MIC2172_06 Datasheet, PDF (14/20 Pages) Micrel Semiconductor – 100kHz 1.25A Switching Regulators

Micrel
Figure 10. Single Point Ground
A single point ground is strongly recommended for
proper operation.
The signal ground, compensation network ground, and
feedback network connections are sensitive to minor
voltage variations. The input and output capacitor
grounds and power ground conductors will exhibit
voltage drop when carrying large currents. Keep the
sensitive circuit ground traces separate from the power
ground traces. Small voltage variations applied to the
sensitive circuits can prevent the MIC2172/3172 or any
switching regulator from functioning properly.
Applications and Design Hints
Access to both the collector and emitter(s) of the NPN
power switch makes the MIC2172/3172 extremely
versatile and suitable for use in most PWM power supply
topologies.
Boost Conversion
Refer to figure 11 for a typical boost conversion
application where a +5V logic supply is available but
+12V at 0.14A is required.
Figure 11. 5V to 12V Boost Converter
The first step in designing a boost converter is
determining whether inductor L1 will cause the converter
to operate in either continuous or discontinuous mode.
Discontinuous mode is preferred because the feedback
control of the converter is simpler.
When L1 discharges its current completely during the
MIC2172/3172’s off-time, it is operating in discontinuous
mode.
April 2006
MIC2172/3172
L1 is operating in continuous mode if it does not
discharge completely before the MIC2172/3172 power
switch is turned on again.
Discontinuous Mode Design
Given the maximum output current, solve equation (1) to
determine whether the device can operate in
discontinuous mode without initiating the internal device
current limit.
I_OUT
≤
⎜⎛
⎝
I_CL
2
⎟⎠⎞V_IN
V_OUT
δ
(1)
δ = ^{VOUT + VF ± VIN}
(1a)
V_OUT + V_F
Where:
I_CL = internal switch current limit
I_CL = 1.25A when δ < 50%
I_CL = 0.833 (2 – δ) when δ ≥ 50%
(Refer to Electrical Characteristics.)
I_OUT = maximum output current
V_IN = minimum input voltage
δ = duty cycle
V_OUT = required output voltage
V_F = D1 forward voltage drop
For the example in figure 11.
I_OUT = 0.14A
I_CL = 1.147A
V_IN = 4.75V (minimum)
δ = 0.623
V_OUT = 12.0V
V_F = 0.6V
Then:
⎜⎛ ^1.147 ⎟⎞ × 4.75 × 0.623
I_OUT ≤ ^⎝
2
⎠
12
I_OUT ≤ 0.141A
This value is greater than the 0.14A output current
requirement so we can proceed to find the inductance
value of L1.
L1 ≤ ^{(VIN δ)2}
(2)
2 P_OUT f_SW
Where:
P_OUT = 12 ⋅ 0.14 = 1.68W
f_SW = 1⋅10⁵kHz (100kHz)
For our practical example:
14
M9999-041806
[phone]