Activating an NPN BC817 for a LED
Assuming that \$V_{BE}\$ is around 0.7V. According to the datasheet, the minimum DC gain is 100. But that's when \$V_{CE}\$ = 1V. You will most likely saturate the transistor, and the DC gain gets really low in saturation. Well let's take the worst case scenario, DC gain = 10.

With 1 volt input and R2 = 1kΩ, the base current will be \$\frac{1V-0.7V}{1000Ω}=0.3\$mA. That means that \$I_{CE}\$ will be up to \$10×0.3\$mA \$=3\$mA.

I'll assume that \$V_{CE}=0.1\$V because it is in saturation.

With VDD = 5V and R1 = 150Ω, the \$I_{CE}\$ current will be \$\frac{5V-0.1V-3.5V}{150Ω}=9.3\$mA

\$9.3\$mA is greater than \$3\$mA. So it looks like you will need to reduce that 1kΩ to say 330Ω at least. I'd go for 200Ω to be on the safe side. This is assuming an absolut worst case scenario gain of 10. Your transistor might have 10.. or 30.. or more, so if the gain is 30, then the 1kΩ will suffice, but if you want to guarantee that it will work, then 200Ω will be enough. Because I've never seen any gain that is worse than 10 while being fully saturated.

Here's the link to a schematic in case you want to play around.