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kivdenn
Nobel Prize Winner

Uganda
535 Posts
Posted - Apr 24 2008 :  10:35:11 AM  Show Profile  Reply with Quote
Actualy I drew the diagram above some time back but now I use BC546 instead of the TIP41C. Talking about that, Is it posible to convert the low battery disconnect part of the circuit into a battery over voltage disconnect circuit.
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audioguru
Nobel Prize Winner

Canada
4218 Posts
Posted - Apr 24 2008 :  12:18:48 PM  Show Profile  Reply with Quote
I think a window comparator using both comparators in an LM393 will be a good overvoltage cutoff and an undervoltage cutoff circuit.
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kivdenn
Nobel Prize Winner

Uganda
535 Posts
Posted - Apr 25 2008 :  07:57:33 AM  Show Profile  Reply with Quote
I have always pleaded to you Audioguru to help me with that kind of circuit but all in vein so I think talking about it here is useless because at the end we shall never have it. Lets go with the cheap one that we have, rermnber a bird in hand is better than two in the bush.
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audioguru
Nobel Prize Winner

Canada
4218 Posts
Posted - Apr 25 2008 :  09:03:31 AM  Show Profile  Reply with Quote
Please look at Window Comparator Circuit in Google. You will find some good circuits with explanations about how to set the threshold voltages.
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kivdenn
Nobel Prize Winner

Uganda
535 Posts
Posted - Apr 25 2008 :  09:53:57 AM  Show Profile  Reply with Quote
I have to done all that but all in vein. All I find are fan speed control circuits and the rest but not low battery disconnect circuits
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audioguru
Nobel Prize Winner

Canada
4218 Posts
Posted - Apr 25 2008 :  11:16:51 AM  Show Profile  Reply with Quote
I entered Window Comparator Circuit into Google. The very first link has this good circuit.

The output is low when the input voltage is higher than reference #2 and the output is also low when the input is less than reference #1.
The output is high (but with low current as determined by the LM393 dual comparator IC) when the input voltage is in the designed window.

Download Attachment: window comparator circuit.PNG
6.47 KB
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kivdenn
Nobel Prize Winner

Uganda
535 Posts
Posted - Apr 26 2008 :  05:45:36 AM  Show Profile  Reply with Quote
quote:
Originally posted by audioguru

I entered Window Comparator Circuit into Google. The very first link has this good circuit.

The output is low when the input voltage is higher than reference #2 and the output is also low when the input is less than reference #1.
The output is high (but with low current as determined by the LM393 dual comparator IC) when the input voltage is in the designed window.

Download Attachment: window comparator circuit.PNG
6.47 KB



I have also seen the same circuit before but it is incoplete. The resistor values are not shown and also I dont believe it can drive a relay directly, it needs a current amplifier circuit which is not added it also doesnt indicate the exact voltages it was built for either 12v or 24V. Lets trey to solve those problems then it can serve its purpose. Otherwise thanks forall your effort.
Dennis
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audioguru
Nobel Prize Winner

Canada
4218 Posts
Posted - Apr 26 2008 :  09:18:46 AM  Show Profile  Reply with Quote
quote:
Originally posted by kivdenn
I have also seen the same circuit before but it is incomplete. The resistor values are not shown

The datasheet for the comparator shows that its input current is extremely low so just about any resistor voltage dividers or zener diodes can be used.

quote:
I dont believe it can drive a relay directly, it needs a current amplifier circuit which is not added

Its datasheet shows that its minimum output current is only 4mA so a darlington transistor or Mosfet must be used to drive your big high current relay.

quote:
it also doesnt indicate the exact voltages it was built for either 12v or 24V

The voltage dividers or zener diodes are simple to calculate for any threshold voltages from 0V to 34.5V.
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audioguru
Nobel Prize Winner

Canada
4218 Posts
Posted - Apr 27 2008 :  09:00:24 AM  Show Profile  Reply with Quote
Hi Juan,
All those old-fashioned transistors can be replaced by two modern Mosfets.

You have the power supply connected backwards on the CD4001. Pin 14 is supposed to be positive and pin 7 is supposed to be 0V.

The CD4047 and CD4001 both should be powered from a 100 ohm resistor in series from the 12V switch then a 16V zener diode to 0V at the ICs.

The ICs do not have enough output current to drive the transistors. Opamps were used in the original circuit because they have higher output current.

The transformer should be 9V-0V-9V for this modified sine-wave circuit so that the average output voltage is high enough.

The output needs to have a high voltage capacitor in series with a power resistor to reduce voltage spikes. They must match the transformer so use an oscilloscope to see what works best.

The power output from this circuit must be limited to 350W because the "modified sine-wave" creates peak currents in the transistors that are much higher than in a square-wave inverter circuit.
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audioguru
Nobel Prize Winner

Canada
4218 Posts
Posted - Apr 28 2008 :  1:33:30 PM  Show Profile  Reply with Quote
I explained what needs to be fixed on your schematic.
I have very complicated circuits for Modified-Sine-Wave inverters that use Mosfets and other parts that are not available in the Philippines.
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audioguru
Nobel Prize Winner

Canada
4218 Posts
Posted - Apr 28 2008 :  11:32:25 PM  Show Profile  Reply with Quote
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.
Edited by - audioguru on Apr 28 2008 11:35:01 PM
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audioguru
Nobel Prize Winner

Canada
4218 Posts
Posted - Apr 29 2008 :  12:19:31 AM  Show Profile  Reply with Quote
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.
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audioguru
Nobel Prize Winner

Canada
4218 Posts
Posted - Apr 29 2008 :  11:16:35 AM  Show Profile  Reply with Quote
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.
Edited by - audioguru on Apr 29 2008 11:17:58 AM
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audioguru
Nobel Prize Winner

Canada
4218 Posts
Posted - Apr 29 2008 :  11:20:45 AM  Show Profile  Reply with Quote
Power the LM358 from the 12V feeding the 100 ohm resistor, not at the zener diode.
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audioguru
Nobel Prize Winner

Canada
4218 Posts
Posted - Apr 30 2008 :  12:34:00 AM  Show Profile  Reply with Quote
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.
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