audioguru2
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Your revised circuit is much better.
1k is so high that the batteries will get hardly any charging current.
The resistor's value is calculated from the input voltage when loaded, the battery's recommended charging current and time. We can assume that the lead-acid batteries each have an average voltage of 6.6V.
1k is so high that the batteries will get hardly any charging current.
The resistor's value is calculated from the input voltage when loaded, the battery's recommended charging current and time. We can assume that the lead-acid batteries each have an average voltage of 6.6V.
Just a few more questions:
The supply is 18V, the batts spesifies less than 1.20A and say the sermons lasts at most 2 hours the resistor will be 30 ohms, correct?
If the batts specify 6V4.0AH/20HR each how quickly will the batteries run down by the laptop?
What are the risks, if any with the way we want to use this?
The supply is 18V, the batts spesifies less than 1.20A and say the sermons lasts at most 2 hours the resistor will be 30 ohms, correct?
If the batts specify 6V4.0AH/20HR each how quickly will the batteries run down by the laptop?
What are the risks, if any with the way we want to use this?
audioguru2
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18VDC is too low to charge an "18V" lead-acid battery. The battery needs at least 21.6V to fully charge.JuniorHP said:
What spec is that? Charging current?
No. The supply voltage is too low to charge the battery.
You calculate the time only if you know the current drain of the laptop.
Just make sure that the battery is never connected backwards.
Having recently acquired a PIC development board that requires 16 v., I decided to build this LM317 PSU from the projects section of this board.
It took less than two hours to construct, using junk box parts, (including the LM317 already on hand).
It took less than two hours to construct, using junk box parts, (including the LM317 already on hand).
audioguru2
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I also made a variable supply with an LM317 IC.
Its regulation is extremely good and its output doesn't have any ripple.
Its regulation is extremely good and its output doesn't have any ripple.
audioguru2
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audioguru2
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Then calculate a current-limiting resistor with the supply voltage that you have so that the max current spec is not exceeded.JuniorHP said:
audioguru2
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You cannot connect two LM317 variable voltage regulators in parallel because each one will have a different output voltage and the one with the highest voltage will power the load and the other one will do nothing or it will be damaged.
You can connect a low value resistor in series with the output of each one if they have a fixed output voltage the same. The outputs of the resistors are connected together to give more current than one regulator. The resistors will ruin the voltage regulation when the load current changes.
An LM350 has a max current of 3A and an LM338 has a max current of 5A but read the fine print in the datasheet for the max voltage from input to output before they reduce the max current.
You can connect a low value resistor in series with the output of each one if they have a fixed output voltage the same. The outputs of the resistors are connected together to give more current than one regulator. The resistors will ruin the voltage regulation when the load current changes.
An LM350 has a max current of 3A and an LM338 has a max current of 5A but read the fine print in the datasheet for the max voltage from input to output before they reduce the max current.
x_dadu,
You do not need more than one LM317 to do this. You can use your LM317 regulator to excite the base of a power transistor. This will allow you more current from your power supply. If you need more current than what this allows, you can certainly parallel the power transistors as long as you add a small resistance from each so that the load is shared equally between all of them.
MP
You do not need more than one LM317 to do this. You can use your LM317 regulator to excite the base of a power transistor. This will allow you more current from your power supply. If you need more current than what this allows, you can certainly parallel the power transistors as long as you add a small resistance from each so that the load is shared equally between all of them.
MP
ok I understand that but if I put a paralel tranzistor, short circuit protection is still active ? What can u tell me about ripple output rejection ? how can I measure that ? I set output for 3Vcc and conect an lihgt bulb who draw 530mA from 317. On the osciloscop I se an ondulation of 15mV. Is it bad ? What level is good for riple ?
audioguru2
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No. The datasheet for a 78xx fixed voltage regulator shows how to add a 2nd transistor for current limiting.x_dadu said:
The amount of ripple on the output depends on a few things:
1) Each LM317 is different. Some are excellent and some are not as good. Look at the range of ripple rejection spec's on the datasheet.
2) The wiring of the resistors could add resistance that reduces the voltage regulation of an LM317.
3) The datasheet shows that adding a 10uF filter capacitor from the ADJ terminal to ground reduces the ripple to 1/3rd.
4) The amount of ripple at the input determines how much ripple is at the output.
ok thank you for answer but there sitll something that I dont like. I measure now the ripple on a 25ohm/18W rezistor with 20Vcc on it, riple was 20mV with 770mA trough the resistor. But when I rise the volotaje to 22 or more (25 is maxim with my 19vca traf) the riple was > 350mV. It seems that after 21 Vcc the ripple is rise imidiatly . What is hapaning with LM317 closer that the maximum output V? I also have the 10uF capacitor on ADJ and a 4700 on it input, but I put an 100uF at it output, is that ok?
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audioguru2
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Look at the datasheet for the LM317:
Its input voltage must be at least 2.5V or 3V higher than its output voltage or it fails to regulate and passes the input ripple to its output.
The factory tests ripple rejection with its input 5V higher than its output voltage.
Its input voltage must be at least 2.5V or 3V higher than its output voltage or it fails to regulate and passes the input ripple to its output.
The factory tests ripple rejection with its input 5V higher than its output voltage.
one quastion pls, is there a problem if I use 390ohm for R1 ? I use that rezistor because I have just a 10kohm potentiometer and a 2,2k one. I use r1=390ohm and r2=10kohm pot. Is that ok ? If I use the 2,2k pot, I must use aprox. 100ohm for R1, it will be ok a 100ohm/0,5W ?
audioguru2
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R1 is supposed to be 120 ohms or less for an LM317. If its resistance is higher then its output voltage might rise when there is no load.
If R1 is 100 ohms and the pot is 2.2k then at max setting the voltage will try to reach 28.75V but it can't, so the range of the pot will be reduced.
There is low power in R1, a 1/4W resistor is fine.
If R1 is 120 ohms then the pot should be 2k for a max output of 22V if your transformer is 19VAC. The power in the 2k pot is 217mW so an ordinary 1/2W pot is fine.
If R1 is 100 ohms and the pot is 2.2k then at max setting the voltage will try to reach 28.75V but it can't, so the range of the pot will be reduced.
There is low power in R1, a 1/4W resistor is fine.
If R1 is 120 ohms then the pot should be 2k for a max output of 22V if your transformer is 19VAC. The power in the 2k pot is 217mW so an ordinary 1/2W pot is fine.
