Electronics > Beginners

Current question in switchmode boost converter

**galpi**:

Hi everyone,

First post from a newbie here ...

I'm making a switchmode boost converter using the MC34063 mentioned in blog #110. Input is 12V, output should be 24V with 350 mA of current. I figured out all the parts I needed using the formulas Dave goes over, and I end up with a peak current of 1.62A with feedback resistors (R1 and R2 in MC34063 specs) having values of 100 KOhm and 1820 KOhm. Dave mentions that you have to be careful when selecting your diode and inductor to make sure they can handle the peak current (here 1.62A), but what about the resistors ? Should I be concerned about peak current going through the resistors ? How do I figure out the necessary power rating required for those two resistors ?

Thanks a lot folks,

Pierre

**TopherTheME**:

There should be very little current going through the resistors so its not an issue (i = V/R).

--- Quote ---having values of 100 KOhm and 1820 KOhm

--- End quote ---

Are you sure this is correct? A value of 1.82M is a pretty big resistance. You may want to cut that down a bit to maybe a 10k and 180ishk resistor.

**galpi**:

--- Quote ---There should be very little current going through the resistors so its not an issue (i = V/R).

--- End quote ---

Right, yet somehow there is mention of a peak current Ipk of 1.62A, and with no load all of it has to go through the resistors for a ridiculous power consumption ... that's what I don't get.

Thanks !

Pierre

**Rhythmtech**:

--- Quote from: galpi on October 09, 2010, 03:41:37 am ---Right, yet somehow there is mention of a peak current Ipk of 1.62A, and with no load all of it has to go through the resistors for a ridiculous power consumption ... that's what I don't get.

Thanks !

Pierre

--- End quote ---

We need to consider the plausibility of where the peak current is actually going. If 1.62A were going through a 100kohm resistor then you would have 162,000 V somewhere in your circuit, which is somewhat impossible. So the peak amps are finding their way to common or ground via another path. Perhaps the peak current is travelling through the load or the output filter cap or is maybe even inrush into the inductor. I would tend to think of it as the largest average current the diode will see over a short period of time until the regulator circuit settles for most cases.

Maybe how transient responses move into steady state would help describe what I am trying to simplify better. This is just my untech way of taking a stab at it without an analytical approach, perhaps somebody else would like to detail out where the peak current comes from actually.

**DJPhil**:

Buckle in for way more mediocre advice than you need, I've had too much coffee!

You get the formula you need by taking the general instantaneous power formula:

P = I * V

and subbing in Ohm's law . . .

V = I * R subbed in for V OR I = V / R subbed in for I

so you get . . .

(P = I * I * R) -> P = I2 * V

OR

(P = V * V / R) -> P = V2 / R

I remember a bit from the very beginning of Art of Electronics. One of the first half dozen exercises is to prove that you can't roast a 1/4 watt resistor over 1k with a 15V supply. If you flip it about, you can figure just about whatever you like.

Bad advice that often works anyway:

Resistors are fairly slow to fail (maybe whole seconds!) compared to other electronic parts (Google for 'silicon fuse') You often only need to worry about average power, especially above audio frequencies, unless you're dealing with something fairly crazy. Carbon composition and carbon film (the cheap ones) tend to behave best under power surge, with the non-wirewound metal resistors tending to fry more quickly. It's all about thermal mass, so higher power resistors (that are naturally larger) have the mass to average out peaks better than tiny 1/8W SMD units. Remember that through hole resistors dissipate most of their heat through their leads.

/end bad advice (use as a rough guide only, some may consider this wreckless [or wrong, if so let me know!])

Now to address the actual problem. The current that's hurling through the inductor and diode is large because their series resistance is low.

Break out ohm's law for this one, just as Topher said:

I = V / R

I = 24V / 100K

I = 240uA (yep, that's microamps)

Let's flip it with the equations above, what's the lowest value resistor you could use at 24V and stay below 200mW (50mW safety margin, woohoo!)?

P = V2 / R ninja flip! both sides times R then both sides divided by P

R = (24V)2 / 0.2W

R = 2880?

Now that's assuming you're shorting out the supply with this resistor, so don't assume you can't have a resistor in the circuit anywhere below that value. Consider R1 in the circuit from the datasheet. It's only 2.2K, but it's got a giant resistor restricting it's max current input.

Whew. Then also, you don't need (also as Topher said) such large resistors. High value resistances are inherently noisy (small problem) and produce tiny currents which are susceptible to interference (much bigger problem). With an inductor nearby throwing an EMI party this could be trouble. You also don't want them too low or they'll be drawing power out of your power supply and dumping it to ground! The datasheet circuit is a good example.

Update cause I'm way gabby and taking too long:

--- Quote from: galpi on October 09, 2010, 03:41:37 am ---Right, yet somehow there is mention of a peak current Ipk of 1.62A, and with no load all of it has to go through the resistors for a ridiculous power consumption ... that's what I don't get.

--- End quote ---

Ah, I see the problem, you've got an incorrect concept of current. Current must flow through something. Ipk is actually Ipk(switch) to express what it's flowing through, and is the maximum current drawn across the switching circuit. What's the switching circuit? The Q1 and Q2 in the datasheet, plus any external transistor used to switch current to the inductor for energy storage. The inductor is always heaving and shoving current around (low internal resistance), and that's where you get your big peaks. The diode has a low internal resistance as well, hence Dave's warning. Co smooths out a lot of the mess, and the optional filter circuit at the end does as well.

The output circuit itself is a voltage source, so it will attempt to deliver a fixed voltage until it can't keep up enough current (falls out of compliance or out of regulation, they say), after which the voltage will drop. In your design it's supposed to be 350mA, so that's the most you'd expect without the voltage dropping. With no load it's still flowing, but consider the resistances. You've got the resistive divider, the leakage of Co and the optional filter cap, flux losses in the inductors, and other such. The feedback network is the highest one there, but work out the current at 24V and you'll see that it's very, very small. The remainder is even smaller still. The balance of the output current doesn't exist, because there's nothing for it to flow through. If you work out the lowest resistance the supply can handle at 350mA (exercise for the reader, muahaha), then you'll actually see all 350mA flowing through that single resistor if you shunt the supply with it.

Good grief, I'm gonna stop now, I need more coffee. Hope that helps. :)

Edit: Included relevant datasheet schematic for reference.

Edit: Included the correct relevant datasheet schematic (sheesh).

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