You'll lose regulation, and the output voltage will be slightly less than the input voltage. The buck converter topology can only reduce voltage, not increase it, because it's essentially just filtered PWM on the power supply.
If you want to have an output voltage higher than the input, you'll need a boost converter, and if you want the input to be able to ...
In this answer I'm assuming that the basic circuit is a flyback converter that regulates to produce a 5 volt DC output. A regular old-fashioned power transformer cannot be relied upon to produce a stable 5 volt output due to the transformer's input AC voltage changing and because of its inability to provide decent regulation against load current changes.
AJN nicely answered (+1) - I'm just putting the following as an answer so i can show detail and pictures.
I find that using Laplace to solve circuit problems algebraically is often times easier than wading through differential equations. I'm much better at algebra than d.e. For example, below i will solve this case for the inductor current:
Transform to ...
With less, smaller external parts, Buck-Boost converters provide a more powerful solution. You can use this minimum number of components for either step-up or step-down voltages and deliver a shorter operating time and a higher performance over a wide range of input and output voltages. Buck boost converters are also much cheaper than other converters.
a) Advantage is if you need 3.3V too and the 1.8V cannot handle 12V input. Also any 1.8V converter can be used as it does not need to handle 12V input. Disadvantage is more power losses and heat production in each conversion, and the 3.3V converted needs to be larger to support also the power for 1.8V converter.
b) Advantage is usually that there are less ...
Your solution (to the homogeneous as well as particular) is correct. If we solve the differential equations as usual, we need to plug in the initial conditions at \$t=0\$ while the initial conditions are given at \$t=\alpha/\omega_0\$ in the book. That is why you seemed to have ended up with an unknown \$(I_0 - I_2)\$ in your coefficients and missing a \$(-\...
This cannot work as on this type of buck converters, the - and the input and the - at the output are shorted. Let me draw those connections into your drawing:
Note how your "-15 V has a direct connection to ground.
You're also shorting (to ground) the + output of the most left converter.
This is the simplest and easiest switching converter to understand: -
The input voltage is modulated by the switch's duty cycle to produce an average value that corresponds with the required output voltage.
The LC acts as a low pass filter and converts the modulated duty cycle from above into a steady DC value.
There are mathematics involved if you want ...
Did you use an autorouter?
@Andyaka I kinda did, yes. Routing a PCB was surprisingly time
consuming and actually wasn't that easy. So I used an autorouter and
edited the traces afterwards to, for instance, remove 90° angles in
The routing, component placement and track widths are pretty bad in very important areas. Don't use an autorouter unless it'...
How do i ensure the output voltage stays constant for a varied input
You use a method called feedback - the output voltage is compared against a stable reference voltage and, if it rises above the reference voltage the duty cycle of the buck converter is lowered. If it falls below the reference voltage, the duty cycle is increased. For this you ...
Well that's easy - the source is connected to the ground plane and the MOSFET is turned on, so the current comes through the MOSFET from the ground plane.
"Wait", you say, "it comes from the ground plane?"
Yes. The left end of the inductor has a voltage below the ground plane. Remember the inductor really wants to keep the current the ...