Download 250W PWM inverter circuit SG3524.

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250W PWM inverter circuit SG3524.
circuit that has all essential circuitry required for making a
switching regulator in single ended or push-pull mode.
The built in circuitries inside the SG3524 include pulse
width modulator, oscillator, voltage reference, error
amplifier, overload protection circuit, output drivers etc.
SG3524 forms the heart of this PWM inverter circuit
which can correct its output voltage against the variations
in the output load. In a non PWM inverter the change in
output load directly affects the output voltage (when
output load increases output voltage decreases and vice
versa), but in a PWM inverter the output voltage remains
constant over a range of output load.
A 250W PWM inverter circuit built around IC SG3524 is
shown here. SG3524 is an integrated switching regulator
Circuit diagram of 250W PWM inverter.
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PWM inverter circuit
About the circuit.
Resistor R2 and capacitor C1 sets the frequency of the
ICs internal oscillator. Preset R1 can be used for fine
tuning of the oscillator frequency. Pin 14 and pin 11 are
the emitter terminals of the internal driver transistor of the
IC. The collector terminals of the driver transistors (pin 13
and 12) are tied together and connected to the +8V rail
(output of the 7808). Two 50Hz pulse trains which are
180 degree out of phase are available at pin 14 and 15 of
the IC. These are the signals which drive the subsequent
transistor stages. When signal at pin 14 is high, transistor
Q2 is switched on which in turn makes transistor Q4, Q5,
Q6 ON are current flows from the +12V source (battery)
connected at point a (marked with label a) through the
upper half of the transformer (T1) primary and sinks to
ground through the transistors Q4, Q5 and Q6. As a
result a voltage is induced in the transformer secondary
(due to electromagnetic induction) and this voltage
contributes to the upper half cycle of the 220V output
waveform. During this period pin 11 will be low and its
succeeding stages will be inactive. When 11 of the IC pin
goes high Q3 gets switched ON and as result Q7, Q8 and
Q9 will be also switched ON. Current flows from the +12V
source (marked with label a) through the lower half of the
transformer primary and sinks to the ground through
transistors Q7, Q8, Q9 and the resultant voltage induced
at the T2 secondary contributes to the lower half cycle of
the 220V output wave form.
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The output voltage regulation section of the inverter
circuit works as follows. The inverter output (output of T2)
is tapped from point’s labelled b, c and supplied to the
primary of the transformer T2. The transformer T2 steps
down this high voltage , bridge D5 rectifies it and this
voltage ( will be proportional to the inverter’s output
voltage) is supplied to the pin1 (inverting input of the
internal error amplifier of the IC) through R8, R9, R16 and
this voltage is compared with the internal reference
voltage. This error voltage will be proportional to the
variation of the output voltage from the desired value and
the IC adjusts the duty cycle of the drive signals ( at pin
14 and 12) in order to bring back the output voltage to the
desired value. Preset R9 can be used for adjusting the
inverters output voltage as it directly controls the amount
of voltage fed back from the inverter output to the error
amplifier section.
capacitor for the voltage regulator IC 7808. R11 limits
limits the current through the indicator LED D2.
Notes.
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IC2 and its associated components produce an 8V supply
from the 12V source for powering the IC and its related
circuitries. Diodes D3 and D4 are freewheeling diodes
which protect the driver stage transistors from voltage
spikes which are produced when the transformer (T2)
primaries are switched. R14 and R15 limit the base
current of Q4 and Q7 respectively. R12 and R13 are
pulldown resistors for Q4 and Q7 which prevents their
accidental switch ON. C10 and C11 are meant for
bypassing noise from the inverter output. C8 is a filter
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Mount the SG3524 on a holder.
All capacitors other than C10 and C11 must be
rated at least 15V.
Preset R9 can be used for adjusting the inverter’s
output voltage.
Preset R1 can be used for adjusting the inverter’s
operating frequency.
Transistors in the driver stage require heatsink.
T2 is a 220V primary, 12V secondary, 1A
transformer.
T1 is a 12-0-12 V primary, 220V secondary,
300VA transformer.
Driver transistors must be isolated from the
heatsink using mica sheets. Mounting kits for these
transistors are easily available in the market.
An optional finned aluminium heatsink can be
attached to the 7808.
If 1A bridge is not available, make one using four
1N4007 diodes.
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Switching (transformerless) DC/AC
150W modified sine wave inverter
12V/230V
The inverter is suitable for battery powering of mains
appliances. It is a switching converter, which contains
no bulky, heavy and expensive iron transformer. The
advantage of small size and weight, precise output
voltage stabilization and among other things, very little
quiescent power consumption, which can not be
achieved using th
Inside:
Repository
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e classical transformer. The disadvantage is perhaps it
is a little more complicated :). Use: Can be used for
most appliances except those capacitive (eg. with high
parallel capacitor) and those that require pure sine for
another reason. Not suitable for capacitive limited mini
fluorescent nigth lights and LED light bulbs. For the
conventional fluorescent (or discharge) lights is
necessary to disconnect the capacitor. Circuit Switching
Inverter with modified sine wave typically consists of two
parts: 1) DC/DC converter that increases the DC input
voltage of 12V (24V, 48V) to the high DC voltage
corresponding to about the mains voltage amplitude (for
mains 230V~ it is 325V=) and 2) polarity switching
bridge that turns this increased voltage into a modified
sine wave. For lower output power switching converters
(up to about 150W) the DC/DC converter can be built
as flyback supply. For more power would be necessary
to use a push-pull supply (or forward, but the
complication is very high secondary voltage). Single coil
step-up inverter is not suitable, because the step-up
ratio is too large. Due to the low power I have my
DC/DC converter finally built as flyback supply. I will
omit the detailed description of the DC/DC section,
because this schematic has already been described in
detail here: DC/DC converter 12V / 325V 150W. The
inverter bridge is composed of the transistors of
MOSFET N type: T3, T4, T5 and T6. They are
controlled by integrated circuits IO2 and IO3 of type
IR2153. The circuit IR2153 is an integrated halfbridge
driver with internal oscillator. IO3 oscillator works here
as a master oscillator frequency 50 Hz. IO2 is a slave,
its oscillator section is used as a phase shifter,
depending on IO3. To make a modified sine wave
bridge that could manage inductive loads, it is
necessary to block (short circuit) the output while the
zero output voltage. It is done by switching two
transistors in the bridge - either both lower or both
upper. For even distribution of losses it is obviously
once the top and once the bottom pair. The best way to
do this is using two rectangular singals with a phase
shift. When the input voltage is 325V, the 230V rms
output voltage is achieved with 90 ° shift, which is 25%
duty cycle. Sometimes in the converters, another
compromise is selected, eg 30% duty cycle with the
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input voltage 297V, or 33. 3% with the input voltage
282V. DC/DC converter is set by the trimmer P1.
DC/DC converter is recommended to be tested and
adjusted independently. If we chose a 25% duty cycle,
set P1 so that the output of DC/DC converter was 325V.
P2 sets the output frequency. We measure the
frequency of the IO3 pin 2 and set to 50Hz (for some
countries it is 60Hz - I recommend to replace the 120k
resistor by 100k it case of 60Hz). Finally, use P3 to set
the correct duty cycle. Warning - voltmeters
(multimeters), which can not measure the effective
(RMS) value of the voltage at the inverter output will not
show correctly 230V, but a little less! Therefore, they
are not suitable to set the duty cycle! Actual voltage
would be higher than the measured! The inverter is
protected against overload and short circuit. The current
through the bridge is sensed by R3 shunt and T7
transistor. If an overload or short circuit, the T7 and T8
subsequently turns on. T7 to T8 keep each other open.
They bring negative voltage on pins 3 of IO2 and IO3,
which activates the shutdown (SD) and the bridge will
remain off until disconnecting and reconnecting input
power of the inverter. Given that most appliances
creates a power surge when switched on, it is
necessary to delay the protection, otherwise it will keep
shut down. The delay is ensured by C9 capacitor. Thus
the inverter can be short term overload without
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protection activating. In case of real short circuit the
inverter but must be able to turn off quickly. The current
threshold of the fast shutdown is much higher than the
one of delayed shutdown and is set using R4. R3
determines the threshold of the delayed shutdown. L2 is
used to limit the dv/dt in case of short circuit. Without it,
the protection would only protect against overload, but
not against short circuit. Under normal circumstances,
the voltage on L2 is negligible. Activation of protection is
indicated by LED1. Inverter efficiency is over 90%.
Note: For 220V~ mains the amplitude is 310V, for
240V~ mains the amplitude is 340V. Warning! The
inverter must be protected by a suitable fuse. Although
the input voltage of the inverter is safe, its output
voltage is mortally dangerous. Capacitors of switching
inverter can remain charged to dangerous voltages
even after shutdown. The inverter can be inappropriate
for some appliances. Everything you do on your own
risk. For any your harm I do not take responsibility.
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