Hero999

The 4N25 has a maximum voltage rating of 60V and a current rating of just 50mA so is unsuitable for driving the motor directly. You haven't provided enough information. How much current does the motor use? Where are you getting the 220VDC from? Is it a smooth supply such as a battery or an SMPS or it it just rectified mains? Does it matter whether the positive or negative side of the motor are switched?

Why were you even considering putting three regulators in series? The output current would remain the same but the drop-out voltage would triple. Assuming the encoder will be damaged by 17V (it might be able to work of that voltage fine) you need a low drop-out regulator, otherwise the battey life will be poor. I've found the datasheet for the nte970 and it doesn't sat what the drop-out voltage is but it's characteristics specified with the output voltage 3V below the input so it's probably similar to the LM317: about 2.5V. This means that when your battery voltage drops to 12V, the output voltage will be 10.5V. http://www.nteinc.com/specs/900to999/pdf/nte970.pdf A more detailed explanation of what the drop-out voltage is can be found in the thread linked below: http://www.electronics-lab.com/forum/index.php?topic=24319.msg100192#msg100192

Post a schematic. What's the: Base current? W/oC rating of the heat sink? Switching frequency? And is the load inductive?

The solution is just to power everything differently off the battery. I don't see what all the fuss is about, you've got a range of appliances which either have built-in voltage regulators (this applies to the answer machine, router and modem) or are designed to run off a wide range of voltages (the audio amplifier inside the speakers). If you open them up, you'll probably find regulator ICs such as the LM7805 or LM7808 which are fine for up to 35V, although they'll probably overheat and shut down if you throw that at them as the heat sinking will be inadequate but note I said shut down, not smoke and blow up as they're thermally protected. The audio amplifier inside the speakers will probably be slightly more sensitive, although I'd be very surprised if it's rated for anything under 25V and it'll certainly take 15V, minimum. Going from what I've read about the USA electrical system, the mains voltage can be as high as 126V in which case even the voltage to the answering machine will be 14.85V which is much higher than the battery when it's hot off charge and the modem will be running off over 19V. In short all of the appliances you've listed should be able to run from 10V to 15V without any problems and will probably run much cooler and efficiently off the battery than the mains adaptors. The audio amplfier will certainly be better as the battery voltage will certainly be cleaner than the mains. The only problem you may have is noise created by a ground loop which will form if a device connected to the speakers' audio signal input is also powered from the same battery as the speakers. The solution to this problem would be either an audio isolation transformer or powering the device off a a supply isolated from the speakers e.g. a DC-DC converter.

The TRIAC will now turn on but you've stlll not paid any attention to the issues raised in my previous posts.

With a 12V battery and a linear regulator, you're wasting over half of the power. You could use a low dropout regulator and a 6V 7Ah battery which is half the size and it will power the circuit for just as long. What's the point of the diode? To stop the battery from injecting power back into the 12V supply? You could easily omit it and just connect the circuit to the battery which is permanently connected to the charger. You need to design the charger so that the battery can be permanently connected though.

You don't need to ground the 0V rail.

Yes, gassing is a problem for lead acid batteries connected to solar cells, even 100mA can cause a 3.2Ah battery to gas. In fact I've had problems with a 14Ah battery gassing when being fed with 50mA continuously. The problem is the voltage. Are you planning to leave the solar cells connected continiously? You need to limit the voltage, rather than the current. If the solar cells are to be left connected to the battery continuously, the voltage should be limited to 13.8V. This could be done with an LM317 or a TL431 with a booster transistor. Unfortunately SLAs will take ages to charge if the voltage is limited to 13.8V. To charge quickly, the current should be applied until the voltage reaches 14.4V then stopped until the voltage drops to 13.8V, at which the voltage can be continuously maintained at 13.8V. NiMH cells are harder to charge than SLAs. The voltage will rise as they're charged, then fall again just before becomming fully charges, after which the current should be reduced to a very low level. It's possible to trickle charge NiMH cells continiously, if the current is kept low. 100mA will probably be too high for 2500mAh AA cells but it probably won't be a problem for 9Ah D cells, it's best to check the data sheet.

What do you want help with? And don't say everything. No one has responded because you've not shown what you've already done. I admit I don't know much about GPS. What's your budget?

Such a solution will require more expensive electronics, be less tolerant of voltage transients and less efficient. I don't think you've thought it through. For a start, you'll need a class D amplfiier otherwise the efficiency will be abysmal. I suggest you do more research.

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We don't know because you forgot to post a schematic of the current circuit.

No, all LEDs have pretty similar voltage/current characteristics and drive requirements, unless you count the type with build in resistors or a regulator. The only difference between your LEDs and the standard type is they're much, much bigger and will most probably have a metal back designed to be fitted to a heat sink. You'll also find that the forward voltage differs widely from one LED to another and is temperature dependant, as the temperature rises the forward voltage will drop. This is why it's undesirable to drive them from a constant voltage supply. Suppose you set the voltage so the current draw is exactly 1A say the voltage is exactly 3V, the LED's temperature will start to rise so the forward voltage will fall, causing the power supply to increase the current, the current will keep rising until the regulator limits it or the LED burns out. A constant current regulator, keeps the current the same, regardless of the LED voltage so the thermal runaway situation won't occur. Do you have the data sheet for the LEDs?

The TRIAC still won't turn on because the the switch is just connecting the gate to A1. Why use a TRIAC when it's only being used as an SCR as a diode is connected in series with it? You should just replace it with an SCR. A resistor can be used to limit the charging current. The switch hardly passes any current because it mostly goes thought the TRIAC and SCR.

I've never seen an LM7803 before. I assume you're talking about a linear regulator, like the LM7812 but set to 3V. This will not work for a couple of reasons: LEDs are designed to work from a constant current source, not constant voltage. You need a current regulator, rather than a voltage regulator. A linear regulator is very wasteful, the input current is the same as the output current so connecting 8 linear regulators, each drawing 1A, will draw 8A in total: a 3V linear regulator wastes 75% of the power supplied to it. You ideally need a switched mode constant current regulator to power the LEDs. You need a more powerful power supply. Although a switched mode power supply is typically 80% to 95% efficient so most of the input power is delivered to the load, you still need to make up for the 5% to 20% power loss so you need a power supply capable of between 25.2W to 28.8W. I assume you're powering this from the mains? If so your best bet is to buy a proper mains powered LED driver. Failing that, a crude solution is to connect two strings of three LEDs in series, each with a 3 Ohm resistor to your 12V supply. This is only 75% efficient but it's better than using 3V linear regulators.

The datasheet covers a whole familly of regulators with the same architecture but differing output voltages and current limits. http://cds.linear.com/docs/Datasheet/1083ffd.pdf Yes, if the regulator has a dropout voltage of 1.5V and an output voltage of 12V, the output voltage will drop to 10.5V. If the output voltage is 5V then, not it'll remain at 5V until the input voltqage drops below 6.5V. The voltage drops and so will the current, assuming the load is resistive. If the load current is constant then the voltage will drop and the current will stay constant. If the load is constant power then the current will increase and voltage will decrease. The dropout voltage is dependant on other factors: mainly the current and the temperature of the regulator chip. The graphs show how the dropout varies depending on the current and temperature.

You don't need a double pole switch, SPDT will do. What's the resistance of the 2.7mH coil? The peak current will be near 200A, if it's nothing which can cause the switch contacts to weld. You should also limit the charging current otherwise the bridge rectifier will be killed as welll as the switch.

The values you've chosen will work will draw a lot of current than necessary, 5mA may not sound like much but it will cost you battery life, if it's sat in standby. I'd recommend multiplying the resistor values by 10, giving 22k, 5k and 390k (the nearest E12 value to 407.4k) and using the BC328 rather than the BC557 as with 100mA you're pusing ti close to its maximum current rating. You can make it more accurate though. You can download some free software to help calculate the values. http://www.opend.co.za/software/rescalc/index.htm For E24 values, the program gives 300R for R1 and 68R for R2. R3 would need to be 5543R but of course you'd use 30k for R1, 6.8k for R2 and 560k for R3. If it's going to be connected to the battery permanently, you can increase the resistors further but you need to take into account that the referance (pin 1) takes 1.5uA from the potential divider so has a resistance of 1.66M which appears in parallel with R2 but it's insignificant until R2's value is over around 16k. EDIT: I've found a website which can do a similar thing: http://awasteofsalt.com/voltage-divider/

The problem you've run into is one of the reasons why video cameras aren't cheap. It requires a powerful computer to compress the image data from the CCD and write it onto the SD. This is because uncompressed standard definition vidio (720x480 25 framse per second with a CD quality sound track) requires 26MB of storage per second so a minute of footage would require over 1.5GB of storage. It will probably cost much more money for you to design and make one than it would to simply buy one. Video cameras are produced in large quantities so the manufacturers get huge bulk discounts on components. For example, you'll probably need a voltage regulator to stabilize the supply to the camera's sensitive electronics. To get decent battery life you'll need a switching regulator. My local distributor sells the LT1111CN8 DC-DC converter chip (RS Components stock number 545-6587P) for

It' good you haven't managed to turn the triac on given that it would probably blow up as it will be short circuited every half a cycle via the diode. Why all the complexity? Surely it's easier to just use a changeover switch so the capacitor charges when it's in one position and discharges when it's in the other position?

A stepper motor and a belt actually seems like the most simplistic solution. Do you need voltage regulation? If so, an open loop control system won't be good enough. You'll need to adjust the variac to maintain the correct output voltage, as the load changes which will require a servo motor and a voltage measuring transformer.

If the command is connected to the same 0V rail as the regulator, ther wil be a short circuit on the 12V supply when the switch conects it to +12V.

CApacitive or ultrasonic sensing is the way to go.

Have you checked that the Command - isn't permanently connected to the same ground as the 12V regulator?

Attached is a circuit which should do the job. The site linked below has the formula to calculate the values of R1 and R2. R5 is just a pull-up, make it 10k and R4 limits the base current to the transistor; 2k2 will do. R3 provides hysteresis, otherwise the relay could chatter, make it 100 times the value of R1 and R2 in parallel to give 130mV of hysteresis. http://www.reuk.co.uk/TL431-Battery-Voltage-Monitor.htm