Background:
I have designed a number of LED lighting products which are manufactured in China.
I have several cylindrical LED flashlights that have a large number of LEDs in them
... Are there ultra-bright LEDs that you can drive directly off of 4.5 volts without a current limiting resistor? Or are there special purpose ultra bright white LEDs made for 4.5 volt supply that have internal current limiting resistors?
No and no, unfortunately.
Many LED lights are constructed as you describe, with multiple white LEDs wired in parallel and connected essentially directly across the battery.
They are junk.
They are not "designed".
They build them this way "because they can" and they work well enough to be able to sell them.
When supplied with 4.5V + the LEDs are driven well above their maximum design rating and their lifetimes are greatly shortened. The LEDs used are typically low lifetime low cost devices.
Follow-up question: Does anybody know if the 12 volt LED bulbs that are in landscape lights have a voltage regulator in them?
The 12 volt LED strips usually use 3 LED die in series plus a series resistor.
Turn on / turn off time is liable to be sub `1 microsecond if capacitors are not used downstream of the switch.
Current is set to be "about right" at 12 Volts so will vary substantially if used in an automotive context where several volts of variation occurs. Many strips use individual LEDs but some use 3 die per package LEDs with all 3 independent die wired in series. It is possible but not certain that strips with individual LEDs will run somewhat cooler due to a lower concentration of Energy per package.
Lifetime of these LEDs may be better as the series resistor means that they are somewhat more properly driven. I have seen very substantial variations in output of similarly appearing strips. The brightness bears no obvious relationship to LED specifications and a brighter strip may simply reflect a manufacturers 'marketing decision'. You can get a range of LEDs per metre but current drain and number of LEDs are not directly related.
White LEDs are typically have a voltage drop in the 3.0 - 3.5V range at rated current.
Current increase tends to be exponential with voltage and at 4.5V almost any LED would self destruct almost instantly. The "saving grace" (if it can be called that) is that the combination of small batteries and many LEDs means that the batteries are unable to produce more than 'vastly too much' current when the batteries are new. Any light constructed in this manner demonstrates a total lack of concern and/or knowledge by the manufacturer.
Adding even a single common series resistor makes a substantial improvement in voltage/current profile and a resistor per LED would greatly assist current balancing between LEDs.
Added May 2016
Harper commented:
OP is asking about LED bulbs, not strips. Those are commonly made as screw-in replacements for incandescents. Some have a resistor, but many have a switching buck converter which will accept a range of voltages from 12-30V or higher. The LED series voltage is quite close to 12V actual, so if voltage drops much below 12V the buck converter will go to 100% duty cycle and simply pass the voltage through, causing the LEDs to dim rapidly.
My answer addressed LED strips as I noted, which the OP did not ask about, as Harper noted :-).
Harper's comments above are correct where applicable. I have not seen a bulb with a buck converter internally, but no doubt they exist. White LEDs have Vf typically in the range 2.8V - 3.5V. 2.8V is unusual and usually only seen in reasonably modern LEDs or ones operated well under full power. At 12V nominal, 4 LEDs have 12/4 = 3V each available. Allowing a small voltage drop in connectors and wiring 4 LEDs with Vf of 2.8V to 2.9V would be able to be operated at full power. In real world situations with Vin able to be somewhat below to substantially above 12V, 4 LEDs in series will often work but 3 x LEDs in series plus a series resistor is 'safer'. Bulbs may not match strips in configuration, but all 12V LED strips that I have seen use 3 LEDs in series plus a resistor.
An extremely simple and low cost constantish current source is shown in the circuit below.
Q1 is initially turned on by R2.
I_LED flowing in R1 produces a voltage per V = IR
of V = I_LED x R1
When V_R1 reaches about 0.6V Q2 is turned on and shunts base drive to Q1, thereby limiting any current increase.
ILED max ~= I_R2 x Beta_Q1 where Beta = current gain.
For say 5V and R2=1k IR2~= (5-0.6)/1k = 4.4 mA.
For Q2 Beta = 100 Imax = 440 mA - ample for most purposes.
This circuit drops a minimum of ABOUT 0.6V
Cost, size and complexity are minimal.
Results are far better than using just a series resistor.
ie Current is not "constant" but is far closer than with a resistor alone.
simulate this circuit – Schematic created using CircuitLab
Aka:
E&OE.
Best Answer
Lowest cost solution from a battery.
Load Power = 3.3V*20mA*17=1.1W @340mA , which would be short operating time for AA's and 5V USB pack has no tolerance specs, cost or other obvious assumptions were given.
until then, this is the cheapest.
simulate this circuit – Schematic created using CircuitLab
If Vf@20mA is matched with 1% ( expected from same batch) and Board has good copper heat radiation on pads for cooling, you can eliminate all the R's and drive with 3.3V Buck regulator rated for >350mA.
Or use a 3.3~3.7V LiPo cell and 3.3 Buck regulator with 90% efficiency max or TPS736-3.3 LDO (0.4A) for $0.75 but same efficiency as 17 * 15ohm R's but depends on 1% matching but gives constant If output to Vbat=3.4.
For a Boost Regulator with 20mA output, you need an array of 17S1P with 17*3.3V = 56.1V.
The problem
You cannot trust any EBAY specs for Ah when there is no datasheet to prove how they perform when loaded faster than 20h discharge rates are used. Anything must be proven with a datasheet and supported by supplier. Not all 18650 cells or USB packs are the same!! Buyer beware!