How do I replace an LED with a brighter LED?
Generally, improvements in brightness have come about a result in improvements in manufacturing processes and materials. The electrical characteristics of the brighter LEDs don't differ drastically from the older generation of indicator LEDs. In a lot of cases, the brighter LED is just more tightly focused.
In almost every case, replacing an indicator LED with a brighter one is a simple, drop-in replacement. You won't need to consider the electrical characteristics other than to make sure the polarity is correct. There are exceptions, however, and they can be tricky！I've had problems with floppy drives with replacement LEDs. They appear to function normally, but won't read disks properly. Computer cases have LEDs for power and hard drive and possibly other functions！these can be swapped out with no problems.
How do I drive LEDs from alternating current?
Without getting into a detailed circuit design, there are a couple things to think about when using an AC source. Diodes will conduct only when forward biased, so with AC driving them, they'll be dark half the time. That isn't generally a desirable situation, so most people prefer to rectify the AC. A full-wave bridge rectifier is an inexpensive way to effectively double the brightness. Make sure your rectifier's peak inverse voltage is larger than the supply peak voltage. The rectifier will cause about a Volt and a half reduction in your supply, so consider that.
The other thing to consider is that AC voltages are usually expressed in Volts RMS (root-mean-square). The concise explanation is that this allows power calculations to give the same results as for DC voltage. But the peak voltage will be higher (1.4 times higher) than the RMS voltage, so when you do rectify and convert to DC, you have to be sure to use the peak numbers.
How many LEDs can I drive in series?
It depends on your supply voltage. You have to be able to overcome the cumulative forward voltage of the series combination in order to get them lit. As an example, with a 12 VDC supply, you could drive up to five red LEDs (2.1V @ 20 mA) in series. The current isn't used in the calculation: 5 * 2.1V = 10.5, which is less than 12, but 6 * 2.1V = 12.6 which is over 12, so five is the maximum series combination for that type of LED at that supply voltage.
When you need more, you can build arrays by combining multiple series strings in parallel. In that case, your load current needs to be calculated and kept lower than the rated supply current. For a given number of LEDs in a series+parallel array, using the largest series combination possible will keep the total load current lowest
