Hero999

What voltage does the solenoid run off? Unless it's mains then it won't have mains rated insulation so isn't safe to use a capacitive power supply. How much current does it draw? Capacitive power supplies are only suitable for low currents. Above 20mA or so, a small transformer becomes more compact and cost effective. There's no decoupling on your current circuit. Follow the suggestions I made in my previous post using a switched mode power supply before going any further,

More information is needed. Where are you getting the 5V 1.5kHz signal from? What's the wave from? Is it 5V peak, peak to peak or RMS? A capacitive voltage multiplier or a transformer with a rectifier on the output would work.

What are you switching with the 555 timer? How is it triggered? Capacitive power supplies are only suitable for low currents and are not isolated from the mains so the whole circuit, including everything connected to it, needs to be treated as mains and heavily insulated. The correct solution of course is to add some filtering so it works from the switched mode power supply. At a bare minimum there should be a 100nF capacitor across the supply rails, 10nF between pin 5 and 0V and some ferrite beads on the power supply rails.

Obviously a 4.5Ah battery won't hold as much charge as a 10Ah battery. The other issue could be maximum discharge current and charge current, if it has a built-in charger.

There are plenty of AM/FM radio circuits about. What parts can you not get hold of? Here's a schematic for a high quality receiver which you should be able to find the components for: http://quazar31.home.comcast.net/~quazar31/index.html Another option is to buy a kit: http://www.elenco.com/admin_data/pdffiles/AMFM-108CK_low-res.pdf Of course, don't think about saving any money. It's much cheaper just to buy a radio off the shelf but see it as an educational exercise.

It appears as though you've confused the non-inverting and inverting amplifier topologies. As we said many times in the thread linked to above, you need to build a non-inverting amplifier with the negative feedback taken from the output side of the buffer. Read the documents linked below. If you don't understand them, read them again. Then if it's still not clear ask here. https://www.onsemi.com/pub/Collateral/SR004AN-D.PDF http://www.ti.com/lit/an/snva558/snva558.pdf

I've told you before on the other forum: you haven't provided enough information; the clearance between two 3.3V traces is as small as the PCB manufacturer can make. http://www.eevblog.com/forum/beginners/what-is-the-formula-for-distance-between-two-trace/msg570584/#msg570584

We had this discussion 7 months ago. Have you forgotten? Here's a link to the thread so you can remind yourself: http://www.electronics-lab.com/forum/index.php?topic=38954.0

Yes, it's an OR network. The diodes make it so the op-amps can only divert current away from the bases of the driver transistors. A1 is the voltage error amplifier and A2 is the current error amplifier and the OR network stops them from fighting one another. The drop-out voltage will be huge, made worse by the ORing diodes. All those transistors are connected together to form a lagre three stage Darlington so the over all voltage drop will be over 3 diode drops, plus the saturation voltage.

Yes, you need an oscilloscope. Spin the motor at a constant speed, perhaps use a drill but don't go too fast as very high voltages will be produced, and measure the waveform using an oscilloscope.

The op-amp will need a negative supply to compensate for the diode voltage drop and that of the op-amp's output stage.

According to the text. The circuit was originally designed for TTL and has been modified to use CMOS. Try using 74HC ICs and a low drop out 5V regulator, such as the LM2936-5.0.

That won't regulate the voltage though, will it? What's the output of the current amplifier when the the non-inverting input > inverting input? You need to design it so the current amplifier can only reduce the output voltage, not increase it.

I've never tried this myself but I'd suspect it's neither square nor sine. It's probably a messy waveform. The number of steps will determine the number of cycles per revolution.

The only real solution to the stability problem is to use an emitter follower on the output.

I've does this before but I used voltage regulator ICs which have built-in over current protection. A transformer needs to be de-rated when used to power a rectifier and filter capacitor so a 1A transformer may overheat if 1A is drawn from the DC side continuously.

As long as Q1 is a common emitter amplifier there's a risk it'll oscillate, especially under certain conditions (a large capacitor connected to the output with a low ESR). This is because you've introduced another pole with a phase shift influenced by the load impedance. http://en.wikipedia.org/wiki/Phase_margin

Here's the schematic, inverted and converted to monochrome to make it more clear.

Fortunately all decent autoranging DVMs allow the range to be set manually (the meter linked to above does), so range hunting it's a non-issue.

The voltage across the capacitor should never exceed its rated value, indeed it's a good idea to leave a safety margin of at least 10%, so it it's rated to 2.7V, keep the voltage below 2.43V. Supercapacitors are polarised and should never be subjected to reverse voltage for any period of time, typical ratings for short term voltage reversal are 10% of the rated voltage or 0.6V, whichever is lower, so if it's rated to 2.7V, the reverse voltage should never be greater than 0.27V and it's a bad idea to leave the capacitor reverse biased for long periods of time.

After correcting liquidbytes's simulation, the steady state power dissipation in R6 is 3.9994µW.

You need to wait until the steady state, before attempting to calculate the power in the load. When the power is first applied, there will be a power surge as the DC bias voltage stabilises; it takes awhile for the capacitors in the circuit to charge. Once the steady state has been reached, the RMS power in the load can easily be calculated: PRMS = Vpk-pk2/8RL

Yes you probably need a new meter and make sure you buy a reputable brand and not something cheap, so it lasts longer.

I don't see how the circuit can work anyway. The problem isn't the power supply voltage. If the op-amp rated to be run off a bipolar 22V supply, then it should be fine on a 36V single rail supply.

I don't understand the question. If both the plates of a capacitor are at the same potential, then there will be no charge on the capacitor.