transistors
I haven't studied electrical engineering however I am currently studying about Moore's law.
I found an answer to why transistors keep increasing but I still can't get my head around why so many are needed for dynamic RAM.
It says 6 transistors are needed for each bit in dynamic RAM.
What exactly do these 6 transistors do? Are the used to actually store and hold the data since it is volatile?
Sorry for the stupid question but I really can't get my head around it.
One reason you see less than 100mA when you connect the display up to a 5V supply through an 8 ohm resistor is because the forward voltage of the back light LED is higher than the 4.2V level. Another reason could be due to the meter that you put in series to measure the current. Meters generally measure current by looking at the voltage drop across a small value resistor and you would have effectively placed that resistance in series with the 8 ohms and thus reducing the current some amount.
For the buffer circuit try connecting it up like this:
This will allow the MCU to switch the NPN transistor on and off successfully. As you had it drawn before the LED back light load was being operated as an emitter follower. In such configuration the load was seeing a voltage that was at least a Vbe drop lower than the voltage the MCU pin was placing on the base circuit of the transistor.
With it redrawn as I've shown it the MCU can saturate the transistor and allow the collector circuit to sink the full current from the supply and through the LED and 8 ohm resistor.
If the MCU is able to place a 5V signal on its output pin then the base current of the transistor with 1K resistor will be:
Ibase = (5 - Vbe) / 1K = (5 - 0.7) / 1K = 4.3mA
The transistor current gain is Hfe. With a desired up to 100 mA of current in the collector the transistor will need an Hfe of at least this much:
Hfe = 100mA / 4.3mA ~= 23.
The transistor part numbers I have shown should have Hfe values that are higher than 23 and so the 4.3mA base current should allow the transistor to fully saturate.
One note is that this circuit is going to turn on the backlight when the MCU pin is placed into the high state. Thus logical '1' = High = On. Nice.
What you have is basically a two stage amplifier - two consecutive amplifiers. In such a circuit configuration the gain of both amplifiers multiply. Since each stage has negative gain in your example, the overall gain is positive again.
So let's say Q1 and R2 have a voltage gain of -10 and Q2 together with R3 create a gain of -10, too. Then the overall gain is 100 which is positive and much larger than the gain of a single stage.
In your example this means the following: If UART_TXD goes High, TX_TTL will go High, too. If you omit Q1 and directly feed Q2 with UART_TXD, then TX_TTL will go Low when UART_TXD is High.
Best Answer
Assuming that by "dynamic RAM" you are referring to DRAM, it is the first time I hear about "6 transistors per DRAM bit".
DRAM's bit is usually implemented in the following way (or some slight variation on it):
As you can see, there is just one transistor and one capacitor here. Since capacitors in integrated circuits also have "transistor-like" structures (at least in MOSFET technologies), you could say that DRAM cell consist of two transistors.
However, in order to support DRAMs structure and functionality there is a standard "framework" which is required: sense amplifiers, repeaters, logic for refreshing the contents of DRAM, bit-line/word-line selectors etc. It is this "framework" which makes DRAM unacceptable for small volume storage - you need to provide the entire "framework" even for DRAM of size 1kB, which makes this 1kB storage too big and too power consuming.
However, if you make DRAM storage large enough, the standard implementation of 2 transistors per bit compensates for the "cost" of the framework.
In order to be completely accurate, we could account for the increase in "framework" size as a result of increased storage area. For example: it is evident that you'll need stronger sense amplifiers if your DRAM is not 1kB, but 1GB. Even then, I'd speculate that the "cost" of one additional DRAM bit is not higher than 3-4 transistors.
As for what the transistor does... In simple words: transistors behaves like an open-closed switch in the above configuration. When the transistor is "open" (i.e. not conducting) - the charge on the capacitor does not change, thus the level of the voltage does not change, thus the logical value of this bit does not change. When the transistor is "closed" (i.e. conducting), the voltage of the capacitor can be either probed (i.e. read the value of this bit), or changed (i.e. write a value to this bit).