Is the fan being switched on/off? If not then I wouldn't worry about it. The capacitor across the rail will take care of it.

 

No just running.

Here are some pictures. ( Don't pay attention to where the waveform is on the scope. I moved it around for the best picture)

Schematic

15V rail before 470uF cap

15V rail after 470uF cap

7v rail

339 output at about 99% duty cycle

Emitters of BJT's at same duty cycle

So any idea how to get rid of the ringing?
Thanks everyone for making this happen!

 

Well Since I am pulsing the LED I ramped up its current to 1A

You might be destroying the LED because it heats extremely quickly as shown in the "absolute max allowed current" in the spec's. Extra current is used for very short duration pulses so that the LED appears as bright as with steady DC at less current.

are you referring to Fig 8 in the datasheet?

The curves in datasheets are for "typical" devices. Figure 8 in the datasheet for the old IRF510 mosfet shows the on-resistance and current for a typical one.
But you can't buy a typical one. There are weaker ones and stronger ones and you get whatever is available. The Electrical Specifications (text) in a datasheet guarantees the spec's for the weakest ones then when you design a circuit for the weakest one, all devices will work.

 

You might be destroying the LED because it heats extremely quickly as shown in the "absolute max allowed current" in the spec's. Extra current is used for very short duration pulses so that the LED appears as bright as with steady DC at less current.

Is the only thing that will destroy an LED heat? If you can control the heat can you pump up the current? Or will the little wires just break from the current? I have it on quite a large heatsink.

The curves in datasheets are for "typical" devices. Figure 8 in the datasheet for the old IRF510 mosfet shows the on-resistance and current for a typical one.
But you can't buy a typical one. There are weaker ones and stronger ones and you get whatever is available. The Electrical Specifications (text) in a datasheet guarantees the spec's for the weakest ones then when you design a circuit for the weakest one, all devices will work.

Ok. So since the gate threshold voltage is when it just starts to conduct, is there a spec that I am missing that signals the end of cuduction? That would be nice. haha

 

You forgot to say which brand of LEDs you have. A Philips Lumileds Luxeon Rebel white LED is rated at 700mA and has a max allowed current of 1A when it is cooled well enough.

The gate of the Mosfet is getting only 6.0V to 6.3V because the complementary transistors have a 0.7V to 1.0V base-emitter voltage drop.

The ringing is probably caused by the inductance of all the long wires on your breadboard. A breadboard should be used only for DC and low frequency circuits.

 

Hey Bill I thought I would let you know,
When using R1 as 100k and R2 of 100ohm when R1 was adjusted to the lowest value the sawtooth would actually overstep its 1/3 and 2/3 values. I didn't try 10k and 100ohm but Now I am using 10k and 1k and it seems to have helped.

 

Yeah, you could actually eliminate the two resistors, or tweak them. 100% is all the way on (no modulation), 0% is all the way off (again, no modulation).

 

You forgot to say which brand of LEDs you have. A Philips Lumileds Luxeon Rebel white LED is rated at 700mA and has a max allowed current of 1A when it is cooled well enough.

The gate of the Mosfet is getting only 6.0V to 6.3V because the complementary transistors have a 0.7V to 1.0V base-emitter voltage drop.

The ringing is probably caused by the inductance of all the long wires on your breadboard. A breadboard should be used only for DC and low frequency circuits.

The gate of my MOSFET is receiving 6.8v

And I got the LED off ebay and just looked to see. It is the WHITE one.
It is supposed to get .3A....I am giving it 1A
haha. Works for now...

 

Yeah, you could actually eliminate the two resistors, or tweak them. 100% is all the way on (no modulation), 0% is all the way off (again, no modulation).

Really? What effect would that have on the circuit? Much higher frequencies?

 

No frequencies involved at this point. 100% is all the way on, 0% is all the way off. Even a little off will have the modulation frequency, say 0.001% (or 99.999%) will be an extremely narrow pulse either way. If you use a pot without the other resistors only 1/3 of the range will be active.