Here's what we know about the constant current buck converter:
It provides a current of exactly 700mA from its output side, through whatever load is connected between \$^+V_{OUT}\$ and \$^-V_{OUT}\$.
It is supplied with a constant voltage of 24V.
It is 95% efficient.
Its output voltage \$^+V_{OUT} - ^-V_{OUT}\$ can be anything between 0V and \$V_{IN} - 3V\$, or 0V to 21V.
Since efficiency E is the ratio of input power to output power, we can represent this relationship as:
$$ P_{OUT} = E \times P_{IN} $$
Using the power law \$ P = I \times V \$, and plugging in what we know, this is the formula we get:
$$
\begin{aligned}
I_{OUT} \times (^+V_{OUT} - ^-V_{OUT}) &= E \times I_{IN} \times V_{IN} \\ \\
700mA \times (^+V_{OUT} - ^-V_{OUT}) &= \frac{95}{100} \times I_{IN} \times 24V
\end{aligned}
$$
Notice that \$+V_{OUT}\$ and \$-V_{OUT}\$ are unknown at this point, since we haven't yet considered the load:
We don't yet know what values they have with respect to each other (the potential difference between them), and
importantly, we don't even know what values they have with respect to 0V (ground) on the input side. They may be completely isolated from the input, information which you must obtain from the driver's data sheet.
Point (1) is a result of the fact that any constant current source adjusts its output voltage to maintain the required current (700mA in this case), and this will be whatever voltage the load develops when passing that current.
For your LED, that seems to be 17V, which will rise as the LED's temperature falls, and vice versa. How much that voltage rises and falls with temperature is a parameter you obtain from the LED's datasheet. As long as the forward voltage of your LED never exceeds 21V, this unit will work fine, and the condition \$I_{OUT}=700mA\$ will be maintained.
From now on, instead of writing \$(^+V_{OUT} - ^-V_{OUT})\$, I will refer to this difference simply as \$V_{OUT}\$
With a value of \$V_{OUT}=17V\$, we can return to the \$P_{IN}\$ vs. \$P_{OUT}\$ relationship, to find the last unknown \$I_{IN}\$:
$$
\begin{aligned}
I_{OUT} \times V_{OUT} &= E \times I_{IN} \times V_{IN} \\ \\
700mA \times 17V &= \frac{95}{100} \times I_{IN} \times 24V \\ \\
I_{IN} &= \frac{\frac{100}{95} \times 700mA \times 17V}{24V} \\ \\
&= 0.52A
\end{aligned}
$$
Now we have all the information necessary to calculate input power \$P_{IN}\$ and output power \$P_{OUT}\$ (power delivered to the LED).
$$
\begin{aligned}
P_{IN} &= I_{IN} \times V_{IN} \\ \\
&= 0.52A \times 24V = \\ \\
&= 12.5W
\end{aligned}
$$
$$
\begin{aligned}
P_{OUT} &= I_{OUT} \times V_{OUT} \\ \\
&= 700mA \times 17V = \\ \\
&= 11.9W
\end{aligned}
$$
Their difference will be the power \$P_{DRV}\$ is wasted in the driver unit:
$$
\begin{aligned}
P_{DRV} &= P_{IN} - P_{OUT} \\ \\
&= 12.5W - 11.9W \\ \\
&= 0.6W
\end{aligned}
$$
All the above serves to illustrate the relationships, and how they fit in with the behaviour of a DC-DC converter. It's long-winded though, and could have been cut down to a very simple:
$$
\begin{aligned}
P_{DRV} &= P_{IN} - P_{OUT} \\ \\
&= \frac{1}{95\%} \times P_{OUT} - P_{OUT} \\ \\
&= (\frac{1}{95\%} - 1) \times P_{OUT} \\ \\
&= (\frac{1}{95\%} - 1) \times I_{OUT} \times V_{OUT} \\ \\
&= (\frac{1}{95\%} - 1) \times 700mA \times 17V \\ \\
&= 0.6W
\end{aligned}
$$
I'll rewrite the LED and driver power equations here, to illustrate my next point:
$$
\begin{aligned}
P_{LED} &= I_{OUT} \times V_{OUT} \\ \\
P_{DRV} &= (\frac{1}{E} - 1) \times I_{OUT} \times V_{OUT} \\ \\
\end{aligned}
$$
Since LED current \$I_{OUT}\$ is constant, and assuming effiency also remains constant (negligible change over the entire range of operating conditions), you can see that power disspipated in each element is a function of \$V_{OUT}\$ only. Therefore, to find the worst case power dissipation in the driver and LED, you need to find the worst case value of \$V_{OUT}\$.
In both cases, maximum power is when \$V_{OUT}\$ is at a maximum, and as I said before, you'll need to refer to your LED's datasheet to find out what this will be. It will occur at the lowest temperature that you expect your LED to experience.
I do not think you will have to worry about driver power increasing very much over the entire LED operating temperature range. However, in the terrible case where total LED forward voltage rises to 21V, the maximum that your converter can provide, power dissipated in the driver and LED would be:
$$
\begin{aligned}
P_{LED} &= I_{OUT} \times V_{OUT} \\ \\
&= 700mA \times 21V \\ \\
&= 14.7W
\end{aligned}
$$
and
$$
\begin{aligned}
P_{DRV} &= (\frac{1}{E} - 1) \times I_{OUT} \times V_{OUT} \\ \\
&= (\frac{1}{0.95} - 1) \times 700mA \times 21V \\ \\
&= 0.77W
\end{aligned}
$$
As for your reference to a MOSFET, I can't see from anything in your question which MOSFET you are talking about. I assume you mean that you wish to switch the LED on or off using an exernal transistor. Here are my thoughts on that:
Why do that when you have a digital on/off input on the driver?
Since I don't know if \$^+V_{OUT}\$ and \$^-V_{OUT}\$ are isolated, or in other words, if there's a common ground and which side it's on, the practicality is questionable.
You had better be very sure of how your driver will behave when the load is connected and disconnected in this way. It might even be discouraged. Again, the datasheet is your friend.
Hello, please shed some light if my analysis is correct regarding the circuit. It is a DC circuit with an LED, LED driver and MOSFET.
