quote:Originally posted by JUAN DELA CRUZ Meaning to say that both IC must have a regulated supply??? The 100 ohm is a BLEEDER Resistor that limit the current across the 16V Zener???
The ICs are very low current and operate from the 13.8V from the battery. The 16V zener diode limits voltage spikes.
quote:Want OP_AMP did you used in the your circuit?? CD4001 is a OP-Amp right???
Didn't you see the 500W square-wave inverter that I helped fix with Rhonn from The Philippines? Look in Google or in www.datasheetarchive.com at the CD4001 Cmos low current quad 2-input NOR logic gate. It is not an opamp.
quote:You mean I need to use a BYPASS Capacitor across the Emitter Resistor??? What type & value must be the BYPASS Capacitor???
No. I did not say a bypass capacitor across the emitter resistor. I said, "The output needs to have a high voltage capacitor in series with a power resistor to reduce voltage spikes." They are in series and are directly across the output of the transformer. Every transformer needs different values. Use an oscilloscope to see which values work best.
Ronnie made a square-wave inverter. It doesn't use the CD4001 logic gate. His inverter uses an LM358 dual opamp to boost the low current from the CD4047 Cmos IC.
You want to make an inverter that produces a modified sine-wave. The CD4001 Cmos makes the modified sine-wave. Its outputs also must have the current boosted with an LM358 dual opamp. The modified sine-wave inverter has less output power and must use a 9V-0V-9V transformer.
I have never seen a transformer made in The Philippines. So I don't know the value of the resistor and capacitor voltage spike filter parts to match it. Maybe they are not needed.
Use Ohm's Law to calculate the power dissipation in the 100 ohm resistor and in the zener diode. The voltage spike might be 24V and occur for 1/100th the total time. So the 100 ohm resistor will have 24V - 16V= 8V across it for 100th the total time. Then its current is 8/100= 80mA for 100th of the time and it dissipates an average power of 0.006W. Use a 1/4W resistor. Or use 1/2W.
The power in the zener diode is about 4 times the power in the resistor so it dissipates an average power of only 0.024W. use a 400mW or 500mW zener diode.
The 4.7k resistors have the input current of the opamps in them. The datasheet for the opamps says that the max input current is only 0.24uA which is almost nothing. Use 1/4W resistors. Or use 1/2W.
You cannot calculate the resistor and capacitor across the output of the transformer unless you measure its leakage inductance and internal capacitance. Just connect an oscilloscope to the output and adjust the values for the best results. The voltage spikes are the highest without a load.
I think you should add a circuit that turns off the inverter when the battery voltage gets low so the low voltage doesn't destroy the battery or blow the fuse.
Please look at the datasheets for the ICs. 1)The max allowed supply voltage for the CD4xxx is omly 18V. The max allowed supply voltage for the LM358 is 36V. 2) The operating current of a CD4xxx is almost nothing. The operating current of an LM358 when it drives 20mA into a transistor is 22mA which causes a voltage loss of 2.2V in the 100 ohms resistor.
The "snubber" is a capacitor in series with a resistor and is connected at the output of the inverter to reduce voltage spikes. The capacitor must have a voltage rating that is higher than the peak output voltage.
Two back-to-back zener diodes in series with a resistor can also be used at the output to clamp voltage spikes.
We have talked about Low Voltage Cutoff circuits in this forum. It should have a flip-flop latch to make it stay turned off when the battery voltage rises when its load is disconnected.
The LM10 has a low current opamp and a low current voltage reference inside. The other circuit uses much more current with a lousy old 741 opamp and a zener diode to do the voltage detection.
The first circuit has a flip-flop latch to keep the battery disconnected when its voltage rises when its load is disconnected. The second circuit will detect a low battery and disconnect it or turn off the inverter's oscillator. Then it will detect the battery voltage rising without a load and connect it. It will turn on then turn off over and over.
The TIP31C is a power transistor that can activate then disconnect a high current relay that applies power to the inverter. The second circuit has a small low current transistor that stops the CD4047 to stop the inverter.
The high current of the 741 opamp and zener diode in the second circuit needs re-design of the 100 ohm resistor and 16V zener diode to power them and protect them from voltage spikes. The low current of the first circuit allows it to be powered from the 100 ohm resistor and 16V zener diode with the CD4047.
Two gates of the CD4001 are used for the modified sine-wave logic and the remaining two gates are used for the flip-flop latch.
The circuit should work as it is. Once the battery is cut off its voltage will have to rise to about 18V before the circuit will cut in again.
Then the circuit will work only one time and never work again. The circuit needs to have a reset pushbutton like my flip-flop circuit has. The pushbutton in my circuit has a low current so a switch with gold contacts should be used. The gold plated contacts cost the same or pennies more than high current silver that corrodes and doesn't work with low currents.
I realize that one of my circuits uses a CD4001 quad NOR IC and the other circuit uses a CD4011 quad NAND IC. It wouldn't be difficult to change one to the other so a single logic IC can be used.
The LM10 in my circuit performs exactly the same function as the circuit with the lousy old 741 opamp. The LM10 doesn't have a 200mV battery. It has a 200mV adjustable voltage reference circuit plus an opamp inside.
The MJ11032 is a darlington that has a much higher voltage drop than a 2N3055 power transistor. The circuit should be re-designed for it. Power Mosfets should be used instead.
You are talking about a lot of power. A 12V car battery will boil or explode when it tries to drive such a powerful load. Where will you get the huge custom-made transformer? Rediculous.
Hi Senahia, We have many schematics for inverters in this foum. Some use old transistors and others use new efficient Mosfets. Some are 100W and others are almost 1000W. Some have a square-wave outputs that don't power many electronic products. Some are modified sine-wave that is like most inverters in stores. Pure sine-wave inverters are too complicated for a forum. You don't say what you want.
Solar power is a separate topic. Battery chargers are also a separate topic.