To calculate the value for R2 that will set the output voltage (VOUT) of the boost converter to 10V, we can rearrange the formula provided:
VOUT=1.233V×(1+R2R1)
Rearranging the formula to solve for R2:
R2R1=1.233VVOUT−1
Given that R1 is 1.8MΩ (1,800,000Ω) and the desired output voltage is 10V:
R2R1=1.233V10V−1=8.104−1=7.104
Now we solve for R2:
R2=7.104R1=7.1041,800,000Ω≈253,363Ω
Since standard resistor values might not exactly match this calculated value, you would typically select the nearest standard value. A common approach is to use a standard E-series value, such as E24 or E96 series, which provides a range of values at specified tolerance levels. The nearest E24 value to 253kΩ is 249kΩ or 255kΩ. You might select 249kΩ to provide a slightly higher output voltage than 10V, to account for possible tolerances, or if precision is critical, you could put two resistors in series or parallel to achieve the desired value as close as possible.
To summarize:
R2 calculated value: ≈ 253,363Ω
Possible standard E24 value: 249kΩ or 255kΩ
Please note that resistor tolerances may affect your output voltage, so if this is a precise application, make sure to use high-precision resistors or adjust the value of R2 as necessary after testing.
2y
4
IC1
R2
Resistance
249kΩ
D1
C2
Capacitance
4.7uF
R1
Resistance
1.8MΩ
L1
Height
1 mm
Inductance
6.8uH
VOUT
VIN
GNDTERMINAL
CFF
Capacitance
4.7pF
C1
Capacitance
10uF
EN
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TPS61040DBVR Boost Converter Module
Boost Converter 1.8V-6V input. 10.15V Output. 350mA max current.
Enable is not internally pulled high or low. Please pull up or down according to your application.
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