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Low-pass Filter?


audioguru

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for backup computer for a short while only.

I don't know if a computer's power supply will work from a square wave.
I wouldn't use a simple low cost inverter on my computer.

the secondary transformer output is 240V,
240V x 2A = 480VA right?
why time the 2A with the primary voltage; 12v?

Your transformer is probably rated as a voltage step-down transformer with a small 24V/2A output. That is only 48VA. A 480VA inverter will draw 40A from 12V and therefore needs a huge transformer.
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  • 2 weeks later...

Hi audioguru,

1. how come the -TRG (pin6) of the CD4047 is not ground like +TRG (pin8)?
2. The output of the Q (pin10) & -Q (pin11) is a square wave right, then what is the function of LM358? square wave generator?
3. what is the function of 2SC1061 and 2N3055?
4. why connect 12VDC to the pin2 of the center tap transformer?

Thanks

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1. how come the -TRG (pin6) of the CD4047 is not ground like +TRG (pin8)?

On the datasheet it shows that it must be wired that way to oscillate.

2. The output of the Q (pin10) & -Q (pin11) is a square wave right, then what is the function of LM358? square wave generator?

The LM358 is a dual current amplifier. The output current from Cmos logic devices like a CD4047 is very low.

3. what is the function of 2SC1061 and 2N3055?

They are current amplifiers.

4. why connect 12VDC to the pin2 of the center tap transformer?

The transformer is used push-pull.
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A CD4047 is destroyed much quicker than a fuse can blow. It is a supply voltage spike that causes the damage. A fuse slowly protects against current overload.

The very low impedance of a car battery as a supply for the project has proven to keep supply voltage spikes down.

The addition of a 100 ohm resistor feeding positive power to the CD4047 and a 16V zener diode across it will protect it. 

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This uses a push-pull topology, so only one half of the primary is active at any point in time. Each half winding has 12V across it relative to the center-tap. Through flyback action, the opposite winding end will have 2 times the input voltage when the switching element on side that's being measured is switched off. This is the reason push-pull primary switching elements have to have a voltage rating of at least 2X Vin. If your measuring 24VAC, your measuring from gnd to one end of the transformer primary winding.

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  • 2 months later...

Hello, i have seen you proyect, and i would like to know if the IC is only to make a square wave (and the negated) to control the two LM358?
what level of signal is at the output of Q and /Q? could it be maded with a simple n555,R and C circuit?
What will happend if i set to the LM358 a  sinusoidal signal?coult it generate a SIN signal at the transformer?or does it work as a comparator and it gets 12 - 0 - 12 - 0 (Q=1, Q=0, Q=1, Q=0)?
and what about to set the Sinsuidal generator after LM358 and driving the MOSFETs?The Sinusoidal must make only one Pi ,i mean  PI=3'14,the half of the period, to generate a half wave each time with each LM358

Thank you Very Much. Great proyect ;)

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Hello, i have seen you proyect, and i would like to know if the IC is only to make a square wave (and the negated) to control the two LM358?

Yes, the CD4047 has a frequency divider so its outputs are a perfect square wave.

what level of signal is at the output of Q and /Q? could it be maded with a simple n555,R and C circuit?

The CD4047 is Cmos so with its very light load its output voltage swing is the supply voltage.
A 555 doesn't make a perfect square wave. The CD4047 circuit is perfect, simple and cheap.

What will happend if i set to the LM358 a
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Hello and Thanks, now i have the question, you said some replies up, you will not use to give power to a computer, because it needs a sin wave, then if the effective value for a wave is ->  Vf / (2^(0'5) and 220V/(2^(0'5) = 156 V

if you use a Transformer 12 - 0 - 12 To 156 will be usefull, isn't it?

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can i use tip142 for the darlington pair in 500wat inverter to rdive 2n3055s?

Then you might need to select them for high gain and also select 2N3055 transistors for high gain.

With a 500W load, the battery current is 50A. Therefore the 4 sets of output transistors pass 12.5A for each transistor. The minimum current gain for a 2N3055 is 5 with a 10A collector current so might be only 4 with 12.5A. Therefore the total base current for 4 output transistors could be as high as 12.5A.
The TIP142 isn't spec'd with a 12.5A load and its minimum gain is not high enough and its saturation voltage is poor at only 10A.
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  • 4 weeks later...

Now Can we use a Sine wave oscillator instead of a square wave osicllator , this would increase performance for runnning systems that need sine waves?

Are you talking about a power inverter?
What about the heat?
In a square-wave inverter, the output transistors alternately switch fully on or fully off to reduce power wasted as heat. A good inverter using transistors is about 80% efficient and one using Mosfets is about 90% efficient. So to produce 1000W output the Mosfet one would heat with only 110W and draw 92.5A from a 12V battery.
A simple sine-wave inverter is just a linear audio amplifier. The output transistors are conducting all the time and therefore waste a lot of power as heat. It would be about only 50% efficient. So to produce 1000W output it would heat with 1000W and melt, and draw 166.7A from a 12V battery.

A Class-D audio amplifier uses Pulse-Width-Modulation so that its output transistors can switch fully on and fully off and remain fairly cool, but still produce a low frequency sine-wave. It would make a good sine-wave inverter but is complicated.
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There are many high power Class-D audio amplifiers available.

I don't think a computer's power supply works with a square-wave mains.

A "ferro-resonant" transformer has an additional winding that is tuned with a capacitor so a square-wave input makes a sine-wave output. Its core saturates in order to regulate the output voltage.
If you tune an ordinary transformer then its output voltage will change when the load changes.

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