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.
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.
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.
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.
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
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.
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.
Here we go again: 1) The max current from the opamps is only 20mA so the max input to each 2SD880 transistor is only 20mA and its max output current might be 300mA. 2) The max input to the 2N3055 transistor is 300mA and its max output might be 4.5A. 3) The max input to the BUT100 is 4.5A and its output might be 45A.
The peak output current is 49.8A so the RMS output current is 35.2A. The peak output voltage is 8.2V so the RMS voltage is 5.8V. The transformer should be 6V-0V-6V for the output voltage to be correct. The power into the transformer is 35.2A x 5.8V= 204W RMS. If the transistors have low gain then the output power and output voltage is less. The transistors have a power loss which makes 95W of heat. The average current from the battery is 24.9A.
If you use a pair of modern Mosfets then the power to the transformer will be about 400W and the voltage will be correct. 2 pairs of Mosfets make 800W.
The LM10 circuit has a low logic output when the battery voltage is low. It is designed to give a low logic input to a NAND gate. It must be re-designed to give a high logic input to a NOR gate. Or you can use the spare 3rd gate in the CD4025 to invert the output from the LM10.
quote:Originally posted by JUAN DELA CRUZ Can I use "2 Pair of IRFZ44 Mosfet" to obtain "800W" output..???? ..........so if I'll use "3 pair of IRFZ44 Mosfet the output power will be "1200W"..????? I think so but I have never tried to blow up a car battery with that much current.
How about this circuit( w/ LM741)??? It will do the same as the LM10 circuit.
quote:Can this circuit can supply "High" current to turn-off ......"TWO NOR GATE of CD4025"????
Its output voltage is low not high. It needs to use the 3rd unused gate in the CD4025 as an inverter to feed a logic high voltage to the CD4025 gates so they can turn off the Mosfets.
quote:Can I connect it directly to series resistor & Zener voltage spike filter???
I think so.
[quote]What is the purpose of "100K Pot."????.....What type??? ...like the Pot. used in a Audio Amp.????? .... How many Wattage is needed???
It adjusts the voltage for the low voltage cutoff to work. It is not a logarithmic volume control, it is a linear pot. Calculate the tiny amount of power it dissipates.
A zener diode limits a voltage spike. A 15V regulator does not work when its input is less than 18V and a voltage spike goes through it because it reacts too slowly.
You found a very old circuit. I think the 1k resistors in series with the gates of the Mosfets should be 47 ohms. The resistors have an extremely low current in them so 1/4W is fine.
I think a transformer from a microwave oven is cheap. It is designed for full power for a short time so it wastes a lot of power if it is not at full power. It might overheat and its output voltage might have poor voltage regulation. The transformer should be 8V-0-8V. With an 800W load the average current from the battery is about 75A. A huge car battery will last for about half an hour if it doesn't boil dry sooner.
The capacitor in the snubber is certainly not 220uF. It is 220nF which is 1000 times smaller. The values of the snubber and where they are placed depends on how the transformer makes voltage spikes. Each make of transformer is different.
Most of your schematics have disappeared so I don't know what you are doing amymore.
Now you have inverted the input to the flip-plop latch so it won't work. You need to invert its output instead. I always inhibit unused inputs by connecting them to the positive supplu or to ground like this:
sir i will be happy if you can send me schmatic diagram of 12vdc to 240vac, and it will supplyed by solar panel of 12vdc and lead acid battery of 12vdc please i need immediate diagram for my projrct work(i need sine wave inverter diagram)