# 34063A Step-Up converter - I need help

Discussion in 'The Projects Forum' started by ke5nnt, Nov 22, 2009.

1. ### ke5nnt Thread Starter Active Member

Mar 1, 2009
384
15
I am working on a project using a 34063A step up converter to output ~18V (17.82 based on the math) from a "12V" vehicle power supply. Note the quotes around 12V... Data sheet for device is HERE

I have some concerns, I've gone over and over the data sheet and there are some things that confuse me, so I have some questions. I'd be very grateful if I got some input.

The data sheet says "Output switch current to 1.5A". If you look at page 5 at the test conditions there, all conditions are tested with 175mA. Does the output current allowed vary based on what your application is? I need right close to 240mA for my load.

Looking at R1 and R2 values for my desired output of near 18V, R1 = 2.49K and R2 = 33K. Are 1/4W resistors ok here? I know the P = I*E formula but I'm not sure how to determine what kind of voltages and currents these resistors would see.

The last problem I'm having in figuring out values & numbers is power dissipation. The data sheet says "maximum package power dissipation limits must be observed"... which I knew already, been down that road before. With an output voltage of 18V and device current needs at .240A, P=18*.240 = 4.32W? or am I using the wrong values to figure that one out (I have to assume I am). BTW, since max power dissipation values are different from package to package, my project is designed with the SOIC pack in mind.

I've come a long way in 6 months of getting into electronics, but not far enough yet apparently. Please help me!

All the best,
Ryan

2. ### SgtWookie Expert

Jul 17, 2007
22,182
1,728

Just to make sure we're on the same page, the datasheet I'm referring to (ONsemi's version) is attached.

File size:
49 KB
Views:
270
File size:
1.4 MB
Views:
68
3. ### SgtWookie Expert

Jul 17, 2007
22,182
1,728
From that same datasheet, on page 11, Figure 17, the formula for the output voltage is:

Vout = 1.25 x (1+ R2/R1)
The existing values are R2=47k, R1=2.2k
So, Vout = 1.25(1+47/2.2) = 1.25(1+ 21.363) = 1.25(22.363) = 27.954.

Just a suggestion, but I think it would be better to increase R1 than decrease R2.

Let's see what happens if we change R1 to 3.5k (roughly 2.2k x 28/18)
Vout = 1.25(1+47/3.5) = Vout = 1.25(1+ 13.429) = 1.25 x 14.429 = 18.04v.

4. ### SgtWookie Expert

Jul 17, 2007
22,182
1,728
Now let's have a look at what you're really asking about, which is related to Rsc

Rsc = 0.3/Ipk(switch)
where:
Ipk(switch) = 2Iout(max)(Ton/Toff+1)
So, if you want up to 240mA out, then:
Ipk(switch) = 2 x .24A x (Ton/Toff+1)
Ipk(switch) = .48A x (Ton/Toff+1)
Now we need to figure out what Ton/Toff will be...
Ton/Toff = (Vout + Vf - Vin(min))/(Vin(min) - Vsat) (are we having fun yet?)
Ton/Toff = (18v + .6v - 12v)/(12v - .6v) (just using 12v for Vin(min)) [eta] Re-calculated using Vf=0.6 from here down
Ton/Toff = 6.6/11.4
Ton/Toff = 0.579 (rounded off)
Going back to Ipk(switch)...
Ipk(switch) = .48A x (Ton/Toff+1)
Ipk(switch) = .48A x (0.579+1)
Ipk(switch) = .48A x 1.579
Ipk(switch) = 0.758 - Hooray, you're under the 1.5A limit!
Back to Rsc:
Rsc = 0.3/Ipk(switch)
Rsc = 0.3/0.758
Rsc = 0.396 Ohms.

Apparently, since you are using a lower output voltage, your Rsc can be increased over what's shown in the example application schematic for 175mA @ 28V.

In the schematic, they're using an 0.22 Ohm resistor to limit the peak switching current.
But it's more complicated than just changing Rsc; the value of L1 comes into play, too.
L(min) = ((Vin(min) - Vsat)/Ipk(switch)) x Ton(max)
L(min) = ((12v - Vsat?) / 0.74112) x Ton(max) .... so now we need to figure out what Vsat is and what Ton(max) will be...
Looks like the output switch is connected in Darlington configuration (pins 1 & 8 connected via L1), so we'll look back up in the datasheet on page 3 under Output Switch, and find out that at 1A, Vsat's typ=1v, maximum is 1.3v. Since we want to make sure the thing is going to work, let's go with the maximum.

So, what's Ton(max)?
Ton = (Ton+Toff)-Toff
Ahhh, nuts! Somebody pulled a fast one.

However, you can pretty much figure that if the peak switching current is actually less than what's required to get the output up to 28V @175mA, and that the 170uH inductor was adequate for the example, it will actually be somewhat overkill for your application. That's OK, we like overkill, right? Why use a 2x4 when a 4x8 will work just fine?

Does that help?

Last edited: Nov 22, 2009
5. ### ke5nnt Thread Starter Active Member

Mar 1, 2009
384
15
Ok, that is no problem on my end. As a follow-up to your answer for this particular question, to help me understand better, what is your reason for suggesting I increase R1 rather than decrease R2?

EDIT: I'm sorry Sarge, you're still posting responses. I'll let you finish.

6. ### SgtWookie Expert

Jul 17, 2007
22,182
1,728
OK, I think I'm done for now.

According to what I calculated (assuming that I've done the calculations properly), it looks like you'll be fine.

I have yet to build something with one of these particular switching regulators.

Note that the MC34063A IC is not rated for operation in an automotive environment!
If you want to use it in a vehicle, you will need to use one that is.
ONsemi's part numbers for similar regulators that ARE rated for automotive environments are:
NCV3063 - Page: http://www.onsemi.com/PowerSolutions/product.do?id=NCV3063
(Holy cow! They have an Excel spreadsheet design tool for the above IC!)

NCV33063AV - Page: http://www.onsemi.com/PowerSolutions/product.do?id=NCV33063AV
[eta]
Ugh - I used 0.34v as the Vf for the 1N5819; I should've used 0.6v to 0.7v. Going back to re-calculate...
[eta]
You're still OK. Rsc could be 0.39 Ohms.

Last edited: Nov 22, 2009
7. ### SgtWookie Expert

Jul 17, 2007
22,182
1,728
Very valid question.

The idea is to keep the current draw on the output as low as possible, in the interests of keeping the efficiency as high as possible.

This may have an effect on circuit stability; the higher the impedance in the feedback circuit, the less stable (and less noise-immune) the entire circuit will be.

If the increasing of R1 to adjust the output voltage proves to make the circuit less stable, then decreasing R2 would be the way to go.

8. ### ke5nnt Thread Starter Active Member

Mar 1, 2009
384
15
Sarge, as usual your insight is beyond simple thanks. I sincerely appreciate your help.
I'm writing those formulas down for future reference and will do the calculations again myself for the hands-on experience (helps me remember). I've done ohms law so many times in the last six months, I swear I'm crunching numbers in my dreams.
I did read in the ON data sheet about the automotive app chips, i assume the value for components will be the same? I'll find one later when I'm not using my phone's web browser and check the datasheet. The phone doesn't like PDFs.
Thank you again.

9. ### ke5nnt Thread Starter Active Member

Mar 1, 2009
384
15
I guess I should ask again just to make sure, because in everything you've shared with me thus far, I still need to be sure that with 18.04V output and .240A, I wont exceed the device power dissipation maximums.

10. ### SgtWookie Expert

Jul 17, 2007
22,182
1,728
Yes, you'll be OK with the power dissipation. The peak switch current will be under 0.76A; which is about 1/2 what the IC is rated for.

I did some playing around with the spreadsheet for the NCP3063; it looks like you should use a larger value of inductance if you want to keep the output ripple down. If you haven't downloaded that Excel spreadsheet tool, you should do so. If you don't have MS Excel, you can download the Excel Spreadsheet Viewer for free from the Microsoft website, or install something like OpenOffice (a freeware MS-Office-like suite) or StarOffice.

670uH looks fairly decent.

However, you could also use an output pi filter like they show in the typical applications. Using two smaller inductors with another cap would help to keep the size down.

11. ### ke5nnt Thread Starter Active Member

Mar 1, 2009
384
15
I'm looking at the data sheet for NCP3063 at page 11 for the typical boost application schematic. It shows Vout at +24V @ 350mA. Of course the component values are all different compared to what we had worked out for the MC34063A, and I notice that R101 is a 1/2W resistor.

I understand increasing the value of R1 vs decreasing R2 for Vout. In the case of the NCP3063:

1.25*(R2/R1+1)
1.25*(18/1+1)
1.25*(19)
Vout = 23.75V

For ~18V
1000*24/18 = 1333.3 = 1.33K
Vout = 1.25*(18/1.33+1) = 18.17V

Now I'm looking at the excel worksheet you linked to. Where I'm getting really confused is ΔIL and IL(avg). It says to select a desired peak-to-peak inductor ripple current, which I gather is ΔIL. It goes on to say For maximum output current it is suggested that ΔIL be chosen to be less than 10% of the average inductor current IL(avg). This will help prevent Ipk (Switch) from reaching the current limit threshold set by RSC. However, I don't need maximum output current.

So... If the design goal is to use a minimum inductance value, let ΔIL = 2(IL(avg)). This will proportionally reduce converter output current capability. I think that's fine, but the data sheet for my inductor (the one I have anyways) only shows values for Irms (max) at 1.2A and Isat (typ) at 1.7A. So I'm not sure where to find the value for IL(avg). That data sheet here. I should mention my inductor is 180uH (because of availability issues with other values near the original choice from the MC34063A of 170uH).

It seems to me that to use the NCP3063 instead of the MC34063A, I pretty much have to change the entire PCB layout and use all different component values. But if I want to use it in a vehicle, it's the best option...

12. ### SgtWookie Expert

Jul 17, 2007
22,182
1,728
Yeah, you need to work out the power dissipation across Rsc.
1/2W sounds like a lot, but remember that's to protect against a dead short/output current too high.

I just used 1.5k and 20k in the spreadsheet; they're standard values, and the output will be 17.92v. Hard to get a lot closer than that without going to E96 values.

What's your tolerance on the ripple current?
With a 180uH inductor, the spreadsheet shows your ripple current will be .16A, or 35%. With your load being .25A, that's not so good; something is not quite right with the spreadsheet.

Basically, what they're trying to say is as the inductance goes down, the more current is required to "charge" the inductor. The inductor will reach saturation current much more quickly. This IC operates at a much higher frequency than the MC34063A; this should allow the use of smaller inductors to achieve the same or higher current output at the same or less ripple.

I don't know what your load is; it may or may not be tolerant to current excursions (ripple).

Il(avg) will be (I believe) the output current.

You could try to use the MC34063A as an experiment, but you have been cautioned that it is not rated for automotive temperature ranges.

13. ### ke5nnt Thread Starter Active Member

Mar 1, 2009
384
15
Ok, I can work that out when I get to the point where I can figure out Rsc.

$RSC = \frac{0.20}{I_{pk (switch)}}$ Ok, that's straightforward enough. My problem is in finding Ipk (switch). $I_{pk (switch)} = I_{L (avg)} + \frac {\Delta I_L}{2}$ At this point, I don't understand ripple current, or how to find it (see below in post).

Ok, that sounds good to me, 17.92V will work well. In the spreadsheet, it asks for worst case input voltage. Since we are talking about a vehicle, should I plan for 11.5V to compensate for when the vehicle is off (assuming a good battery) when the voltage might be a little under 12V, or will 12V be ok?

Ok, this is one point where my brain is having trouble comprehending some things. First, I have no idea what my tolerance on ripple current is, or even what that means to be totally honest with you. So far in my dive into this hobby, I've never encountered this before so I'm new to it. I'm not sure how much less inductance you might think I can get away with, but this inductor looks promising. 150uH. I'm clearly doing something wrong in my spreadsheet, because after entering the information, it tells me that my minimum inductance value is -3228uH.

Ok, I understand that. Thank you for explaining it to me, it helped.

My load is 60 LEDs that have a forward current of .024A each. There are 10 Series each with 5 LEDs. This relates to the same project I've been working on for a while. I have reasons for setting up the series in the way I did (10 x 5) as opposed to using 24 or 28V and doing more LEDs in a series.

Ok, not sure what kind of temperature ranges you mean. If you're referring to something like, if I were mounting this in the engine compartment where it gets hot, or if you mean that the kinds of input loads from an automotive system would cause the component to get hotter.

14. ### SgtWookie Expert

Jul 17, 2007
22,182
1,728
Yeah, that's a toughie.

I'd use 11v for the minimum. If you want the thing to work when you're cranking the engine, you'd better use 9v, or have a big cap on the input with a diode to prevent the starter from drawing current from it.

When you're getting negative values for inductance, something is definitely wrong!

OK, so with a load of LEDs, you'll probably want a fairly low ripple in the output. One easy way to lower the ripple is to add on a pi filter to the output that consists of one inductor and one capacitor. Such output filters are shown in the datasheets.

This really does need further qualification.
Commercial grade temp range is 0°C to 70°C or so.
Automotive grade temp range is -40° to 125°C.
If you use commercial temp range parts in an automotive application, they are very likely to fail. Automotive environments are quite severe.

MC34063A temp range is 0°C to +70°C - this wouldn't be good.
NCP3063 temp range is 0°C to +70°C - this wouldn't be good.
NCP3063B temp range is -40°C to +140°C - this would be fine.
NCV3063 temp range is -40°C to +140°C - this would be fine.

15. ### SgtWookie Expert

Jul 17, 2007
22,182
1,728
Here are some numbers for you:
Rsc = 0.133 Ohms or greater, 1/2W min (increasing Rsc will decrease efficiency.)
Cin = 100uF
Ct = 2.2nF (2200pF)
R2 = 1.5k
R1 = 20k
L1 = 180uH
Cout = 330uF 35v low-ESR type (160mOhm or less)
Output ripple voltage will be < 84mV, or about 0.47% - that's pretty darn good. You shouldn't need an additional pi filter.
Efficiency: 84%

16. ### ke5nnt Thread Starter Active Member

Mar 1, 2009
384
15

Holy cow. You're the man Sarge! I can't thank you enough.

Cheers!

Jul 17, 2007
22,182
1,728
18. ### spacewrench Member

Oct 5, 2009
58
1
Hi, I've been following this thread with interest. I've used the x063 boost controllers in various projects, but usually with minor modifications of the sample circuits, and very little design analysis or testing. (Hey, they work, why bother? )

Anyway, I don't really understand the inductor ripple current very well. I wonder whether it makes sense to relate it to the output current, at least in any direct sort of way.

On the load side (downwind from the output diode, at the top of the output filter cap), any ripple current should be related mainly to any non-constant current demand of the load, and should be accompanied by ripple in the voltage across the output cap. This has little to do with inductor current: as long as the IC and everything on the other side of the diode can supply enough average current to keep the output cap charged up to the output voltage, then the only output ripple current it might cause is when globs of flyback current overcharge the output cap slightly, raising its voltage and causing more current to flow in the load.

On the other side of the diode, though, things are much different. At the beginning of a PWM cycle, the switch turns on and current begins to flow in the inductor. Its magnetic field builds up (hopefully short of saturation). Eventually (when the inductor current reaches the max set by Rsc, or the pulse width that's proportional to the output voltage is reached) the switch is turned off. The magnetic field begins to collapse, and the inductor forces current into the node where the (now open) switch and output diode anode connect. Since there's only a little stray capacitance at this node, its voltage shoots up rapidly. Shortly after the switch opens, the voltage at that node exceeds the diode Vf plus the current output voltage, and current begins to flow through the diode and into the output cap (and output circuit). As the magnetic field continues to collapse, this current decays, and eventually there's not enough to keep the stray capacitance charged to Vout + Vf, so the diode stops conducting. After that, and until the next PWM cycle, the charge on that node probably just hangs around, keeping the potential near Vout + Vf, or leaks through the switch, or rings back to the input source.

Later, another PWM pulse begins, the switch closes, current begins to flow, and the cycle starts again.

I think -- but I'm not sure -- that the inductor ripple is the difference between the high current at the end of a PWM pulse when the switch opens, and the low current at the end of a PWM cycle when as much energy as possible has been transferred to the output and the circuit is just waiting for the next pulse to start.

Incidentally, I think this fits in with continuous / discontinuous mode operation like this: in continuous mode, the switch closes before the inductor has quite finished dumping all its magnetic energy through the diode, while in discontinuous mode, the inductor current drops to 0 before the next pulse starts.

Does this make sense to you guys?

Mar 1, 2009
384
15

20. ### ke5nnt Thread Starter Active Member

Mar 1, 2009
384
15
I was comparing the typical application schematics between the MC34063 and the NCP3063. The NCP3063 shows pin 8 (driver collector) is NOT connected while 34063 does. Also, the application shows 5 total capacitors for the NCP, only 3 for the 34063.

Why do you need a .1uF capacitor in line with polarized electrolytic caps? That's where the 2 extras are coming from.

Why is the driver collector not connected in the NCP version?

One last question, should the 12V input ground be independent from the (in my case) 18V output ground?

Thanks.