Hero999

What are you using it for?

How much current does the solenoid need? What's the current rating of the push button switch? You might be able to omit the MOSFET and drive the solenoid directly from the switch.

It depends on what power supply you want to use? The simplest way is to run the 3V device from 3.7V via a low drop out regulator or for greater efficiency use a switching regulator. Another way would be to use a boost converter to run the 3.6V device off 3V which would probably give the best battery life as AA cells have a capacity of 2500mA, rather than 130mAh.

No because it's not practical. Take a typical transformerless power supply circuit: For 3A@12V C1 would need to be 47uF.

Are you sure you don't have a shorted turn? Did you touch the primary? You should leave it as is. Does it get hot off load? Measure the primary and secondary current. Is there DC in the secondary? You should always use a full wave rectifier with a transformer powering a high power circuit, never a single diode or a voltage doubler to get a bipolar output. Please post the schematic.

I ahd a quick look for the bandpass filter in RS and Farnel without any luck. The other components shouldn't be hard to find or make.

You generally don't bother, buy a new one. If you want to bother then check the main filter capacitor, then the switching transistors, followed by the rectifier on the output stage, then the bridge on the primary side.

Don't create duplicate threads. Where do you live? It's impossible to help you without knowing the suppliers in your area.

It's a matter of opinion. I like using a 10 turn pot so I can have course and fine control with one pot but having to turn it ten times to go from one end of the range to the other might annoy some people.

There are plenty of ways to do this. It depends on what components you have access to? You can use the 555 timer, some inverting gates (NOT, NAND, NOR), CD4013, 74HC7474, a comparator/op-amp (LM311, LM393, uA741) a microcontroller or even a couple of transistors. Here's how to do it with a 555 timer which has the advantage of being fairly simple and can provide up to so will directly drive a relay. Unfortunately it also uses lots of power so you'll need to use the 7555 with higher resistor values or another CMOS solution if it's going to be connected to a battery continuously.

Just use a mains transformer and relays. Using triacs means having ro use a transformerless power supply which isn't very safe because the whole circuit is at mains potential. With a transformer only the primary of the transformer, relay contacts, light and fan are at mains potential. If you don't understand how to use a TRIAC then you probably don't have enough knowledge and experiance to build the circuit with a transformerless power supply.

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I don't see any need for more diodes. It depends on what you want to do. I forgot to remove the 9V zener which is no longer needed.

I told you before: you only need one relay to power the light, a total of two relays. Yes, a 9V 600mA transformer will do. The current consumption of the relays can be calculate using Ohm's law. There are many other redundant components in the circuit . I've removed them all. Here's the latest schematic. The LEDs will be very dim and hardly light at all so I's recommend replacing the CD4017 with the 74HC4017 and reducing the LED series resistors (R7 to R12) to 220R.

The transformerless PSU won't be able to provide enough power to drive the relay coils. Why are you using two relays for the light? Yes, either use a power transformer based PSU or replace the relays with TRIACs.

Could yoube confusing peak voltage and peak to peak voltage? If the V/division setting is 1V the signal should occupy two squares. You might also be using complementary outputs on the signal generrator (180 degree phase shift) so you'll see double the voltage, than you'd expect from the signal generator. Failing that it could be due to inductance in a poorly/uncomensated cable causing ringing.

Why are you running a motor from a regulated power supply? An LM7805 will stop regulating (the output voltage will fall below 5V) when the input voltage falls below 7V and the battery will still have plenty of charge left in it. Why do you need a limting resistor? A resistor will slow the motor down so is not a good idea. To be picky that circuit won't work because it's an open circuit. The positive end of the battery should be connected to ground. THe votlage across R1 and R2 can be calculated using the potential divider formula, see Wikipedia. Is that with or without the 120R series resistor? If the motor has a DC resistance of 18R the starting current will be 278mA. It sounds like you're doing it the wrong way. It depends on what kind of motor you have? If it's a standard DC motor. The most sensible solution is to power the motor directly from the unregulated 9V battery. The speed can be reduced using PWM. Don't worry if 9V is higher than the maximum rating, the power dissipation can be limted by reducing the duty cycle. If the starting current is too high, you can connect a resistor in series with the motor and bypass it using a transistor which is turned on once the motor is up to speed.

Thye voltage is dependant on the temperature, current and materials. 0.7V wasn't chosen, it's due to the band gap of silicon. Germanium has a lower bandgap so the forward voltage is much lower 0.3V. Here's a link to an experiment you can do yourself. http://www.phys.csuchico.edu/~eayars/publications/bandgap.pdf

You're paying all that money for one good reason: safety. It's designed to be reliable so it doesn't fail exposing you to harmful cancer causing UV radiation. The product you've linked to seems like good value for money, it will be guaranteed to reliably and safely filter dangerous UV radiation.

Contact your local college. I'm not aware of any specific courses. The only real way to learn is to do an apprenticeship but doing a generic electronics course (A-level, HND, degree level) and some independent study will certainly help.

That's easy to get around: use an SSR made with two MOSFETs connected back to back rather than a TRIAC.

You haven't made your requirements clear, please post a schematic. Is the power or signal? CMOS switchs are lower power signal devices, only work with signals within the power supply rails and shouldn't be confused with solid state relays.

You didn't help us to help you much in the first place by not making your requirements clear. In future, post drawing, schematics, graphs and charts and you'll get the answer you were looking for much more quickly. To be honest, I'm still not sure whether the circuit 'you've designed will meet your requirements but if you build it and it works then I'd be very happy for you.

My design has not been tested but it consists of sub-circuits which have been tried and tested. The mains transformer is just a standard 9V >100mA or >1.2VA unit. Just make sure it's reasonable quality and has short circuit protection or is fitted with a thermal fuse. If you're worried about the hazards associated with mains, simply replace the transformer and BR1 with a 9VDC wall brick mains adaptor, rated to above 100mA. Tr1 is generic NPN transistor, BC338 will do, Tr2 is a general purpose PNP transistor, B328 will do. The resistors can all be 0.25W 5% carbon film but higher-specification 0.5W 1% metal film is fine though overkill. It's possible to simplify the circuit by replacing the constant current source circuit with a resistor but the battery won't charge as well. The circuit can also be scaled up by using a 12V transformer, a power transistor constant current source to charge the batteries, six AA cells in series and a switched mode constant current source to power a high power 300mA LED on a heat sink.

This is a totally silly idea. Why hash a dangerous transformerless power supply together with a 12V relay when it would be cheaper, safer and more reliable to use a relay with a 240VAC coil in the first place? 9V is too lower voltage to power three white LEDs connected in series properly. It isn't hard to design a circuit to trickle charge rechargeable the batteries when the mains power is on. The propper way of doing this is to use a transformer to charge the batteries via a constant current source (Tr1, R1, R2, D1 & D2) and use a transistor to switch on the LED when the mains power fails (Tr2, R3 & R4). A toggle switch can be added to disconnect the batteries when the device is not in use.