can see the startup waveform showcasing how this voltage behaves. 

BR,

Haroon

Hi Haroon,

I want to deeply understand this circuit from the point of view of design, but not just replying on the help of Pspice.

Best regards,

Genie

Hi Genie,

I will get back to you next week with some more information regarding this, please allow me 3 to 4 days.

Thank you

BR,

Haroon

Hi Haroon,

Just a soft remind that I am still looking forward to your reply, thank you.

Best regards,

Genie 

Hi Genie,

the FET Q100 works like an LDO and regulates down the input voltage from the supply voltage to ~ 15V (from D1) + V_GS from Q100.

This is used as the input VIN_HV in this reference design can be higher then the allowed supply voltage for the LM5022.

I am sure with that as starting point you will be able to calculate input design.

Note: R4, R5, R6 R121, C121 Q101 do not impact the voltage at VIN of the LM5022 in a considerable degree.

Best regards, 

 Stefan

Hi Stefan,

Do you have the equation to elaborate on your explanation? Thank you.

Best regards,

Genie

Hi Genie,

sorry, I also do not have this ready right now.

Best regards,

 Stefan

Hi Stefan,

1. Sould VIN of LM5022 equal to  ~ 15V (from D1) - V_GS from Q100, but not  ~ 15V (from D1) + V_GS from Q100?

2. What 's the function of Q101? According to your state, FET Q100 works like an LDO and regulates down the input voltage, Q101 seems to be redundant.

3. What's the function of R106 and REN?

Best regards,

Genie

Hi Genie,

1. your are completely right here - good catch - thanks.

2. this is generating the UVLO input signal - this circuit implementation does prevent a higher signal on UVLO then on VIN.

3. REN is the pull down for UVLO - so to set a proper 0V when not enabled through Q101.

  R106 will take some part of the voltage as the FET is only rated for 100V but the input can go up to 400V.

  See also the startup scope plot in the report of the PMP20745 (Chapter 5.1) 

Best regards,

 Stefan

Hi Stefan,

Question 2: When VIN_HV down to what level will trigger UVLO? How to calculate it?

Question 3: How much voltage will R106 take when VIN_HV is 400V? How to calculate it?

Best regards,

Genie

Hi Genie,

2. UVLO threshold = 1.25V  (LM5022 Datasheet)   +   BE Voltage of Transistor  = 1.95V   => starts if voltage at B of Transistor is > then 1.95V

-> note: please consider also the min and max level and hysteresis if required.

3. U = R * I   =>  R = 2.49k    I = 3.5mA (LM5022 Datasheet)  + VIN / REN

Best regards,

 Stefan

Hi Stefan,

Question 3: According to your equation, does it mean that Vce of Q101 is 0V?

Best regards,

Genie

Hi Genie,

No, that will not be generally the case. This depends on the supply voltage as the Q101 needs to take the remaining voltage.

Best regards,

 Stefan

Hi Stefan,

If Vce of Q101 is not 0V, may I say that the current flowing through equals (Vd2 - BE Voltage of Transistor) / REN?

Best regards,

Genie

Hi Genie,

sorry, mixed Q101 and Q100 - i always need to scroll up to the first message to see the drawing and the thread is already quite long.

So once enabled the Voltage at Vce of Q101 can be seen as 0V.

Best regards,

 Stefan

Hi Stefan,

I am confused. 

1. VIN of LM5022 equal to  ~ 15V (from D1) - V_GS from Q100.

2. Vce of Q101 can be seen as 0V (according to your last reply). So Ve of Q101 equal ~ 15V (from D1) - V_GS from Q100..

3. The base voltage of Q101 is 5.1V

Combined 1, 2, how Q101 can be turned on?

Best regards,

Genie

Hello Genie,

As we did not hear back from you, I expect that the answer was sufficient or you solved your issue yourself. I will close this thread now. If my assumption is wrong, please post below or create a new thread, if this one is already locked.

Best regards,
Brigitte