Hero999

The question doesn't make any sense. a) What's the leakage current of the capacitor? b) How good is the ammeter? c) What's the speed? Logic family?

Draw what you mean. The most important thing is the wire is rated to carry the current.

It's the energy stored, rather than the capacitance which is important. E = 0.5CV2 Doubling the voltage quadruples the energy storage, given the same capacitance. A 5mF capacitor charged to 100V will be storing twice the energy as a 10mF capacitor charged to 50V and it's likely it will be physically twice the size. Using higher voltages also means when the voltage drops too low for the buck converter, more energy will be extracted from the capacitor so you're able to use the storage more effectively. It's worth keeping the capacitor shown on the datasheet, even if you're connecting a large capacitor across it because smaller capacitors have superior high frequency characteristics.

A fuse is mandatory and the cable must be rated to carry the current rating of the fuse. The box must be vented when the battery is charged, to avoid the build up of hydrogen gas which is explosive.

Why not use a 100V capacitor, then you can utilise the power more efficiently and don't need any over-voltage protection?

It's not an LED display. It's a vacuum fluorescent display. The chance are you won't be able to find the data sheet. You'll need to experiment to find the pin out. http://rctractor.proboards.com/thread/60

How long does it need to last for? I hope you're aware that you will not be able to charge the capacitor and draw 5V at 20mA at the same time. Before you said the transformer charges the capacitor to 90V when nothing is connected to it, then it drops to around 50V when a 22k resistor is connected to it. Using Thévenin's theorem the internal resistance of the transformer plus the stepper motor can be calculated as 17.6k. Maximum power transfer occurs when the load resistance equals the source which is when the output voltage is half the open circuit voltage so it can be calculated to be 452/17.6k = 115mW. The datasheet for the says the LTC3639 says it's 85% efficient so the power required to give 5*0.02 = 100mW out is 0.1*0.85 = 117.6mW. This means you can charge the capacitor as long as the buck regulator is very lightly loaded.

You didn't say anything about a buck converter before. How much current do you need to draw? What's the point of increasing the voltage to 50V then reducing it to 5V? According to the datasheet, the LTC3639 can work up to 150V so as long as the capacitor is suitably rating you don't need to limit it to 50V. The data sheet shows you how to calculate component vales for a given output voltage, http://cds.linear.com/docs/en/datasheet/3639fd.pdf

The TL431 is not an ordinary zener diode. It's an op-amp with a buit-in voltage reference. Unfortunatly its maximum voltage rating is 0nly 36V so a transistor is used to buffer it. http://www.ti.com/lit/ds/symlink/tl431.pdf

Not all UV lamps are equal. Ideally you need a black light, like the type used in insect killers but the short used in discos will do. Ambient light, including the shorter visible wavelengths (blue and violet) can expose the board. Once exposed, the PCB shouldn't be exposed to any light at all and ideally it needs to be developed in total darkness. The yellow and red part of the won't expose the board so you can use a photography safe light or make your own with a load of red and yellow LEDs.

The zener diode only wastes power when the capacitor has charged to 50V. You may be able to make a precise 50V zener diode from the TL431. The problem is it's only rated to 36V but a common base amplifier can be added to boost it to 50V. The extra gain may cause oscillation though. I haven't tested it. The efficiency of this circuit will be poor anyway.

Sorry to hear about your mum. A transformer is a good idea but bear in mind the reading on the meter is the RMS voltage, not the peak, so if you rectify 50VAC you'll get 50*sqrt(2) = 70.7V minus the losses in the bridge rectifier. Use a 230V:24V transformer or a 120V:12V transformer.

It isn't a comparator. A common ground is needed because all voltages in a circuit are relative to one another. Measuring potential difference is like measuring distance: you need to start from somewhere.

No, I don't know what you're talking about. The circuit doesn't have a common 0V reference. Yes, the current through Q is zero because the two batteries are only connected together via one wire and there needs to be two wires to complete the circuit.

How on earth is that a comparator? It's a potential divider. A comparator compares two inputs and turns on when one is greater than the other. It has a digital output and two analogue inputs.

It's difficult to analyse circuit without a 0V node because all voltages are relative. Ignoring that. The negative side of the capacitor gains electrons and the positive side loses electrons.

That's not a comparator circuit. This is a fairly simple comparator circuit.

A comparator. Look up low battery cut off circuit.

Your circuit is working, at least in simulation. Increasing the values of the capacitors in the voltage multiplier circuit will speed things up. Try replacing the 10uF capacitors with 100uF capacitors. The maximum voltage on the capacitor will always be lower than expected, due to the diode losses. Using Schottky diodes may help this but the reverse leakage current is normally a bit higher. In reality there's also a limit on how fast the capacitor will charge as the stepper motor will have an internal impedance which will slow it down.

It's a trick question. An SCR might not even switch on and stay on at 20mA and it will certainly drop less than 2V, at that low current. SCRs don't work like BJTs. The gate current doesn't need to be on continuously. You need a small gate current, for a short length of time to trigger the SCR and it will remain on until the current falls below the minimum holding current. A schematic would help.

A normally closed relay. https://www.google.co.uk/search?q=normally+closed+relay&ie=utf-8&oe=utf-8&gws_rd=cr&ei=uHWlVKf_E4jjaoSwgaAN

What do you think? A boost converter increases the voltage of a DC supply. The voltage multiplier circuit I linked to converts AC to DC, as well as boosting it, unless you're talking about the CMOS charge pump.

A fairly good description of a voltage multiplier can be found on Wikipedia: http://en.wikipedia.org/wiki/Voltage_multiplier The charge rate of a capacitor can be calculated by rearranging the following formula: Q = CV Q = It

That may be true but capacitors don't store much energy, compared to batteries and will only charge as quickly as the energy is supplied to them. If your 5V generator only gives 1mA once it's been boosted to 50V, then it will take 500s to charge up, assuming the current is constant.

What do you intend to power with the capacitor? How about gearing it up to increase the speed?