LED Driver LM3409

[abi0987654321]

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

 

[KlausST]

Hi,

why did you post in the "IC design" section? Do you want to design an IC on your own?

How it is Vcst= 2.48mV?

the formula/explanation is given in "section 1"

How did they derive Vcst=238mV?

Where do you find this?

Klaus

 

[abi0987654321]

Hi,

why did you post in the "IC design" section? Do you want to design an IC on your own?

the formula/explanation is given in "section 1"
But op amp sees with ground ref. In that case Voltage will be 1.24 right. How op-amp sees 248mV. what principle they hav eused?

Where do you find this?

Sorry it is 248mV. I have attached an image.

Klaus

 

[danadakk]

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

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

 

[abi0987654321]

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

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

 

[std_match]

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

 

[danadakk]

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

 

[abi0987654321]

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

Could you please explain how the current path. Since vcsp is the high potential how current flows from 5R to Vcst ,R ?

 

[abi0987654321]

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

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

 

[D.A.(Tony)Stewart]

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

 

[abi0987654321]

dear abi

If IT = max, then 1.24V = 5*Vcst due to single current path and FET is 0V drop. at threshold.
Since CSN pin has no current, VSNS=VCST = 1.24/5

View attachment 197413

Sorry, Im not understanding your explanation.could you please elaborate?

 

[KlausST]

Sorry, Im not understanding your explanation.could you please elaborate?

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

 

[D.A.(Tony)Stewart]

Sorry, Im not understanding your explanation.could you please elaborate?

Can you elaborate what you do understand?
Virtual null in Op Amp inputs, controlled by Vin+
Voltage controlled current?

 

[D.A.(Tony)Stewart]

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

 

[abi0987654321]

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

 

[danadakk]

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

 

[KlausST]

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

 

[D.A.(Tony)Stewart]

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.