Datasheet

LMP8646EB User’s Guide
November 2011 Page 6 of 20 SNOU010
a) Using the formula above, calculate for the LMP8646 output accuracy knowing that V
SENSE
=
(1.5A)(55 mOhm), R
G
= 50 kOhm, V
OFFSET
= 1 mV, and Gm_Accuracy = 2% (LMP8646
datasheet, electrical characteristics table).
  
      
   
%27.3100
825.0
852.0825.0
852.0
100/21200/1
50100555.1
825.0
200/1
50
555.1
_
_








xAccuracyOutput
V
V
kOhmmVmOhmA
V
V
V
kOhm
mOhmAV
CALCOUT
THEOOUT


b) After figuring out the LMP8646 output accuracy, choose a tolerable system error or the output
current accuracy that is bigger than the LMP8646 output accuracy. This tolerable system
error will be labeled as I
ERROR
, which has the equation I
ERROR
= (I
MAX
– I
LIMIT
)/I
MAX
(%). In this
example, we can choose I
ERROR
= 5%. This value will be used to calculate for ROUT in the
next step.
5. Step 5: Choose the output resistor, ROUT, for LMP8646
At startup, the capacitor is not charged yet and thus the output voltage of the LM3102 is very small
(V
O_3102_MIN
, see Figure 3). Therefore, at this point, the output current is at its maximum (I
MAX
). When
the output voltage is at its nominal, then the output current will settle to the desired target value.
Because a large current error is not desired, ROUT needs to be chosen to stabilize the loop with
minimal initial startup current error. Follow the equations and example below to choose the
appropriate value for ROUT to minimize this initial error.
a) Target current I
LIMIT
= 1.5A
In the previous example, we chose the maximum tolerable system error I
ERROR
as 5%.
Now, let’s calculate the maximum tolerable current, Imax:


  AAI
III
ERRORLIMIT
575.1
100
5
15.1
1
max
max










b) Calculate for ROUT, assuming that the minimum output voltage for the LM3102 is 0.6V (see
figure 3).