Electrical – LED lights wired in series will not light

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.

Firstly, you need to go back to school to learn the basics. What you are trying has no hope of working.

An LED is a current driven device, and it has a fixed forward voltage drop. Quite frankly, you are lucky the LEDs are even working at all any more - you must have LEDs that have a forward voltage at (or around) the 3V the batteries are supplying.

You have to run the LEDs with some form of current limiting device. At the moment that current limiting device is the batteries themselves being in the non-linear portion of the LEDs IV curve, which is very fortunate for you, or they would have died.

So your LEDs have a forward voltage of (around) 3V. Putting two in series makes that forward voltage 6V, so you will need at least 6V to drive them in series. You also need to limit that current to the right value to run the LEDs (no idea what that should be for your specific LEDs, but 20mA is common for normal LEDs).

So running from 9V would be sensible in this situation. Assuming 20mA and 3V forward voltage, you will need a resistor, of:

$$ V_{LED}=9V - 6V = 3V $$ $$ R_{LED}=\frac{V_{LED}}{I_{LED}} = \frac{3}{0.02} = 150\Omega $$

^{simulate this circuit – Schematic created using CircuitLab}

You could also run from 4.5V (3xAA) and have as many LEDs in parallel as you wish. Each LED must have its own resistor in this case:

^{simulate this circuit}

The resistor of 75Ω is worked out in the same way:

$$ V_{LED}=4.5V - 3V = 1.5V $$ $$ R_{LED}=\frac{V_{LED}}{I_{LED}} = \frac{1.5}{0.02} = 75\Omega $$

As a side note, it is more common to put the switch in the + side of the circuit, not the - side. It doesn't really make any difference, it's just that's where others would expect it to be who don't know the circuit you have designed.