currentinductanceinductor
I need some help with this basic inductor.
If I have the current in that direction and those values of voltage and inductance, what is the value of $$\frac{di}{dt}$$ ?
I know that:
$$V_L=L.\frac{di}{dt}$$
So:
$$ (2-0) = 2.\frac{di}{dt}$$
So:
$$\frac{di}{dt} = 1$$
but that's wrong because the answer is
$$\frac{di}{dt} = -1$$
What is wrong??
Thanks
Choosing an inductor value for a buck regulator comes directly from V = \$\frac{\text{L di} }{\text{dt}}\$ . Where V is the voltage across the inductor, and i is the current through it. First, you want to design for the case where the inductor is in continuous conduction mode (CCM). This means that energy in the inductor doesn't run out during the switching cycle. So, there are two states, one where the switch is on, and another where the switch is off (and the rectifier is on). Voltage across the inductor during each state is essentially a constant (although it is a different value for each state). Anyway since the voltage is a constant, the inductor equation can be linearized (and rearranged to give L).
L = \$\frac{V \text{$\Delta $t}}{\text{$\Delta $I}}\$ this is the basis for the equation you saw in the app-note.
\$\text {$\Delta $I}\$ is something you define, not determine.
You will want to maintain CCM operation, so define \$\text {$\Delta $I}\$ as some small fraction of inductor current (I). A good choice is 10% of I. So, for your case \$\text {$\Delta $I}\$ would be 0.24A. This will also define the ripple current in the output capacitors, and less ripple current means less ripple voltage on the output.
Now you can choose an optimal value of L using \$V_{\text{in}}\$ and \$V_o\$ (and hence the duty cycle D = \$\frac {V_o} {V_ {\text {in}}}\$). But you can also make a quick over estimate for the inductance where you don't consider \$V_{\text{in}}\$ using L ~ \$\frac{10 V_o}{I_o F_{\text{sw}}}\$ (for more on this look here How to choose a inductor for a buck regulator circuit? ). An over estimate can be worthwhile, especially if you are early in development or uncertain exactly how much the output current will be (output current tends to end up higher than expected usually).
Since you are looking at Linear Tech you should (as Anindo Ghosh pointed out) also look at using their CAD support.
The way you drew it, I and \$\frac{dI}{dt}\$ are both positive in the direction of the arrow and the correct expression is
\$V_L = L\frac{dI}{dt}\$
The negative sign depends on the direction you define your voltage and current.
simulate this circuit – Schematic created using CircuitLab
Best Answer
The current direction in the diagram is from lower potential to higher potential but in applying the formula you have assumed the current to be from higher potential to lower. The formula becomes:
$$(0-2) = 2\times\frac{di}{dt}$$
Apply KVL in the direction of current you are assuming
In this particular question if you rig the circuit up, current will always from from higher potential to lower (2V-0V), hence it is opposite to the assumed direction and therefore negative.