Hello,
My customer is using R2=10Kohms and R1=56Kohms to get 3.3V output at their production released before the PCN.
How much would the drift performance with those resistors at tj=125oC?
Best regards,
K.Hirano
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Hello,
My customer is using R2=10Kohms and R1=56Kohms to get 3.3V output at their production released before the PCN.
How much would the drift performance with those resistors at tj=125oC?
Best regards,
K.Hirano
Hi K. Hirano,
The reason that R2 is restricted is to improve accuracy over temperature. At higher temperatures the feedback current increases enough that we are not able to ignore the effect on the resistor divider (as is done in most LDOs).
The change in the output voltage would be:
∆VOUT = IFB x (VOUT / VREF) x [(VOUT / VREF) x ∆R2 – ∆R2]
Where ∆VOUT = VOUT(R2=30k) – VOUT(R2≠30k) and similarly ∆R2 = 30k – R2 (divide ∆VOUT by VOUT(nom) if you want it in percent like the datasheet spec). This accuracy would have to be added to the accuracy range specified in the datasheet.
Very Respectfully,
Ryan
Ryan,
Thank you for your response.
Let me confirm.
In my customer case,
Vout=3.3V
Vref=0.5V based on datasjeet.
ΔR2=30K-10K=20E3
IFB(max)=1uA based on datasheet.
So, added accuracy % to the datasheet spec at high temperature is calculated as 22.4% as follows?
ΔVout(%)=(1E-6 x (3.3/0.5) x ((3.3/0.5) x 20E3 -20E3) x 100/3.3 = 22.4%
Best regards,
K.Hirano
Hi K. Hirano,
I must apologize, the provided equation had two errors. Please find the correct equation here:
∆Vout = --Ifb x (Vout / Vref) x [(Vref / Vout) x ∆R2 – ∆R2]
This corrected equation will yield a much lower delta Vout. I apologize for the confusion.
Very Respectfully,
Ryan