ML4828
SETTING THE OSCILLATOR FREQUENCY
ERROR AMPLIFIER
The ML4828 switching frequency is determined by the
charge and discharge times of the network connected to
The ML4828 error amplifier has a 10MHz bandwidth and
a 10V/µs slew rate. Figure 4 gives the Bode plot of the
error amplifier.
the R and C pins. Figure 3 shows the relationships
T
T
between the internal clock and the charge and discharge
times.
100
80
60
40
20
0
180
135
90
45
0
RAMP PEAK
2.5V
GAIN
PHASE
RAMP VALLEY
1.25V
t
CHARGE
DISCHARGE
t
INTERNAL
CLOCK
–20
100
1K
10K
100K
FREQUENCY
1M
10M
100M
Figure 3. Internal Oscillator Timing.
The frequency of the oscillator is:
Figure 4. Error Amplifier Open-Loop Gain
and Phase vs. Frequency.
1
f
=
OSC
(1)
OUTPUT DRIVERS
t
+ t
CHARGE
DISCHARGE
The ML4828 has four high-current CMOS output drivers,
each capable of 1A peak output current. These outputs
have been designed to quickly switch the gates of power
MOSFET transistors via a gate drive transformer. For higher
power applications, the outputs can be connected to
external MOSFET drivers.
The ramp peak is 2.5V and the ramp valley is 1.25V,
giving a ramp range of 1.25V. The charging current is set
externally through the resistor R :
T
2.5V
I
=
CHARGE
(2)
R
T
The output phase delay times are set by charging an
internal 6.7pF capacitor up to the REF voltage (2.5V) via a
current that is externally programmed through R and R ,
while the discharging current is fixed at 1.4 mA. The
charge and discharge times can be determined by:
A
B
for the side A and side B drivers, respectively. The
charging current and delay time for side A are given by:
C ×1.25V C ×R
T
T
T
t
=
=
=
(3)
(4)
CHARGE
I
2
CHARGE
2.5V
I =
A
(6)
(7)
R
A
C ×1.25V C ×1.25V
T
T
t
=
DISCHARGE
I
1.4mA
t
= 6.7pF ×R
A
DISCHARGE
DA
The oscillator frequency can then be found by substituting
the results of equations 3 and 4 into equation 1. This
frequency activates a T flip-flop which generates the
output pulses. The T flip-flop acts as a frequency divider
(÷2), so the output frequency will be:
The same equations can be applied to R . For example,
B
with R = 33kΩ:
A
(8)
t
= 6.7pF ×33kΩ = 220ns
DA
f
OSC
2
f
=
(5)
OUT
6
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