# DC-DC Converters & KCL?

@meBigGuy , @Baluncore My apologize for posting a terrible question, I do that when confused about a certain point but let me clarify with an example:
Let's say that I'm using a DC boost-converter to control the speed of a motor like so:

Naturally, as the motor speeds up to higher RPM's there is an induced EMF that reduces the supplied current to the motor(due to the decrease in applied voltage). How can the back-EMF reduce the applied voltage (##V_o##) when it's parallel to the capacitor? That capacitor is just troubling my analysis... hence my confusion and the effects of series/parallel components.

I know for sure that this formula is true: ##V_o - V_e## = ##V_l## , but how with respect to the capacitor being parallel...

Where ##V_l## is voltage at load.

meBigGuy
Gold Member
First we need to make an assumption. We need to assume that Vo is a fixed voltage due to regulation in the converter. Otherwise the question is too open ended to answer. (Vo becomes a function of clock frequency, inductor size, load resistance, motor speed, capacitor size, and anything else you have conveniently neglected to mention yet again.)

So, given that, the motor back emf does not does not reduce the voltage at the Vo node. What does change is the voltage across the load resistor, which I assume represents the motor resistance.

If you try to solve this with anything other than fixed Vo, you need to meticulously specify every circuit component and signal. Note that in a real/practical converter, Vo is fixed (or at least controlled to specific voltages)

PhiowPhi
What does change is the voltage across the load resistor, which I assume represents the motor resistance.
Why only across the resistor? Not also at the node Vo?
Wouldn't that output be "reduced" due to the back emf?

meBigGuy
Gold Member
First off, Vo can't change because we are assuming fixed Vo. It is a constant voltage supply.
Anything other than that, and I will not address it since it is extremely unrealistic.

Let's just talk about the motor model.

Assume Vo is 10V. Assume the motor is spinning very fast (little load) so the back emf is 9V. That means there is 1V across the resistor, which causes a relatively low current. The power consumed in the load represents the motor output power. It is lightly loaded and spinning fast.

Now, assume we load the motor. It now requires more power, so it slows down. This causes the back-emf to drop, say to 5V. Now, we have 5V across the load, which represents 25 times more motor output power. (P = (E^2)/R).

I'll leave it to you to figure out what would happen if you changed Vo to a fixed 20V given the same motor load that previously required 1V.

Again, for now always assume a fixed Vo. Letting Vo vary based on the load in not realistic. The converter is a constant voltage supply. You can assume scenarios where is is V1, or another where it is V2, but never assume it is load dependent. If you want to consider wire resistance, then add a resistor after the Vo node.

PhiowPhi