LED Driver LM3409

A

abi0987654321

Newbie level 4
Joined
Feb 18, 2025
Messages
7
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Activity points
49
I am having doubt on peak current control circuit (Rsns) section. Since op-Amp can ref voltage with ground. In that case voltage at non-inverting terminal should be 1.24v. How it is Vcst= 2.48mV? How Vcst=248mV will be the ref voltage to the comparartor? How did they derive Vcst=238mV? In what basic principle is used in that circuit analysis?


Page N0=12

This is the circuit
 

Attachments

  • Screenshot 2025-02-17 123352.png
    226 KB · Views: 49
Last edited:

Sort by date Sort by votes

Upvote 0 Downvote

Sorry it is 248mV. I have attached an image.
KlausST said:
Klaus
Click to expand...
 

Upvote 0 Downvote
The calculation for Vcst is shown in your posted image of the circuit.....?
 

Upvote 0 Downvote
danadakk said:
The calculation for Vcst is shown in your posted image of the circuit.....?
Click to expand...
Yes. What my doubt is, As per my understanding Vcst should be 1.24v with ref ground.so non inverting terminal reference should be 1.24v. But they calculated as Vcst=248mV which is reference for non inverting terminal. could you please explain how non inverting takes Vcst=248mV as reference? Or else proof or provide calculation that vcst=248mV will the reference not 1.24v
 

Upvote 0 Downvote
Vcst is at the top of the figure, next to the CSP pin, and it isn't referenced to ground. It is the offset/reference for the upper comparator.
1.24V is the reference for the lower opamp. The current through the 5R resistor also goes through the R resistor next to the CSP pin.
1.24V over the 5R resistor will give 248 mV over the R resistor = Vcst.
248 mV = 1.24V / 5
 

Upvote 0 Downvote
The OpAmp creates a current source at output of MOSFET it drives of
1.24V / 5R. That current flows thru R of the upper comparator creating
a Vcst thru divider action of R/5R x 1.24
 

Upvote 0 Downvote
danadakk said:
The OpAmp creates a current source at output of MOSFET it drives of
1.24V / 5R. That current flows thru R of the upper comparator creating
a Vcst thru divider action of R/5R x 1.24
Click to expand...
Could you please explain how the current path. Since vcsp is the high potential how current flows from 5R to Vcst ,R ?
 

Upvote 0 Downvote
How current flows form low to high potential.since Vcsp is at high potential.current flows from vcsp to R to 5R.rit
 

Upvote 0 Downvote
dear abi

If IT = max, then 1.24V = 5*Vcst due to single current path and FET V drop. depends on Vin - 6/5*1.24
Since CSN pin has no current, VSNS=VCST = 1.24/5

 

Attachments

  • 1739893800841.png
    172.2 KB · Views: 60
Last edited:

Upvote 0 Downvote
abi0987654321 said:
Sorry, Im not understanding your explanation.could you please elaborate?
Click to expand...
What exactly don´t you understand.
All parts are marked, you see the 1.24V refeerence, the 5R, the R, .. also Tony showed you the current path ... so I can´t imagine what is unclear.

Klaus
 

Upvote 0 Downvote
If IT = max, then Vsns= max

The CSN pin has no input current, so VSNS=VCST

If the same current in 5R = 1.24V as in R for Vcst then the voltage across 5R is 5*Vcst
then Vcst=V(5R)/5

V(5R) =1.24V must be the same as the Zener if feedback produces zero error voltage across input +-
Thus Vcst=1.24/5

 

Upvote 0 Downvote
Hi ,

That was clear. I have one more doubt. Why Upper comparator is not taking Vadj as a Reference?, since it can see voltage with respect to ground, In that case Vref should be 1.24v . It should compare Vadj =1.24v with Vsns. Why it is taking Vcst=248mV as reference and comparing Vcst with Vsns?.

Regards,

Abishek
 

Attachments

  • Screenshot 2025-02-19 152042.png
    90 KB · Views: 25

Upvote 0 Downvote
Because the comparator is implementing a high side current sense
circuit, so eliminating CM voltage errors downstream from its measurement ?
 

Upvote 0 Downvote
Hi,

yes, Dana is correct:
* V_Adj is referenced to GND, while
* V_Cst is referenced to CSP = V_IN

So it is able to react referenced to GND as well as referenced to V_In

Klaus
 

Upvote 0 Downvote
Using a controlled current is a common way to shift a differential voltage down to ground with R ratios for gain or attenuation. This is called a high-side current sensor with a low-side controller..

Another level shifter example is to make a low-level logic drive to a high voltage gate on a Pch FET and use a low-side current sensor feedback with “ground-shift” to the load.
 

Upvote 0 Downvote
You must log in or register to reply here.

Similar threads

Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…