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I am using an IC as a step-Up converter in my PCB. It is supposedly to increase voltage from 12V to 30V. This is the chip that i used.

https://datasheet.lcsc.com/lcsc/1809251831_Diodes-Incorporated-AZ34063UMTR-G1_C99659.pdf

The effects that i notice are these two:

  • The chip gets overheated a lot.
  • The output voltage is 12V instead of 30V.

In other words, it just outputs the input voltage.

This is the schematic that i should implement (according to the datasheet - page 9):

enter image description here

And this is my actual implementation:

enter image description here

I should also note that according to the datasheet (page 9), the output voltage is determined by this equation: Vout = 1.25 * (R1 + R2) / R2 In my circuit, these resistors are: datasheet -> mySchematic R1 -> R3 R3 -> R5 and their values are R3 = 23.2K and R5 = 1K.

I should also note that without load, i hooked up a power supply at 12V, and i saw that I had 110mA consumption.

It boggles my mind as to what the problem is, as it should be working but it is not! I should also note that this IC can be used a step-Down (with a slightly different circuit configuration), which I have also incorporated in my PCB and this works fine!

EDIT 1: I should also note that I noticed that C7, the electrolytic decoupling capacitor, is placed far enough in the circuit. Could this be the reason?

EDIT 2: This is the layout of the PCB. I only kept the parts that are relevant to the Step-Up, so it's easier for viewing. What you see in the picture are all relevant components.

enter image description here

EDIT 3: This is the inductor I used: https://www.lcsc.com/product-detail/Inductors-span-style-background-color-ff0-SMD-span_TDK-NLV32T-221J-PF_C89015.html And its datasheet: https://datasheet.lcsc.com/lcsc/2304140030_TDK-NLV32T-221J-PF_C89015.pdf

user1584421
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2 Answers2

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The TDK NLV32T-221J-PF is a ferrite core 220 uH inductor with a 50 mA rated current. The inductor will probably saturate before the current limit mechanism of the 34063 turns the switching transistor off. When the core saturates, current will rise very quickly. If the current rises quickly enough, and the 34063 is not quick enough in responding and shutting off the switching transistor, there could be excessive currents for a brief time. These excessive currents could be responsible for the overheating of the 34063. The core saturation could be responsible for the low output voltage.

Choose an inductor that is sufficiently rated for the maximum current of the 34063 is expected to handle. With a 0.22 \$\Omega\$ resistor between Vcc and Vpksense (pins 6 and 7) Ipk will be 0.3/0.22 = 1.36A.

Alternatively, choose a sense resistor that is sufficiently large that the current will be limited before the inductor reaches its rated current.

To be honest, I never really understood what the SENSE does in the datasheet so i left it like that..... Could you please explain its function?

Briefly, the 34063 has a timing capacitor that determines the maximum \$t_{on}\$ for the switching element (i.e. \$t_{on(max)}\$) and the minimum \$t_{off}\$ for the switching element (i.e. \$t_{off(min)}\$. Note that the inductor may stop conducting before the \$t_{off}\$ period is complete. If the feedback voltage \$V_{fb}\$ is above 1.25 V, the actual \$t_{off}\$ will be greater than \$t_{off(min)}\$. If the instantaneous voltage between the \$V_{cc}\$ pin and the \$I_{pk(sense)}\$ pin is above about 0.3V, then the actual \$t_{on}\$ will be less than \$t_{on(max)}\$. Thus, these three parameters, the capacitance of the timing capacitor, \$V_{fb}\$, and the potential difference between the \$V_{cc}\$ pin and the \$I_{pk(sense)}\$ pin together regulate the duty cycle and frequency of the converter as well as the peak current through the switch.

Further details: The 34063 generates a triangular current wave through the inductor in the following manner:

A timing cycle begins when the voltage on the timing capacitor is less than or equal to 0.75V. When the cycle starts, the chip begins charging the timing capacitor. If \$V_{fb} \le 1.25V\$ during the charging of the timing capacitor, the 34063 turns the switching transistor on. A rising current begins to flow through the switch, the inductor, and the sense resistor. The rising current through the sense resistor causes a rising voltage across it. The \$t_{on}\$ phase of a cycle ends when either the voltage across the timing capacitor reaches 1.25 V, or when the potential difference between the \$V_{cc}\$ pin and the \$I_{pk(sense)}\$ pin reaches approximately 0.3 V. Thus, the timing capacitor determines the maximum \$t_{on}\$ for the switching element, but if the \$I_{pk(sense)}\$ pin reached the voltage \$V_{cc} - 0.3 V\$, or if \$V_{fb} \ge 1.25V\$ at the start of the \$C_T\$ charging phase of a cycle, then the actual \$t_{on}\$ will be less than \$t_{on(max)}\$. It is possible, due to \$V_{fb}\$, that the switch does not turn on at all during a timing capacitor cycle.

If the potential difference between the \$V_{cc}\$ pin and the \$I_{pk(sense)}\$ pin reaches 0.3 V before the voltage of \$C_T\$ reaches 1.25V, then the timing capacitor is quickly charged to 1.25V. In either case, the switching transistor is turned off at this at the end of the timing capacitor charging phase.

At this point, the chip starts to discharge the timing capacitor. With the switching transistor off, the current through the inductor begins to fall. When the timing capacitor discharges to 0.75 V, the chip will stop discharging the timing capacitor, and begin a new timing capacitor charging phase. Thus, the timing capacitor also determines the minimum value of \$t_{off}\$, but if \$V_{fb} \gt 1.25 V\$, the actual value of \$t_{off}\$ will be greater than \$t_{off(min)}\$.

[Note that the 34063 uses different levels of current to charge and discharge the timing capacitor. Thus, \$t_{on(max)} \ne t_{off(min)}\$, but rather \$t_{on(max)} \approx 6 t_{off(min)}\$]

The \$I_{pk(sense)}\$ pin controls the peak current through the switch. This is useful in limiting the peak current through the inductor so that it does not saturate. However, the relationship between the feedback provided by \$I_{pk(sense)}\$ and the average output current (or the average current through the switching transistor or the average current through the inductor) is not so straightforward, and depends upon whether the converter is operating in Continuous Conduction Mode (CCM) or Discontinuous Conduction Mode (DCM), among other things such as the topology of the converter. For that reason, I will not go into detail about the relationship between peak switching current, and average output current. Suffice it to say that they are sufficiently different to require separate analysis.

[A converter is said to be in DCM if the inductor current reaches 0 before the end of the \$t_{off}\$ phase. If there is still residual current flowing in the inductor at the end of the \$t_{off}\$ phase, then the converter is said to be operating in CCM.]

Math Keeps Me Busy
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  • My God... Thank you for noticing this... As i stated, I have another 34063 in the PCB that acts as a Step Down.... This datasheet of the 34063 notes i use a 220uH inductor for the Step Down and an 150uH inductor for the Step Up. Accidentally, I used the same 220uH inductor in both cases.... I should have an 150uH one instead for the Step Up..... – user1584421 Jul 20 '23 at 20:13
  • To be honest, I never really understood what the SENSE does in the datasheet so i left it like that..... Could you please explain its function? – user1584421 Jul 20 '23 at 20:16
  • @user1584421 The 150 uH inductor, if it is an NLV32T-151J-PF still only has a rated current of 65 mA. If you use that inductor, you will need to substantially increase the value of the current sense resistor. – Math Keeps Me Busy Jul 20 '23 at 20:17
  • Will it influence the output voltage of the Step up (should be 30-36V).. Will it influence the max current that it can feed? I need about 0.5A... – user1584421 Jul 20 '23 at 20:20
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    @user1584421 if you need about 0.5A out from the circuit what gave you the idea that an inductor rated for only 50mA would be suitable? To get that much current out from a 12V-30V boost converter you're probably going to need an inductor that can handle about 1.5A! – brhans Jul 20 '23 at 20:25
  • Thank you very much! Do you think the step Up IC is dead? – user1584421 Jul 20 '23 at 20:28
  • I have added an explanation of the use of the current sensing resistor to my answer. Also brhans is correct. In a boost converter, the input current is larger than the output current by the boost ratio (or slightly more). – Math Keeps Me Busy Jul 20 '23 at 20:30
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    @user1584421 The 34036 might be dead. It could also be functional with a decreased lifespan. Or it could be unharmed. To test whether it is still functional, you could replace the sense resistor with a 5 \$\Omega\$ resistor, and power your board up again. But if you want 0.5 A output at triple the voltage, you will ultimately need an inductor rated for more than 1.5 A. BTW, 1.5 A is the max the 34063 can handle. (There are many brands, and I haven't checked your specific datasheet, but 1.5A is pretty standard), so that chip is on the border of not being suitable for your project. – Math Keeps Me Busy Jul 20 '23 at 20:39
  • But if that chip can handle 1.5A, then isn't one third of that (0.5A) a really lower value, this making this chip suitable? Why you say it is not suitable? In trying to understand what Sense does, what will happen if I keep the inductor and place 5Ω resistor? – user1584421 Jul 20 '23 at 20:45
  • If you use the 5 Ohm resistor, the peak current will be limited to 0.3/5 = 60 mA oops, you need, say a 6.8 Ohm resistor. 0.3/6.8 = 44 mA. You should then be able to safely power your board. If you can test the board with a light load, you should see the correct output voltage. If you probe pin 3, you should see an approximately triangular waveform. You should see a triangular waveform of 0.3V amplitude and 12V peak at pin 7. *** If the output is 0.5A, the input needs to be 0.5*30/12 = 1.25 A. That needs to flow through the 34063. – Math Keeps Me Busy Jul 20 '23 at 21:04
  • Since max current of the IC = 1.5A and Rsense = 0.3/1.5 = 0.22Ω, I will keep the value of 0.22Ω. Thus what will be changed is the inductor. Should only the current be the only consideration? I am thinking between the range of [1.5-2] Amps. Is 1.5 too low or 2 too high? And something else, does the DC resistance of the inductor play a part on what should I choose? Should the frequency play a role as well? – user1584421 Jul 21 '23 at 14:07
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    I was explaining how to test whether your IC is dead. – Math Keeps Me Busy Jul 21 '23 at 14:13
  • Should I only care about the Rated Current (2A), or should I select based on Saturation Current as well? – user1584421 Jul 24 '23 at 09:32
  • It would be odd to find an inductor that is rated as 2 A, but which has a saturation current less than 2 A. However, if the datasheet specifies both a "maximum" current, and a different saturation current, you want the lower value to be greater than the current you will actually be working with. (In general, don't choose components that have a current rating exactly equal to the actual current. Leave some margin). – Math Keeps Me Busy Jul 24 '23 at 10:37
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The inductor you have linked is a 220 μH with an internal DC resistance that could be as high as 21 Ω: -

enter image description here

This is a seriously bad component choice for this type of design.

Andy aka
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  • I have another 34063 in the PCB that acts as a Step Down.... This datasheet of the 34063 notes i use a 220uH inductor for the Step Down and an 150uH inductor for the Step Up. Accidentally, I used the same 220uH inductor in both cases.... I should have an 150uH one instead for the Step Up..... – user1584421 Jul 20 '23 at 20:19
  • So you suggest i select an inductor also based on the Max DC Resistance? What is an ideal on for my case? – user1584421 Jul 20 '23 at 20:21
  • You need to choose an inductor that has much lower DC resistance; probably less than 1 ohm. It depends on how much output current you intend to supply of course. – Andy aka Jul 20 '23 at 20:23
  • The max output current is around 0.5A. – user1584421 Jul 20 '23 at 20:28
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    That means your average inductor current will be around 0.5 x 30/12 amps = 1.25 amps and, 1.25 amps through 21 ohm (inductor resistance) is more voltage dropped than your input supply hence, it's a poor design choice. And, that's just the average current; the peak might be 2 amps. Think about it. – Andy aka Jul 20 '23 at 20:35
  • Do you think the Step Up IC could be saved? Or did I kill it? – user1584421 Jul 20 '23 at 20:36
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    I don't think you will have killed it for the reasons above because you can't draw the current into it to to cause significant damage. Having said that, I can't realistically tell you if it has broken for some other reason. – Andy aka Jul 20 '23 at 20:39
  • Since max current of the IC = 1.5A and Rsense = 0.3/1.5 = 0.22Ω, I will keep the value of 0.22Ω. Thus what will be changed is the inductor. Should only the current be the only consideration? I am thinking between the range of [1.5-2] Amps. Is 1.5 too low or 2 too high? And something else, does the DC resistance of the inductor play a part on what should I choose? Should the frequency play a role as well? – user1584421 Jul 21 '23 at 14:07
  • They all play a role but, we're in the area of asking new questions in comments and, this is a media that doesn't support answering multiple question. This is a Q and A site (as you know) and some people would suggest that to answer your comments would be a chapter of a book so, consider asking a new question and do some research on boost converters (another chapter of another book). – Andy aka Jul 21 '23 at 14:17