Lidmotor's Penny circuit help needed.

[Dark Alchemist]

Well, there you go then, this confirms what I said. Lowering this current value effectively moves the graph to the right, in other words raises the forward voltage.
Don't make a new model, just put 2 or 3 LEDs in series in your sim circuit, this will effectively raise the forward voltage. Think of it like raising the height of a dam. The blocking oscillator needs to have a voltage swing, and when your LED turns on and passes current this limits or clips the voltage level of the swing. If you put two in series you are raising the height of the "dam" so that the voltage swing when the dam overflows is greater, and contains more power.

Am I only looking to see if it lights all of them up?  I was hoping to put a 300ma and/or a 700ma bright white LED in the real circuit but I need it to simulate that first.

[Dark Alchemist]

www.youtube.com/watch?v=0xEyg7LOCMk

View that, please.

[Dark Alchemist]

I am up to 4 LED and 670pF.

[TinselKoala]

Good for you. Now you are making higher voltages with your circuit, and you should be seeing more total light output.

I can't quite make out the scope trace in the video. It looks to me like your timebase is set to 1 millisecond per division, is that right?

I'd like to see the scope display the actual signal, three or four peaks across the screen, instead of the "comb" you are showing. Try changing the timebase to 0.1 ms or even 1 us per division. We should also look at the voltage signal directly at the base of the transistor on the other channel of the scope. The max rated Vebo for your BC337 is only 5 volts.

I hope you realize what I was saying about the LED current/voltage relationship. To see what the _actual_ current is that you are putting into your LED stack, you can put a 1R resistor in series with the LEDs/capacitor at the cathode end, and look at the voltage drop across this resistor with one of the scope channels. By Ohm's Law, the current in this resistor is I == V/R, and since R is 1, the current in amps will be equal to the voltage drop in volts that you see on this resistor.

[TinselKoala]

Consider what a "300 mA" LED means. This means that you take the LED, and a DC voltage source that is comfortably above the fwd voltage of the LED, and you use a current-limiting resistor to limit the current through the LED to 300 mA, and then you get your rated light intensity output from that LED. The average current the LED draws will depend on the average voltage you supply to it! It will _not_ always be 300 mA!

A typical 1 Watt LED (300 mA) has a fwd voltage of 3.4 volts or so. The problem is that your circuit cannot supply 300 mA at over 3.4 volts for any length of time, so the LED won't reach its full brightness during the brief flashes that the oscillator is sending it.
If I had one of these 300 mA LEDs I would not hesitate to try it in the circuit. It's not going to draw an average of 300 mA, though!