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 12/120V inverter again
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audioguru
Nobel Prize Winner

Canada
4214 Posts

Posted - May 03 2008 :  11:22:50 AM  Show Profile  Reply with Quote
Juan,
If you use a single CD4001 or CD4011 then the logic circuit must be re-designed for it.

Please look at the datasheet for the MJ11032.
With an output current of 25A its input current is only 250mA for it to have a max saturation voltage loss of 2.5V and it will dissipate a max average power of 31.3W. The 0.1 ohm emitter resistor will have a voltage loss of 2.5V and also dissipate an average power of 31.3W.
Four MJ11032 darlingtons on each side will have a max total input current of 1A and can be driven with a TIP31 or TIP41. They will add to the saturation voltage loss.
So nearly half the power from the battery is wasted.
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audioguru
Nobel Prize Winner

Canada
4214 Posts

Posted - May 04 2008 :  12:07:37 AM  Show Profile  Reply with Quote
quote:
Originally posted by JUAN DELA CRUZ
If I will use only one MJ11032 (each side) to simulate the use of "8" 2N3055
And drive with TIP41.

Four 2N3055 transistors are good for about 50A. One MJ11032 darlington is good for about 50A. The TIP41 will have a max base current of about 50ma so another transistor is needed.

The total max saturation voltage loss will be about 5.7V which is too high. The max output power will be only about 340W if a lower voltage transformer is used. a lot of battery power is wasted if the transistors have low gain.

quote]The remaining components will be the same excluding the EMITTER Resistor?[/quote]
Yes, the emitter resistors are used to match the conduction of different transistors and are not needed with only a single transistor on each side.
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audioguru
Nobel Prize Winner

Canada
4214 Posts

Posted - May 05 2008 :  11:50:28 AM  Show Profile  Reply with Quote
Welcome to 1978. Transistors are a poor choice for a high current switching circuit compared to modern Mosfets.
The BUT100 is the most powerful transistor I have ever seen. The two major electronics parts distributors in America have never seen it either. Maybe it is no longer made.

The transistors are not saturated switches, they are in a darlington configuration where their base-emitter voltages add to create a total voltage loss:
The max VBE for a BUT100 is 1.4V at 50A.
The max VBE for an NTE43 is about 1.2V at 2.5A.
The max VBE for a 2N2222 at 250mA is about 1.4V.
The total max voltage loss is 4.0V.

All transistors are different. You might be lucky and get some with high gain. Then the voltage loss is about 3.0V.

A "12V" battery is 12.8V to 13.8V when fully charged. So the pulses to the transformer will be 8.8V to 10.8V.
When the battery voltage drops to about 12.0V when it is nearly dead then the pulses to the transformer will be from 8.0V to 9.0V.

The transformer might be good and not have much loss, or it might be cheap and have a high loss.
In order for it to produce a modified sine-wave the low voltage winding must be rated for 0.707 times its ordinary voltage rating.
Use an 8V-0V-8V transformer.

The BUT100 transistors do not need an emitter resistor because there is only one transistor on each side.
The other two resistors and diode are needed on each side.
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audioguru
Nobel Prize Winner

Canada
4214 Posts

Posted - May 06 2008 :  12:52:21 AM  Show Profile  Reply with Quote
You need to re-design the set-reset latch so it can use the unused gates of the CD4001. It is designed for a CD4011 now.
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audioguru
Nobel Prize Winner

Canada
4214 Posts

Posted - May 06 2008 :  08:54:57 AM  Show Profile  Reply with Quote
The NTE180 is PNP not NPN so won't work in this circuit. The NTE181 is an NPN but its spec's stop at only 7.5A.
NTE transistors do not have full spec's because they are "replacements" for standard transistors that have full spec's.
The Japanese transistor will be fine as a pre-driver transistor.

The low battery cutoff circuit doesn't need the TIP31 and relay. Its output can stop the oscillator in the CD4047.

I don't think the set-reset latch should connect to pin 5 of the CD4047 because then one output will keep the transistors turned on and blow the fuse. I remember in January we discussed using a CD4025 three input NOR in the modified sine-wave circuit and the 3rd inputs are used to turn off all the transistors.

The transformer needs only one resistor in series with a capacitor at the output. The values depend on the transformer since every transformer is different. You attach an oscilloscope to the output and adjust the values so that the voltage spikes are reduced without a load.
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osama madni
New Member

Pakistan
3 Posts

Posted - May 07 2008 :  09:44:53 AM  Show Profile  Send osama madni a Yahoo! Message  Reply with Quote
how i change transformer plz help me and capacitor
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osama madni
New Member

Pakistan
3 Posts

Posted - May 07 2008 :  09:59:25 AM  Show Profile  Send osama madni a Yahoo! Message  Reply with Quote
sir i have a problem with transformer and capacitor in my envirement there is not found such capacitor
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audioguru
Nobel Prize Winner

Canada
4214 Posts

Posted - May 07 2008 :  12:21:36 PM  Show Profile  Reply with Quote
Hi Osama,
Which circuit has a transformer you want to change? Why do you want to change it?

Which capacitor in what circuit are you talking about?
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audioguru
Nobel Prize Winner

Canada
4214 Posts

Posted - May 08 2008 :  12:16:53 AM  Show Profile  Reply with Quote
The NTE line of ICs are replacements for real ICs. So their spec's are not detailed like real ICs. The datasheet lists spec's only at 7.5A. How much base current is needed at 20A and what is its max VBE??

I am not going to guess how well or how poorly it will perform.
i think there will be a high voltage loss.

The CD4025 is a NOR gate like a CD4001 except it has a 3rd input on each gate. When the 3rd input goes high from the low battery circuit then its Mosfet is turned off.

There are millions of opamps. Some have low current outputs and some have high current outputs. I have never looked for nor used one with high current outputs. Have fun looking for one.
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audioguru
Nobel Prize Winner

Canada
4214 Posts

Posted - May 08 2008 :  09:03:57 AM  Show Profile  Reply with Quote
I was wrong when I talked about pin 5 of the CD4047. It will stop the oscillator then one side of transistors will be continuously conducting and then the fuse will blow. Use a CD4025 instead of a CD4001 to turn off both sides.

The peak current in the transistors of a modified sine-wave inverter is 1.414 times higher than the peak current in a square-wave inverter with the same power output. So if you want 500W output then the peak current in the transistors must be 50A x 1.414= 70.7A. A perfect job for Mosfets.

You don't have detailed spec's for the NTE181 transistors so you are just guessing about how much base current they need and how much voltage loss they will have.
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audioguru
Nobel Prize Winner

Canada
4214 Posts

Posted - May 09 2008 :  10:37:35 AM  Show Profile  Reply with Quote
How would I know anything about which parts are available in The Philippines?
Here in Canada, any electronic part is available. There are thousands of Mosfets available to do the job.

I would not re-design the circuit so it can use many 2N3055 transistors.
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JUAN DELA CRUZ
Mad Scientist

Philippines
476 Posts

Posted - May 09 2008 :  11:31:19 PM  Show Profile  Reply with Quote
quote:
Originally posted by audioguru

How would I know anything about which parts are available in The Philippines?
Here in Canada, any electronic part is available. There are thousands of Mosfets available to do the job.
quote:


THANK YOU MR.AUDIOGURU FOR YOUR REPLY....
......I think just name Commonly used MOSFET..or the one you've used maybe..Please just to have an IDEA

I would not re-design the circuit so it can use many 2N3055 transistors.

quote:


How is "4" 2N3055( in each side) used in the modified inverter capable of supplying "350W" output power even the Total output current in that "4" 2N3055 is just "20A"............
( 20A x 9V = 180W) because the output current from the 2N3055 driver is "2A" only that driving the four.For the reason that the output current from 2SC1061 is merely "200mA" (since the output current from the OP-AMP was "20mA").
???

P.S.
What is the output Voltage & Current of CD4001(driving the dual OP-amp LM358)?????

juan dela cruz
Penniless INVENTOR

Edited by - JUAN DELA CRUZ on May 10 2008 05:45:22 AM
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audioguru
Nobel Prize Winner

Canada
4214 Posts

Posted - May 10 2008 :  05:57:31 AM  Show Profile  Reply with Quote
I have the datasheet for the IRFZ44 Mosfet on my hard drive. When it is turned on and is not too hot, its resistance is 0.028 ohms max. So with a current of 30A its voltage drop is only 0.84V and it heats with only 25.2W. There are better Mosfets.

The max output current of the opamp is 20mA. The max output current of the pre-driver transistor is 400mA. The max output current of the 2N3055 driver transistor is 8A. The max output current of the four 2N3055 transistors on each side is 60A if they have enough gain and if they don't get too hot.

A 2N3055 transistor has a max voltage drop of 3V when it has a collector current of 10A and a base current of 3.3A. Four of them in parallel will have a max voltage drop of 3V when the collector currents total 40A and the base currents total 13.2A.

The output is 350W so the input is about 420W. The peak current is 420W/9V= 46.7a which is 11.7A in each output transistor. So some 2N3055 transistors with low gain won't work in a 350W modified sine-wave inverter.

The CD4xxx series of ICs are Cmos. Their supply current is zero at low frequencies when they are not working hard charging and discharging stray capacitance.
Opamps have a very hin input resistance and a very low input current so the Cmos oscillator has nearly no load. The supply current of the LM358 opamp is low but has the input current of the first transistors.

It doesn't matter what the ouput voltage of the opamp is because it limits the current into the base of the first transistor. The total of the base-emitter voltages of half of the transistors limits the output voltage of the opamp.
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audioguru
Nobel Prize Winner

Canada
4214 Posts

Posted - May 10 2008 :  11:42:58 AM  Show Profile  Reply with Quote
The max Vbe of the 2SD880 is about 1.0V, the two 2N3055 transistors is about 1.4V and the BUT100 is 2.0V. Then the voltage loss is 4.4V. Nearly half the power from the battery is wasted.
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audioguru
Nobel Prize Winner

Canada
4214 Posts

Posted - May 11 2008 :  11:49:42 AM  Show Profile  Reply with Quote
Juan,
Do a couple of simple calculations:
1) Each 2N3055 transistor needs a max base current of 3.3A when its collector current is 10A. The max allowed current for the 2SD880 is too low.
2) The input current of whatever higher current transistor replacing the 2SD880 transistors is too high.
3) 100A through the 0.1 ohm resistors is a 10V loss!
4) Where are you going to find a 12V battery that can supply 200A without blowing up?
5) Your transformer is missing its center-tap. Where are you going to find such a huge transformer?

The voltage loss of the transistors depends on their gain. Each one has a different gain. So the output voltage will be too high with some transistors and too low with other transistors.
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