Hero999

Does it vibrate and make a loud humming sound? It could be because you've not secured the laminations and the vibration increases the heating.

Just looked at the last circuit with two transistors. Whn the power is applied C2 will charge causing the horrizontal transistor (I'll call it Q2) to turn on. The vertical transistor (I'll call it Q1) will then turn on as C1 will charge. The relay will be activated shorting out C2. When C1 is fully charged, Q1 will turn off, de-energising the relay causing it to open. The only wat to reset the circuit is to press the start then the on reset or the reset then quickly press the start before C2 recharges.

Never heard of it so used Google. If you buy it, I'd recommend you don't buy any ebooks from Amazon as they can delet them whenever they want.

Threads merged. Don't create multiple topics about the same subject - it wastes our time and yours. We still don't know what you want. An electromagnet is inductive not capacitive. You've posted a coupled of standard 555 timer schematics, not your problem.

Why? What are you trying to do? Your discription is very vague. The current though a relay's contacts will depend on the load. Or are you saying you need a current to activate the relay, i.e. when the curerent exceetds 10mA, it turns on? You'll also recieve more help if you post a schematic.

It's not possible for both the output voltage and current to be constant, either one or the other will vary as the load impedance changes. Even specifying a power supply requires some electronics knowledge, let alone designing one. I think you need to do more research and aquire the knowledge required to know what you want.

Have you tried filtering the power supply to each of the devices? Try adding ferrite beads and ceramic capacitors. I still wouldn't rule out a grounding problem. If so using an isolated DC-DC converter for two of the devices should solve the problem. Have you tried running the preamplifier from 7.2V? If it's designed to work from a 9V battery, it should be able to work all the way down to 6V. If not, crank up the voltage on the 7.2V supply to 8V. The camera will be fine at 8V or even 9V, as it'll be designed to work from a NiCAD/NiMH pack which could be as high as 8.4V to 9v when hot off the charger. The monitor may also be able to withstand up to 16V, check the manual. Reducing the number of different voltages as much as possible will simply things, increase the efficiency and make it easier to find an isolated SMPS, if required. I know you can by isolated 12V power SMPSes but I don't know about 8V.

No one makes 1.22W resistors, it's common practise to just use the next power rating up.

That sounds about right to me. I calculated 349R, the nearest standard E24 value is 360R but any value between 330R and 390R will do. I = P/V = 1.1/24 = 45.83mA R = (Vin-Vout)/I = (40-24)/0.04583 = 349R Your power calculation doesn't look right to me but fortunately your result was conservative. P = I2R = 0.045832*349 = 0.73W So use a 1W resistor, minimum. Of course, if the resistor value is slightly higher the current and therefore the power dissipation will be slightly lower.

Yes, put a resistor in series with it. Measure the current in the coil and calculate the required resistance using Ohm's law. Hint: it's the same formula as that used to calculate the series resistor for an LED (this can be found using Google) but replace the forward voltage with 24V.

Those are both constant current sources so would work but they're step down converters, which means the output voltage will always be below the input. The LEDs will need to be connected in parallel as strings of two. Because LEDs don't current share well you should include current balancing resistors to help share the current, unless you derate them by half.

LTSpice, as I said above.

That's not true, LEDs ideally need a constant current source. The inefficiency of powering LEDs from a car battery via a resistor is the power lost in the resistor. The varying voltage just means the brightness won't be constant. The best and most efficient way is to use a switched mode constant current source. A DC-DC converter doesn't increase the power, it increases or decreases the voltage. If the voltage is increased the output current will be less than the input current and if the voltage is decreased the output current will be greater than the input current. Yes, if you don't want a fire if a short circuit occurs. Why would that blow the DC-DC converter? The battery oltage is lower when cranking not higher. Ideally the DC-DC converter should have a constant current output which is usually inherently short circuit proof. The only exception is a boost converter consisting of a single inductor which will short circuit when the output voltage falls below the input voltage. Only one of those links worked for me and it's not ideal because it's a constant voltage source not a constant current source. http://www.musclecars.net/parts/Powerful-DC-12V-to-24V-10A-240W-Step-up-DC-DC-Converter-For-Sale_200527576514.html Well the maximum forward voltage is 4V and you want to connect four in series making the minimum acceptable open circuit voltage should be 32V. You need three constant current sources, one for each LED and the current should be set to a level which will give white light when all LEDs are fully on. The brightness can be altered by either varying the current or PWM but the latter is preferable as the brightness will be more predictable.

And then the rest should be obvious. Think about it, you don't need any knowledge of electronics to know the answer.

Ideally as high as possible but there comes a point when the element is too big. I'd say 50W is good for a general purpose iron.

I've never heard of Livewire so I Googled it. Maplin sell it for an overinflated price. http://www.maplin.co.uk/livewire-circuit-simulator-33100 When you mean it blows out, do you mean Livewire thinks the absolute maximum rating of a component has been exceeded so shows it exploding?

The 555 timer is the obvious one but it depends on what you're doing.

I doubt it will last a year but good luck with it. You really need a temperature controlled iron. You can substitute this with a cheap plug-in lamp dimmer but the iron's wattage may be too low.

For cheap wireless headphones buy an FM transmitter an a tiny FM radio.

It's possible but is a silly idea. You need to be very careful when connecting batteries in parallel because they need to be balanced to ensure each cell has the same level of charge. You should use a large lead acid battery which will work out cheaper. If it really needs to be compact use four large lithium ion cells in series but you need a special charger. If you must use standard small batteries then use C or D cells which can be found in capacities of up to 5Ah and 10Ah respectively but be careful to check the capacity as some cheap C/D cells are smaller cells with extra packing.

Sounds like a silly idea to me. Email is a better method.

You're right about the comonent values and capacitor suffix. Tolerance isn't important for capacitors either. You've got the diode right. Just pick an 74HC14 in a through hole DIL package. You'll also want some strip board to assemble it on.

The IRF750 springs to mind. http://www.ic72.com/pdf_file/i/49649.pdf The hardest part of building a switched mode inverter will be controlling the MOSFETs across the isolation barrier. You can use a pulse transformer for this but it needs to be able to work down to 50Hz.

It looks like a good article. I like the way you figured out you need to use a modified sinewave and that the peak voltage needs to be the same as the mains. If you use a ready made transformer, you can use 9V-0-9V for 12V input and 18V-0-18V for 24V input. The only thing I have to pick you up on is you say it's 25% duty cycle which is not quite true. It's really 50% dyty cycle, as current flows when the current is both positive and negative. Have you considered making a switched mode inverter yet? You need an isolated DC-DC converter to convert the battery voltage to a higher DC voltage equal to the mains peak voltage, followed by an h-bridge to convert it to AC. This is obviously much harder to do but it has the advantage of being smaller, more compact and efficient than using a bulky transformer.

You've got the value of R7 wrong there. In the previous schematic you posted, you got the values of R1 and R5 wrong. You need to be more careful with your decimal suffixes. Resistors are always specified in Ohms. If there's no suffix, or the letter R is used, it means *101 so a resistor with a 10 or 10R next to it, is just 10 Ohms. The decimal point often replaced with the suffix, so 1k2 is 1.2k or 1200 Ohms, 1R5 is 1.5 Ohms and 0R75 is 750m or 0.75 Ohms. Be careful with your use of upper/lower case, m means 10-3 or milliOhms so 10m is 0.01 Ohms and 10M is 10*106 Ohms or 10,000,000 Ohms.