For 16 levels it's easy to do a simple look-up table "by hand" and convert the 4 bit value in an 8 bit value to pass to PWM controller: this is the component i've used in my FPGA led array driver.
For an 8 bit level controller, you'll need at least 11-12 bit output from the look-up table.
library IEEE;
use IEEE.Std_logic_1164.all;
entity Linearize is
port ( reqlev : in std_logic_vector (3 downto 0) ;
pwmdrive : out std_logic_vector (7 downto 0) );
end Linearize;
architecture LUT of Linearize is
begin
with reqlev select
pwmdrive <= "00000000" when "0000",
"00000010" when "0001",
"00000100" when "0010",
"00000111" when "0011",
"00001011" when "0100",
"00010010" when "0101",
"00011110" when "0110",
"00101000" when "0111",
"00110010" when "1000",
"01000001" when "1001",
"01010000" when "1010",
"01100100" when "1011",
"01111101" when "1100",
"10100000" when "1101",
"11001000" when "1110",
"11111111" when "1111",
"00000000" when others;
end LUT;
Contrary to initial intuition, it's actually to increase the brightness.
LEDs can be driven at a constant current, or they can be driven with a pulsed current.
With constant current you have to limit the current to a relatively low value - for instance many common small LEDs are limited to a constant current of say 20mA. That gives good brightness for indication purposes, but it's not that great.
LEDs, when driven with pulsed current, can be driven with a considerably higher current - maybe 5 to 10 times as much, or even more. That could be say 100mA for what would normally be a 20mA LED. However, there are restrictions on what the pulses can be - typically with limits on the frequency and duty cycle - maybe as little as 1% duty.
The end result is that the higher current increases the perceived brightness of the LEDs, since more photons are being emitted when they are on, but at the cost of some flicker, which is only really noticed when the LEDs are in motion.
A research group at Ehime University developed a pulse drive control method to make LEDs look twice as bright by leveraging the properties of how people perceive brightness.
The group was led by Masafumi Jinno, an associate professor of Dept of Electrical and Electronic Engineering at Graduate School of Science and Engineering of Ehime University.
When a short-cycle pulse voltage with a frequency of approximately 60Hz is applied to an LED at a duty ratio of about 5%, the LED looks about twice brighter [sic] to human eyes than that driven by a direct voltage, the research group said.
- Nikkei Technology - Human Perception Studied to Double LED Brightness
So you get more perceived brightness from smaller and cheaper LEDs without using more current (often less current) on average than if they were on constant.
The report above goes on to explain the effect in more detail:
There are two principles, the Broca-Sulzer effect and the Talbot-Plateau effect, involved in how human eyes perceive brightness. The Broca-Sulzer effect refers to a phenomenon in which light looks several times brighter to the eye than it actually is when exposed to a spark of light, such as a camera flash.
In addition, the Talbot-Plateau effect is a principle where human eyes repeatedly see flashes and sense the average brightness of the repeated lights. Thus far, "it has been believed that, due to the Talbot-Plateau effect, the brightness perceived by human eyes would not change even if an LED is pulse driven," Jinno said.
"The Talbot-Plateau effect is a principle found in the days when fluorescent mercury lamps and other light sources driven by a power supply with a longer voltage cycle of about several hundred milliseconds were used," Jinno said.
Thus, the group decided to drive the LEDs using a power supply with a shorter voltage cycle of about several hundred microseconds. As a result, the group discovered that, when a pulse voltage with a frequency of approximately 60Hz is applied at a duty ratio of about 5%, the impact by the Broca-Sulzer effect becomes greater than that of the Talbot-Plateau effect so that the light emitted from the LED looks brighter to human eyes.
Best Answer
Your problem is pretty straightforward. Your current limiting resistor is much too large. If your LEDs are in fact allowing as much current as you assert, the voltage across the resistor will be .08 x 30, or 2.4 volts. This leaves (at most) 0.9 volts across the LEDs, and that is not enough to allow them to produce much light at all.
You should resize your resistor, taking into account the forward voltage (Vf) of the LEDs, to allow maximum current with transistor fully on, and a transistor voltage drop of about 0.1 to 0.2 volts. Either that, or increase your source voltage.
Once you do that, you're still likely to have problems. With 5 LEDs in parallel, whichever one has the smallest Vf will hog current and glow more brightly than the others. In the worst case, this will cause it to get hotter, its Vf will drop, and it will hog even more current and get even brighter. At this worst-case limit, it will draw nearly 5 times as much current as you expect. If this level is too high, the LED may fail open, leaving the process to repeat in turn with the other 4, then the other 3, etc.
Finally, you need to examine the data sheet for your transistor and determine its current gain. This is the hfe which Ignacio referred to his comment. To make life more difficult, gain changes with current level, as you will see if you pay attention to the data sheet. But let's say that the gain is 100, which is a decent starting point for modern NPN signal transistors running at less than 100 mA. Keep in mind that, due to your large limit resistor, the current will never approach the 80 mA you think it will. Let's say 10 mA, just as a start. Then any base current above (10 mA / 100) will make no difference to the LED current, since the transistor is pulling as much current as it can, and the current is limited by the resistor and the LEDs. 10 mA/100 is 0.1 mA, or 3% of your nominal drive, and is entirely consistent with what you see.
In order to check this, fire up your circuit and connect the collector of your transistor to ground. Now measure the voltage across the 30 ohm resistor, and divide by 30, to give your total, maximum current. Divide this by 10 or so to get the base current you need. To understand why you divide by 10 rather than 100, start learning about transistor saturation.