Would an SST20 in a T9R out-throw a convoy C8?

I’m kind of guessing yes…but not sure

The FF T9R has a fairly large reflector, so an SST20 will result in great throw, and might very well outthrow a C8.
That is however, when the C8 does not have a CSLNM1 or CULPM1.

Hi CRI SST20? Or 6500k?

95CRI would yield roughly 150kcd, and CW would be 220-230ish. An NM1 in a C8+ can hit as much as 230kcd on average across BLF, I’d expect closer to 200 unless you have some top binned emitters.

I probably should have clarified, same emitter… there is only really one high cri choice on the market.

the convoy has an uncoated lens afaik, so there’s also that.

Well then yes. C8 with 4000k sst20 should hit around 70-75kcd.

I would rate usable throw for the T9R at ~390m and ~270m for the C8. These values ate just half ANSI spec.

From what I have been told, you would end up burning an SST20 if run on Turbo in the T9R. Even 6 amps is pushing it to its limit where it has been tested to die at 7amps.

6A should be fine. If determined to lower that a smidge, a sense resistor mod is easy.

Osram version has no turbo, since only Jack has FA4 I would want to have him custom it, but no reply yet.

How would one do the sense resistor mod, is it even necessary for the buck driver? I got a Hakko, but no programming skills/tools, nor much electrical knowledge. I think over 4-5a is a bad idea for sure though given the chart’s data and my C8 overheating experience

In the case of reducing current, you would just find the ‘sense resistor’ on the driver, desolder it, and replace with a higher value resistor. No programming needed - all hardware based changes.

The sense resistor is usually very easy to spot because they are typically large, low value (R010 or R005 is common on high end drivers), and typically near the solder point of the LED- wire.

Their function is to create a feedback signal to the driver controller. The resistor is in line with all the current going to/through the LED. There is intrinsically a voltage drop over this resistor. Ohms law says V=I*R so I=V/R. The controller regulates the switching frequency/duty cycle to keep a steady and targeted voltage across this resistor.

Hypothetically, let’s say they used an R010. This is 0.010 Ohms (the R denotes a decimal btw).

Stock driver output is 6A. We first solve for V to learn what voltage this driver controls to. So V=I*R= 6*0.010 = 0.060V or 60mV.

Now we work backwards with new constants… V = 0.06 and whatever your target current is, let’s pick 4A. Solving for R this time so R = V / I

R = 0.06 / 4 = 0.015

So a R015 resistor is what you’d go for to target 4A in this hypothetical situation.