Average battery current in battery charger

[Antoha1]

Homework Statement

A battery with a 12V EMF is charged using a 5V source $U$. Charging takes place according to the diagram below. The inductance of the coil $L$ used is 1 H, a diode and a switch that is turned on and off for $t$ 0.01s. Determine the average battery current $I_{B}$.

Relevant Equations

$\varepsilon=-L\frac{dI}{dt}$
$U=IR$
$R_{L}=\omega L=2\pi fL$
$f=1/T$

Hi, having difficulties in this problem. Looking for ideas.
For now, I think the combined voltage of source and coil should be bigger or equal to EMF, for the battery to get charged:
$U+L\frac{dI}{dt}\ge EMF$
the current going through the coil, I think, would be:
$I=\frac{U}{R_{L}}=\frac{U}{\omega L}=\frac{U}{2\pi fL}$
for frequency I think it is:
$f=\frac{1}{T}$ and T is 0.01s
How to get to average current flowing through the battery I am not sure yet. Maybe need some explanations first. Thanks.

 

[berkeman]

Have your learned about Boost DC-DC Converters yet in your class? You need to use the equations for the current output for a Boost Converter:

https://en.wikipedia.org/wiki/Boost_converter

 

[berkeman]

BTW, if the switch duty cycle is 50%, you will not be able to boost the 5V source voltage to 12V or above. A 50% duty cycle for an ideal boost converter (no forward diode voltage drop, no losses, etc.) gives a voltage multiplication of 2x...

 

[DaveE]

If you are modeling the components as ideal in the simplest version, I don't think the current is defined since there are no lossy elements. For example, if the duty cycle is set to make 12.1V into a 12V battery, there will be no steady state, each cycle the inductor current will increase with no change in any of the circuit voltages. It will never get to the 12.1V required for steady state. Try it with a resistive load first.

This is never a problem IRL, it's a problem with poorly designed HW questions.

Anyway, as @berkeman said, study a bit online. It's an essential bit of knowledge for analog design. There are tons of great explanations out there, many better than what we will give you.

My favorite is the text book "Fundamentals of Power Electronics" by Erickson & Maksimović

PS: I guess you could put a more realistic diode or transistor model in to stabilize it, but that is more of a semiconductor problem than a SMPS problem.

 

[haruspex]

If you are modeling the components as ideal in the simplest version, I don't think the current is defined since there are no lossy elements.

Is it? Doesn’t it just mean the net voltage is zero, so with the switch closed $L\frac{dI}{dt}=5V$ and with the switch open it is $5V-12V$?
But I'm not sure what "turned on and off for 0.01s" means. Is that the whole cycle time or just one phase (if so, which?)?

 

[DaveE]

Is it?

The average current in the inductor is not defined. The triangular waveform can be calculated for any switch cycle. But for the steady state solution the end points of each cycle's waveform must match the next/previous cycle. This only happens at the duty cycle that exactly transforms the input voltage to the output voltage. This is an unstable equilibrium, any miniscule change in the duty cycle will will cause the current to fall to zero or rise to infinity. At equilibrium, the solution doesn't depend on the DC current value, any value will be maintained.

This is easy to show, but I don't want to do it in the HW forum. Anyway, boost converter analysis is in the literature everywhere.

Doesn’t it just mean the net voltage is zero, so with the switch closed LdIdt=5V and with the switch open it is 5V−12V?

I'm not sure I understand. Those are the instantaneous voltages across the inductors, which determine the current slope. The average, or steady state, voltage across any ideal inductor in any circuit is zero. That is what determines the output voltage, meeting that requirement. In SMPS jargon it's called "volt-second balance".

But he's asking for the current.

But I'm not sure what "turned on and off for 0.01s" means. Is that the whole cycle time or just one phase (if so, which?)?

I stopped paying attention to the details of the switch periods, since as @berkeman pointed out, they don't make sense. That is the least of the issues at hand, IMO.

 

[DaveE]

Oops! I was completely wrong, as was @berkeman. This circuit does work and is solvable.

I was assuming a high current "continuous conduction mode", where the inductor always has current flowing. But this circuit is operating in the "discontinuous conduction mode". It will apply more discharging volt-seconds to the inductor than charging volt-seconds, so the inductor current will ramp up, then ramp down, and reach 0 before the end each cycle. Energy is put in the inductor when the transistor is on, and then completely removed before the transistor turns on again.

So the circuit will have 3 switch states, not two. In order, they are:
1) Transistor ON, Diode OFF -> Inductor current ramping up
2) Transistor OFF, Diode ON -> Inductor current ramping down
3) Transistor OFF, Diode OFF -> Inductor current zero

Very sorry for the misdirection, I didn't read carefully enough.

So you wanted a hint: Start with an inductor current of 0 at the beginning of the cycle and draw the current waveform vs. time.

 

[DaveE]

One easy way to solve these discontinuous conduction SMPS is to calculate the energy stored in each "on" phase, since the energy is depleted in each cycle. For a 100% efficiency assumption, that energy is all put into the load at a know pulse rate.