Which is a good question, what exactly are those characteristics?

Only the TS knows that.

 

The TS said the current draw is 2.6 amps peak for the full string, I'm assuming at 130Vdc meaning each LED draws 2.6 amps.

I think not. That would be 338W. More likely the draw at the power source.

 

Your delay in turn-on is caused by the way V1 is defined; it is NOT a constant voltage source. This is the plot of JUST V1, with the rest of the circuit disconnected.


If V1 is replaced with a 130 DC voltage source, the circuit reaches 130 volts in 1 second with an asymptote response. This is undoubtedly caused by the circuit impedance and the Cgs of the two FETs in the circuit forming an RC time constant. Since the circuit is a concept circuit only, there is no point in pursuing the cause or a fix for the remaining delay.

Since LEDs are really a constant current type of device, it is best to drive them with a constant current source if you want predictability. While you can drive them with a voltage source and a current limiting resistor, the LED characteristics can vary quite a bit between semiconductor lots, and that is in addition to how much a given LED will change with temperature of the die. I agree with others that your approach, while it could be made to work, is fundamentally flawed.

I assume you are using 130 volts worth of LEDs (9 x 3 LEDs) because you want bright flashes. I would drive the series LED string with a constant current source that provides the current each LED requires (typically 200 ma for high intensity white) that is gated on or off according to your timing requirements.

 

Thank you @crutschow for the sim, interesting, but I think I may give @sghioto version a try and see if that works out.
This is a DC powered device, not a battery operated, but efficiency and low current are prime concerns.
LEDs flash for 20ms at 1Hz, the reservoir cap is used to feed the current to the LEDs.
@Parkera V1 IS set as such for the LTspice to simulate the voltage delay/rise in the particular sim I provided originally - its not the power supply.

 

I think I may give @sghioto version a try

What version?

 

Post #18, or just give up on the idea of using a mosfet to bypass.

 

A single LM393 comparator can handle the the turn ON and turn OFF portion of the circuit.
A high voltage current limiter made from a couple of transistors should suffice.
How much current does the LED string actually draw by itself ?

 

o speed the startup until flashing starts, reduce the capacitance being charged. I have in mind a dual time constant scheme as a possible solution.

 

Post #18, or just give up on the idea of using a mosfet to bypass.

That shows a concept, not a complete circuit.

What about the circuit I showed in post #17.

 

@sghioto, yes, but I was trying to minimise the number of chips, but I may have to end up doing that. The LED will draw about 2.6A peak and drops off quickly after that during the 20ms.
@MisterBill2, reducing cap is not an option. Dual cap, i.e. as in your post #14, how would I switch the larger cap in circuit after the initial startup?
@crutschow, yes, that's worth a try too, thanks.

 

@sghioto, yes, but I was trying to minimise the number of chips, but I may have to end up doing that. The LED will draw about 2.6A peak and drops off quickly after that during the 20ms.
@MisterBill2, reducing cap is not an option. Dual cap, i.e. as in your post #14, how would I switch the larger cap in circuit after the initial startup?
@crutschow, yes, that's worth a try too, thanks.

The dual cap arrangement would require a series resistor to limit current for charging the large capacitor and a series diode to allow high current discharging after it was charged. The guess being that it is the low voltage while charging. If the delay in the voltage rise because of capacitor charging is not he cause of the delay then that is not required.