Hero999

You need a low value pull-down resistor on the input, otherwise when there's no signal Q1 and Q4 will turn on, allowing current through Q2 and Q5 which will cause Q3 and Q6 to turn on and short circuit the power supply. There will be switching waste because Q3 and Q6 will both turn on simultaneously when the input voltage changes from 0V to 3.3V. Here's another idea which you should be able to cobble together on the existing PCB.

Here are a couple ways to do it. V1 is more complicated and it has the disadvantage of both output transistors being on simultaneously for a point during switching but it'll have virtually no saturation loss i.e. you'll get +/-15V out. V2 is simpler and has the advantage of not having both transistors on during switching but you loose 0.7V on either supply rail so with 15V in, you'll only get +/-14.3V out.

The transistors are biased so when there's no signal, they're all on. It looks like you've forgotten that a transistor's base is just a diode which will pass current. Look at the current flowing hrough the PNP transistor bases to the NPN transistors.

That will do, there's no point in complicating things.

The relay being powered continiously will not reduce its life. It's switching the relay on and off many times which reduces its life.

The load can be connected in parallel with C1 which can be increased in size, if needs be. The circuit you posted is a bad idea because the potentiometer also increases the impedance of the power supply so the current and voltage to the load will also fall. The brightness of the LED will vary depending on the load connected to it.

The switch does the reverse of what you want, it's off when the temperature exceeds 40oC and turns off when the temperature falls below 25oC.

Here's another way to control the brightness of an LED. It's slightly more complicated but the current through the LED is regulated (it doesn't change much as the voltage varies) and the filter capacitor is much smaller.

I know, I disagree, it's not small at all. The PIC12F508 comes in a large through hole DIP8 which can easily be soldered to strip-board. You can get it in smaller surface mount SOIC8 package which is still easy to solder, compared to the tiny DFN8 package which is only 3x2mm. Unlocked, you shouldn't be allowed to do that, it's a bug in the forum software. Moderator

What do you mean? It comes in a through hole DIP which is the same package as a 555 timer, uA741, LM393 etc. It's not hard to learn, especially if you program in BASIC. With an MCU that can be done in software without any need to change any comkponent values and the delay can be controlled pretty accurately so there's less experimentation. You could bought some PIC12F508s and a cheap USB programmer for mot much more than that or saved even more money by making your own parallel port programmer.

Sounds good, please post the schematic and source code.

I don't know, you drew the circuit. They look like SCRs to me. Seriously, use a microcontroller, the PIC12F508 could easilly do what you want or better still save up or take a bank load to buy the propper welding mask.

You could use a boost converter to get +35V and a buck boost to get -35V but it's probably more efficient to use a transformer based topology i.e. flyback or forward converter.

Have you Googled? As it only neds to transmit over 1m why not drop the laser idea and use a standard IR LED and remote control receiver IC?

It's easy, use phase control. http://en.wikipedia.org/wiki/Phase_control To adjust the brightness, alter the delay of firing of the thyristor, the later the firing, the dimmer it will be.

Then you need a DC-DC converter, not an inverter. You could wire six 12V batteries in series to give you +/-36V or you could use a pridge amplifier to give the same output power from a single 36V power supply.

The resistor in series with the LED is 1k and the load is 1k, giving a total load impedance of 500R, ignoring the LED's forward voltage drop. The output voltage can be calculated using the potential divider formula which can be found on Wikipedia. http://en.wikipedia.org/wiki/Potential_divider

Please post a schematic. You can't use a potentiometer to vary the voltage to a large load because the output impedance of the potentiometer is too high so the output voltage will drop. Furthermore, the potentiometer is probably only rated to something like 0.25W and will burn out if the current is too high.

No because it's not the right IC to use for that application.

I've never used OrCAD before but on most PCB packages you can turn off layers you don't need and print the board for toner transfer.

I'm not sure what you're asking. Just open the file and print the copper layer. Turn the silk screen and solder resist layers off.

Buy an inverter and add a transformer to the output. If you can't get a 45VAC transformer then connect a 30V and 15V transformer in series.

The lamp won't light with a 1k resistor connected in sereis with it.

The original poster has opted for a 230VAC relay, he's has the sense to realise using a DC PSU to power a DC relay was a silly idea. Using an LM7806 to power a relay, are you crazy? Totally silly idea. In developed countries the mains is regulated to within +/-10% of the specified voltage and you can buy relays designed to work over +/-20% of the specified voltage.

You're wrong about how MOSFETs work. They have other modes of operation than just on and off. The gate voltage just needs to be high enough to allow enough current to flow through the load and a small voltage drop across the MOSFET. The maximum gate voltage depends on the device and normally varies between 10V and 20V depending on the device, see the datasheet.