Hero999

It's probably just a bad ballast, just make sure you order exactly the same type. As the tube is black when unlit, it's a black light blue tube which is perfect for etching PCBs. For a decription of colour temperature and rendering index, see the following links: http://en.wikipedia.org/wiki/Colour_temperature http://en.wikipedia.org/wiki/Colour_rendering_index

It might be that you used the wrong tube for the ballast. What colour what colour is the tube when off and when lit? A black light will be white when unlit and bright blue when lit, a black light blue with be very dark blue, almost black when unlit and deep violet/blue when lit: either a black light or black light blue will do. The article uses high pressure mercury vapour lamps which are totally different to fluorescent lamps which are low pressure mercury vapour lamps. Regardless of what type of lamp you use, the ballast needs to be matched to the tube otheriwise either the ballast or tube could be damaged.

When you get round to improving it, try the AD780 which gives 3V or 2.5V

I had a feeling you'd done that. Have you measured the power supply voltage? Bear in mind that most regulator ICs such as the LM7805 have a poor voltage tolerance and will vary by 1% or so depending on the ambient temperature and load current. If the supply voltage is too low, the ADC will read a slightly higher voltage than what you would expect. I wouldn't be surprised if it's in the region of 4.8V. You should use a real precision voltage reference with a close tolerance preferably ≤0.25%.

Yes, fluorescent blacklights will work but are nowhere near as powerful as high pressure mercury vapour lamps. To be honest, I think high pressure mercury vapour lamps are overkill, fluoroscent blacklights should be more than powerful enough. That doesn't make any sense, a colour temperature and colour rendering index can't be applied to a blacklight.

I haven't seen the original article. Here's my guess at the meaning, going from the poorly worded paragraph you've quoted. The project uses high pressure mercury vapour lamps which take five minutes to warm up and produce full light output.. This is because the light producing discharge requires mercury in the gaseous state and when the lamps are cold, the mercury is still a liquid. Unfortunately the lamps can be damaged if started when the mercury is in a gaseous state and the lamp pressure is still high so they need to be left to cool for five minutes before they are restarted. Most modern high pressure mercury lamps don't produce much UV because they are encased in a phosphor coated shield which converts most of the radiation to useful visible light. The shield can be removed but this might be a bad idea as some lamps have a safety device which stops them from working with the shield removed. Even with the shield removed, most of the UV radiation from a high pressure mercury lamp is in the UV region which isn't optimal for exposing photosensitive bulbs. You can buy special black light high pressure UVA lamps designed for blacklight cannons (also called beehive lamps) which are used for theatrical effects and can be purchased from, stage, DJ and audio supply shops. Here's an example of such a lamp: http://www.planetdj.com/i--LL-400BLB Note that you'll need a suitably rated ballast and igniter to use the lamp, although you can buy lower powered lamps with it built-in.

I don't know much about MCUs, especially C so I can't help you with the coding. I don't it's a coding problem, it's more likely to be a hardware problem. Where does the ADC get its reference from? I'd recommend checking that before anything else.

Good, you found out what the problem was. It's just really back luck that both your DVM and ADC are wrong.

How well water proofed is the sensor? Could it be that moisture is getting in?

That's lucky it worked for you but some mercury lamps stop working if the shield is damaged or removed.

What is your room temperature? You still haven't said. How did you measure it? There's no point in calibrating it unless you accurately measure the temperature of your room.

I don't know what you could be doing wrong. Obviously you need to know the exact temperature of the LM335Z to know it's working correctly. My guess is that it's working and the temperature of the LM335 is pretty close to 40

A quick look at the datasheet tells me that the voltage shouldn't vary much as the current changes so the resistor value won't make any difference - I was wrong earlier. Does the voltage increase when the temperature sensor is warmed? Have you had your multimeter and thermometer calibrated recently? If not then how do you know it isn't your multimeter that's out or if your thermometer's reading is wrong? I would suspect the multimeter is wrong, an output voltage of 3.14V would indicate a temperature if 40

Some photo papers are wax coated or actually plastic and are water resistant, did you try normal copier paper? I know I said use the highest heat setting but I've changed my mind since then, lower heat settings seem to be better because the toner doesn't smear so much. That's right, it doesn't matter with the photo method, so long as the ink is opaque to UV: most are. It depends on the photoresist, normally it's positive - black lines create copper. Yes, of course you can. If you can't get hold of transparent film then use tracing paper - it works just as well.

There is nothing wrong with using emitter followers since current gain is required, not voltage gain. The circuit won't work anyway because IR remote controls flash the LED at an ultrasonic frequency so you won't be able to hear it. I would recommend using an LED to indicate the output, a tone decoder such as the NE567 or CD4046 might also be a good idea to block DC and mains frequency light sources.

I can't read your rough ASCII schematic, please post a clear drawing in either.GIF or .PNG format, avoid .JPG which can be fuzzy. Look at the datasheet for more information on how to use the LM335. Use ≤1% tolerance resistors - 5% won't be accurate enough. http://www.national.com/ds/LM/LM135.pdf

The absolute maximum voltage rating is 7V so you'll probably get away with it but it isn't recommended. The trouble is the battery voltage will drop below the minimum voltage rating before the the cells fully discharged. Why are you using an old power hungry TTL lC? You should use a low current CMOS IC such as the 74HC32 which has the same pin-out as the crappy 7432 but with a wide power supply voltage range of 2V to 6V and a much lower current consumption. An alternative is the CD4071 but note the different pin-out and voltage range of 3V to 15V.

I've never understood why people incorrectly copy circuits from datasheets and pretend they designed them when it makes it obvious that they don't know how it works.

Why do you need to simulate it? There shouldn't be any need to because the circuit is so simple, just use Boolean algebra to work out what will happen. Even if you simulate it and it works it doesn't mean it'll work in real life. The simulation program doesn't know the ambient lighting conditions, the brightness of the LEDs or how well the phototransistors are shielded. Providing the LEDs are bright enough to cause the phototransistors to saturate at a low enough voltage and the phototransistors are well enough shielded from ambient light to turn off enough, it will work. A simple circuit like this won't be very reliable because the ambient lighting can interfere with it. A real beam break detector circuit pulses the IR LEDs at a specific frequency and has a bandpass filter or PLL (Phase Locked Loop) after the phototransistor to reject constant or mains frequency light sources. I would use an astable multivibrator (555 timer or logic gates) to pulse the LEDs at about 40kHz and a tone decoder (NE567 or CD4046) for the IR receiver section. I wouldn't bother with a logic gate after the receiver, I'd connect the phototransistors in series. To make it more efficient, I'd connect the LEDs in series.

No one's going to do your work for you. If you make an attempt at solving the problem then we might be able to assist you.

I already responded in the other thread. Please don't create multiple threads asking the same question. http://www.electronics-lab.com/forum/index.php?topic=19195.msg88251;topicseen#msg88251

It depends on where you live. Here in the UK, Maplin, RS Components, Farnel and Rapid are the most popular suppliers. Farnel are good because they don't charge for delivery, even for small orders. RS also don't charge for the delivery of small orders if you have a credit account with them, otherwise there's a delivery charge for small orders. Maplin are handy because they have shops in most towns but don't offer a great range and are expensive. Rapid are the cheapest UK supplier I know of but charge for delivery of orders under

Oh you're talking about the output voltage, I thoguht you were refering to the voltage on the filter capacitor.

Most potentiometers are linear. A linear potentiometer's resistance is directly proportional to the wiper's position. There are also logarithmic potentiometers which is have an exponential/logarithmic (depending on which side of the wiper is being measured) resistance-wiper position relationship and normally only used for volume controls.