| T O P I C R E V I E W |
| liang2408 |
Posted - Feb 03 2008 : 06:57:20 AM
Hi Sir,
I have try to build a DC to AC inverter using 555 timer and a transformer that is 9-0-9V/230V and rated at 36VA. I managed to drive a light bulb of 11W. The attached file shows the schematics drawing of the inverter.
But when i didn't connect to any load and measured the input and output of the transformer, i got an input of 4.98Vrms on the primary side and 122Vrms on the secondary side. The output voltage is far below my requirement of 230VAC.
I hope you can give me some advises on the circuit and show me which part of the circuit to modify so as to achieve the desired output of 230VAC without any load connected. (Maybe i can use a center tap to boost up my input circuit.)
Sorry for your time and really appreciated for your help. Many thanks.
Download Attachment:  DC to AC Inverter.JPG
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| 15 L A T E S T R E P L I E S (Newest First) |
| audioguru |
Posted - Jan 04 2016 : 8:16:44 PM
Why use a 7805? Its output voltage is only 5V so the output transistors or Mosfets barely turn on and the output power from the inverter will be very low. The CD4001 needs to be powered from 12V but a 7812 12V regulator needs an input of at least 14.5V to work properly. |
| sb5_csn |
Posted - Jan 03 2016 : 11:18:47 PM
sorry,it was 47 ohm resistor
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| sb5_csn |
Posted - Jan 03 2016 : 11:16:07 PM
sir,can i use lm7805 without using 470 ohm resistor |
| audioguru |
Posted - Jan 03 2016 : 1:29:02 PM
Measure the output voltage of the 12V regulator. If it is higher than 15V then it and the CD4001 are destroyed. The resistor, zener diode and capacitor prevent their destruction and should always be used in that circuit.
The CD4001 is feeding the gates of Mosfets that use no current so the CD4001 should be cold and not be hot. Maybe the voltage regulator, cD4001 and also the Mosfets are destroyed because you did not use the resistor, zener diode and capacitor. Maybe the CD4001 is destroyed because it had its 12V connected backwards. |
| sb5_csn |
Posted - Jan 03 2016 : 07:52:06 AM
sir, i used lm7812 voltage regulator without using 47 ohm resistor,0.1 uf capacitor and 16volt zenar diode.so,is this condition responsible for over heating of ic cd4001. |
| audioguru |
Posted - Jan 02 2016 : 09:50:10 AM
The circuit that was posted in this thread 8 years ago uses a CD4001 Cmos logic IC powered from a 12V battery and it drives Mosfets that have no input current. The frequency is low and the current is extremely low so the CD4001 IC does not get warm.
Maybe the circuit is missing the resistor feeding a zener diode and capacitor as shown here and voltage spikes have destroyed the IC?
Download Attachment: modified sinewave inverter.png
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| audioguru |
Posted - Jan 02 2016 : 09:23:08 AM
quote:
Originally posted by sb5_csn
sir, due to connecting"CD4001 ic" over heating in ic is occured. please, give a solution in this page.
I do not know the circuit you have that uses a CD4001 IC. Please post the schematic.
The datasheet from Texas Instruments shows the typical output current.
When the IC uses a 5V supply its maximum output current is only a few mA, maybe 5mA per output when shorted. Then the IC heats with 5mA x 4 x 5V= 0.1W and it will be warm. ARE YOU SHORTING ALL 4 OUTPUTS?
When the IC uses a 10V supply its maximum output current is about 22mA per output when shorted. Then the IC heats with 22mA x 4 x 10V= 0.88W which will burn it.
When it has a high impedance load (another CMOS input) then it does not heat up. What is it driving?
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| sb5_csn |
Posted - Jan 01 2016 : 10:37:24 PM
sir, due to connecting"CD4001 ic" over heating in ic is occured. please, give a solution in this page.
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| abdulgtm |
Posted - Jan 17 2010 : 02:14:33 AM
Dear <
Still have u completed ur project or pending , let me know may I becmae help u in this regard
Abdul |
| wasssup1990 |
Posted - Jan 15 2010 : 8:08:13 PM
We're just walking in circles now. |
| audioguru |
Posted - Jan 15 2010 : 6:20:34 PM
We speak English here. I don't know why people from the other side of the world come here when they "no speeky zee engrish".
The power oscillator that steps up the 12V to 340V is at a high frequency so so a small ferrite high frequency transformer is used.
The high frequency is modulated by a 50Hz sine-wave. The modulated high frequency drives the Mosfet H-bridge and an LC filters out the high frequency switching carrier frequency. |
| wasssup1990 |
Posted - Jan 15 2010 : 12:53:28 AM
OMG Jeffrey go and read some electronics books. Do you not understand how PWM works? You're modulating the width of a pulse with a sine 50Hz. The cycle length of a single PWM cycle should be several times smaller in order to reproduce the modulating wave form in PWM. The filter is designed to oscilate at 50Hz and will not pass the PWM frequency of which should be several times higher than the 50Hz modulating signal.
Is it a language barrier for you? I don't know. |
| jeffreywong |
Posted - Jan 14 2010 : 9:04:24 PM
quote:
Originally posted by audioguru
It is easy to make a sine-wave oscillator that has low output power. If you feed it to a power amplifier then the amplifier will make a lot of heat and most of the battery power will be wasted.
It is easy to convert a square-wave to a sine-wave at low power.
A sine-wave inverter uses a high frequency oscillator, a small high frequency transformer and Mosfets to make a suitable high voltage then rectifies and filters it. Then it uses a microcontroller to make pulse-width-modulation at 50Hz or 60Hz and feeds an H-bridge with high voltage switching Mosfets that produce the sine-wave with many steps in it, then two small high frequency filters. Negative feedback is used to regulate the output voltage. A shutdown circuit works during over-current, over-temperature and low battery voltage.
I have some confuse.
In the last part, you said that it use a microcontroller to generate a 50HZ PWM to drive the switches, so why it still need a high frequency filter to filter it since the PWM generate is only 50HZ?
thanks... |
| wasssup1990 |
Posted - Jan 14 2010 : 07:02:47 AM
I think you mean 340VDC to 240AC RMS.
A change in voltage on the primary will cause all secondary coils to produce a derivative voltage. If you single out a 10% PWM duty cycle you will see that for 10% of a PWM cycle the primary coil is being energized - current is increasing in the primary so voltage is being induced in all secondaries. If you put a near 100% duty cycle on the primary, each cycle should only just reach the saturation point of the core material used in the transformer - for maximum efficiency.
In case you didn't understand what I just said in the middle of my last paragraph. The DC part can be confusing to Noobs because it isn't a constant DC. The current on the primary goes from 0 Amps up to V/R where R is the resistance of the coil. So even though you have pulsed a constant voltage across the primary, the primary DC is increasing from 0 Amps over time which in so doing induces a voltage in all secondary coils. I suggest you go and open an electronics book. You'll learn a lot more and a lot quicker than any of us can teach you. Just like everything in the universe, a change of some sort is needed to do any kind of action. The same with transformers - a stepped constant voltage causes a change in current in any kind of medium which conducts electricity until it reaches a maxima of V/R. I could go on and on...
Happy thinking  |
| jeffreywong |
Posted - Jan 14 2010 : 04:10:48 AM
quote:
Originally posted by audioguru
It is easy to make a sine-wave oscillator that has low output power. If you feed it to a power amplifier then the amplifier will make a lot of heat and most of the battery power will be wasted.
It is easy to convert a square-wave to a sine-wave at low power.
A sine-wave inverter uses a high frequency oscillator, a small high frequency transformer and Mosfets to make a suitable high voltage then rectifies and filters it. Then it uses a microcontroller to make pulse-width-modulation at 50Hz or 60Hz and feeds an H-bridge with high voltage switching Mosfets that produce the sine-wave with many steps in it, then two small high frequency filters. Negative feedback is used to regulate the output voltage. A shutdown circuit works during over-current, over-temperature and low battery voltage.
I have some confusing about it.
What u mean is that a pure sine wave inverter have to transform a 12VDC to 240VDC then use mosfet switches to invert 240VDC to 240VAC.
My question is how can the DC power can run in transformer?
thanks... |
modified sinewave inverter.png