Flashing Leds operate in one of two ways inside. The first, I was told by my electronics class teacher a decade ago. The Led has a thermal wire or metal inside, that when heated up, flexes & breaks the connection. Once the connection is broken, the wire cools down, and flexes back, making the connection again. This makes the led flash. Since the Led Junction itself produces heat (and light) when power is turned on, only the thermal wire is needed. Works like a miniature version of a car turn signal flasher. These might not require a resistor, I am not certain.
The second version uses an embedded microcontroller/ic to do the flashing. Similar to the self flashing rgb leds.
In either case, these both provide one other function. They break the connection between the two leads of the led. Any regular led in series with them will flash at the same rate, because power is physically cut off. In this case, since the led is tied to ground, once the led makes the connection, the base of the q1 and q2 transistors are affected, disabling them.
As for replacing them, you could not do it easily, without a microcontroller atleast. The Flashing Led is a critical part of a FLED Solar Engine. Here is a wiki with a more detail explanation on how FLED Solar Engine's work.
I suspect that they use a 9 volt battery since it is a handy, readily-available power source.
The electrical characteristics of LEDs (and diodes in general) are a bit odd - they don't follow Ohm's Law. The voltage across an ordinary ("bare") LED depends primarily on its colour, and will vary only slightly with current (for a red LED, the voltage will be about 1.8 volts). Therefore it is necessary to include something in an LED circuit to limit the current to a safe value - the 1K resistor in the circuit you linked to serves as the currrent limiter.
An LED advertised as "12 volts" or "5 volts" will include a series resistor or other current-limiting device, and can be connected directly to the advertised voltage with no other components.
A "bare" red LED will have a voltage drop of about 1.8 volts, yellow is about 2 volts, and green is about 2.2 volts. Blue and white are about 3.3 volts (white LEDs are really blue LEDs with a yellow phosphor).
Best Answer
lets start with Specs. and assumptions;
CR2032 Nominal Voltage: 3.0 Volts Typical Capacity: 240 mAh (to 2.0 volts) (Rated at 15KΩ at 21°C)
Battery ESR ~ 3kΩ
Red & Yellow LED = 1mA@1.9V , 20mA@ 2.2V +/-10% Voltage thus from 1~20mA, ESR= 0.3V/20mA = 15Ω
let Flash time duration = 100ms RC=0.1 , R=15Ω ∴C = 0.1/15 =6,600 μF, ∴ consider lower duration as this will cause problems for charge time
Resistance ratio = R_bat/R_led = 3,000Ω/15Ω = 200 which means charge time is 20 seconds and thus either one needs to reduce battery reistance ( bigger battery ) or raise LED resistance ( lower current from 20mA to 2mA ) or reduce flash duration (preferred).
(Take 2) Let Flash duration = 10ms, C= 660 μF and consider raising LED current to 30mA
Compute Charge time from discharge Vmin to Vmax voltage:
Vmin = LED threshold , e.g. 1.6V for Red,Yellow Vmax = 3V max, or less depends on charge time, eg RC rise time, ∆V=63%
T = R_bat*C = 3kΩ * 660 μF =2 seconds. where T is defined as ∆V=63%
thus after T charge time Vcap= where 3-1.6V = 1.4V ∆V=63%*1.4=0.9V
∴Vcap=0.9+1.6=2.5V
Simple solution might be a 5Hz CMOS gate relaxation oscillator (Schmitt inverter with 10MΩ feedback R, 1uF cap to gnd) driving a MOSFET in series with LED on Drain to 660uF cap which is in parallel with CR2032 battery
You can make your own Parts List and schematic for Q, two C's, one R1, one LED, one battery.
Random ? go digital.
I wanted to show the analog design process, not the solution.
A digital solution can be done by others.