# Questions on selecting components for buck circuit

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#### thepaan

Greetings all,

I'm planning on building a circuit to convert ~48V DC @ 350mA to 5V DC @ 700mA, but I have so many questions even after burying my face in the interwebs for several days. One of my primary goals with this circuit is to make it with very low-cost parts. I was hoping for about $3-5, so please keep that in mind when considering the choices I've made. I have attached a schematic for the circuit here. View attachment raspipoe.pdf And this is the parts list since, apparently, save-to-pdf doesn't save hyperlinks. Regulator TL783CKCSE3 C1 HMK316B7105KL-T C2 CC0805ZKY5V6BB106 D1 BAV23S-7-F Q1 IRFML8244TRPBF R1 MFR-25FBF-24K9 R2 CFM14JT51K0 R3 CFM14JT510K R4 CFM14JT82R0 R5 RNF14FTD249R Z1 DDZ33BSF-7 So, let me ask my questions. 1. The section labeled under-voltage lockout was borrowed from another circuit I found online. Z1 was originally 28V but I swapped it for a cheaper part, R3 was 200k-ish but I read somewhere it should be ~10x R2, and I removed Z2 (zener 15V) completely. What function was Z2 serving, is it OK that I took it out, and do the values for the parts I chose look OK? 2. I have no idea what I'm doing in regards to choosing Q1. I mean, I read a bunch of pages explaining what all the values are, but none of them really said if I can be higher or lower than the rated value. The suggested Q1 from the reference circuit was IRF530NSTRLPBF, but it is a bit more expensive than I'd like. I don't need 100V Drain to source, and I only need 700mA, so I figured the one I picked should work... but I'd really like to be sure before I start buying parts. How can I determine if Q1 is OK? 3. The voltage regulator states that the output current is 700mA and has instructions on controlling the output voltage via resistors R4 and R5. Do I need to add anything to make sure the current doesn't drop below 700mA? 4. Capacitors C1 and C2 are the values in the reference diagram in the voltage regulator datasheet. Are these sufficient? How can I find out? I think thats all the big ones. I'd appreciate any feedback. Thanks. #### BradtheRad ##### Super Moderator Staff member Re: Buck circuit questions. At 700mA, to drop 48V to 5V will require a device to dissipate upwards of 30 W of heat. Picking up on the title of your thread, it might pay you to construct a buck converter, meaning a switched-coil converter. Your schematic is a simple resistive drop. It can be made to work, however it would be much more efficient if you start with a lower incoming voltage, in the realm of 7 to 10 VAC. About your component questions: 1) The values for the under-voltage lockout are not critical. Z2 is to safeguard the gate from high V spikes or static, etc. 2) IRF530 is a popular low-cost mosfet. Have you looked at mail-order electronics houses, to compare prices? 3) The regulator IC is the device that will likely overheat. It will need massive heatsinking to dissipate tens of watts of heat. As to output volt level, you may need to adjust of R4 & R5, to get the output you want on your load. A potentiometer (say 5k) will make this part easy. 4) The schematic shows small values for your capacitors. To filter out ripple voltage, it is typical to install a smoothing capacitor after the rectifying stage. You can start with 1000 uF and see it it suits your needs. Increase as necessary. thepaan ### thepaan points: 2 Helpful Answer Positive Rating V points: 2 Helpful Answer Positive Rating #### godfreyl ##### Advanced Member level 5 Re: Buck circuit questions. I see a couple of problems here. Most importantly, you said you want: ...a circuit to convert ~48V DC @ 350mA to 5V DC @ 700mA... but you're using a linear regulator. With a linear regulator you can't get more current out than you put in. If you want the output current to be higher than the input current, you need to use a switch mode regulator. If you do use a linear regulator, the efficiency will be bad. For example if the output is 5V at 700mA and the input is 48V, then the useful power delivered to the load = 5V * 700mA = 3.5W. However another 30W will be wasted as heat dissipated by the regulator. That can be done but you'll need a fairly large heatsink. Do I need to add anything to make sure the current doesn't drop below 700mA? I'm not sure what you meant here. The power supply should deliver a constant voltage to the load, but it's up to the load how much current it wants to draw. What is the purpose of the two bridge rectifiers? Is that just to stupid-proof it so a 48V battery can be connected either way around? Even then, why two? thepaan ### thepaan points: 2 Helpful Answer Positive Rating #### BradtheRad ##### Super Moderator Staff member Re: Buck circuit questions. What is the purpose of the two bridge rectifiers? I got the idea there are two parallel AC supplies, contributing 350 mA each. That yields 700 mA for the load. #### sky_123 ##### Advanced Member level 4 Re: Buck circuit questions. For the switched mode route, how about TPS40200? It is quite a nice IC, ideal for high input voltages. It is about$1.30 in qty, and then you need a
MOSFET less than $1 in qty, and an inductor at less than$1 too, and schottky diode $0.50. That is a total of$3.80, leaving $1.20 (admittedly not a lot) for capacitors and sense resistor, keeping it to$5 total. You can use SwitcherPro to design it, and it draws the PCB outline guideline too.

thepaan

### thepaan

points: 2

#### thepaan

##### Newbie level 6
Re: Buck circuit questions.

Thanks for the replies. I'm going to look into the suggestions you guys gave here and put up my changes later.

I didn't realize the difference between a linear regulator and a switched one - I thought it was all the same.

godfreyl,
The power is supplied by a compliant PoE switch. It is nominally 48V, but can be as low as 44 or as high as 57 (after detection) and no more than 350mA. The rectifiers are needed because power could come in on any pair with any polarity. I included them because I read somewhere that the reverse breakdown voltage should be dobule the constant reverse voltage in the circuit (57*2 for me), but when I read the datasheet for those diodes that notion seemed a bit off.

sky_123,
I don't think I'm ever going to make more than 10 or 20 of these, so I think bulk pricing is out of the question. I looked up this part and noticed it is listed under voltage regulator switching controllers. What is the difference between that and a voltage regulator switching regulator?

Again, thanks for all the feedback, you guys are great.

#### sky_123

Re: Buck circuit questions.

I think the main difference is that the controller needs an external MOSFET. If it is not in bulk, it will probably cost twice that (about $10) if you go the TPS40200 route. I think it is hard to build a switch mode converter for less than that in low quantities. Alternatively, you could try a ready made module if it is really cost-sensitive, e.g. this one (I don't know if it is any good, I've never tried it) which is less than £2. It says it can handle up to 50V input. Or this one with adjustable output and up to 60V input, a bit more expensive but still within the$5 budget.

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#### thepaan

##### Newbie level 6
Re: Buck circuit questions.

This is what I have now.

I changed the regulator to TL2575HV-05 which is a switching regulator having about 70% efficiency. I looked at the TPS40200, but it only accepts 52V input where my power source could be as high as 57V. To find the value of the inductor, I used Power Stage Designer. I tried to simulate the circuit in switcher Pro, but the program doesn't support the part. I calculated that my application would need to dissipate around 4W of power as heat, which is very doable from what I can tell.

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##### Super Moderator
Staff member
Re: Buck circuit questions.

Unfortunately your Attachment 78540 brings up an 'invalid attachment' message.

There is more than one way to drop a few volts from incoming 57V. Or limit it to 52V. Etc.

- - - Updated - - -

If you are considering a buck converter then that is probably the most efficient way to go.

#### thepaan

##### Newbie level 6
Re: Buck circuit questions.

Ok, the link should be fixed now. I'm not sure why the inline attachment isn't working now, but it worked in the first post. Also, yeah; I mistakenly thought the initial voltage regulator was a buck. The TL2575 is definately a buck though. One idea I was toying with in an effort to reduce the cost, was a transformer that would allow me to use a cheaper regulator, but transformers are apparently expensive. Another idea was to use a voltage divider, but everything I read said it's a bad idea to use that configuration in a power supply.

One thing I'm still not clear on is how the current is increased. Is that a function of the inductor?

##### Super Moderator
Staff member
Re: Buck circuit questions.

Another idea was to use a voltage divider, but everything I read said it's a bad idea to use that configuration in a power supply.
A voltage divider would be equivalent to a transistor or regulator dropping 43-odd volts. A lot of heat generated.

One thing I'm still not clear on is how the current is increased. Is that a function of the inductor?
Strangely, the coil's purpose is to prevent too much current from flowing during the switch-On cycle.

The coil itself does not increase current, because as an inductor it tries to maintain current at the same level that was going through it at the moment switch-Off occurs.

During switch-On, the coil takes electrical current and builds a magnetic field. At switch-Off, the field collapses and the coil generates current for a while, helping maintain voltage on the load and load capacitor.

You set the switching device to control the 'On' time, so the net current flow will result in 5V on the load and load capacitor.

- - - Updated - - -

And now it suddenly becomes obvious...

If the power-over-ethernet can only provide 350 mA, then that is like putting a 120 ohm resistance inline.

There is no duty cycle possible, that will maintain 700 mA to your load.

I'm afraid we have miscalculated. You need a different topology.

thepaan

### thepaan

points: 2

##### Super Moderator
Staff member
Re: Buck circuit questions.

Just to see if we can find a working solution, here is a transformer-based converter.

The link below is to an animated simulation that will run on your computer.

Click it to access the falstad.com website.

Click Allow to permit the connection and to run the Java applet.

http://tinyurl.com/8w7drxz

It's only an experiment. Resistors are added to limit current flow to realistic amounts.

Hover over components to examine their specs.

The transformer will need to meet particular specs in order for this system to work.

thepaan

### thepaan

points: 2

#### thepaan

##### Newbie level 6
Re: Buck circuit questions.

That program is great. That topology looks like a flyback, but what is the reasoning for the rectifier? I reconfigured it to this and the output seems very similar. One thing I read yesterday is this pdf. On page 15 it says you can add an output to a buck through a transformer, but I only need one output so I was wondering how to hook it up without the buck output capacitor. It seems to me that the transformer becomes the load and it is driven directly from the regulator, but when I configure that topology through the java circuit simulator, I get only a very small output on the load resistor. I was previously looking at the TL2575HV-05 voltage regulator, so I made a new circuit diagram to show how I thought it could work with the coil before the regulator (as my flyback circuit simulation above), but I'm not sure if its possible for the regulator to operate this way. Finally, the TL2575HV-05 switching frequency is 52kHz, but the simulator doesn't allow anything over 25. When I set the frequency to 25kHz I was unable to find a combination of parameters for the transformer that would produce a reasonable output. It seems to me then, that the transformer choice has a huge impact on the output. How can i find out how to choose the right transformer?

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#### merry1shah

##### Newbie level 1
Re: Buck circuit questions.

Most common dc-dc power application circuits are designed for a positive buck or a boost circuit. However, certain designs may require a negative buck or boost topology. The problem usually arises because most board designers aren't accustomed to thinking in terms of negative voltages.

##### Super Moderator
Staff member
Re: Buck circuit questions.

That topology looks like a flyback, but what is the reasoning for the rectifier? I reconfigured it to this ['URL'] and the output seems very similar.
Yes, come to think of it, I did build a flyback.

I used a full-wave bridge because I saw unexpected behavior when pulsing the transformer. At first I tried a single diode, but I could not obtain sufficient current to the load.

Your single diode is oriented to allow current flow in the secondary when the primary is off. In my recent experiments I saw this happen onscreen, but I did not believe it could happen for real... so I tried different arrangements trying to get simultaneous current flow in the primary and secondary.

This is a gap in my knowledge.

To find out what works best, you'll need to try both hookup polarities on the secondary, both diode orientations, etc.

>>>> Also >>>> notice you cannot leave the secondary unconnected. This would leave the primary to act like a single coil, liable to generate high-voltage spikes at the end of each pulse. These could damage components in your project, as well as at a distance down the ethernet cable. Normally we hear we should put a snubber diode across a coil to absorb these surges. However a simulation shows that a snubber diode will eat current, and deprive the secondary in the process. This needs more research.

You may get away with no snubber diode if you keep a load permanently hooked up to the secondary. Or you might use a small value capacitor instead.

One thing I read yesterday is this pdf. On page 15 it says you can add an output to a buck through a transformer, but I only need one output so I was wondering how to hook it up without the buck output capacitor.
Yes, good article. It brings out some of the magic that is possible with these switch-mode converters. It seems you can also install the buck converter too, perhaps to power another device.

Or if not, then since you are using only the flyback portion, you probably can omit the buck capacitor.

It seems to me that the transformer becomes the load and it is driven directly from the regulator, but when I configure that topology through the java circuit simulator, I get only a very small output on the load resistor.
I have not used switching regulator IC's, so I can't be sure what arrangement will work.

I was previously looking at the TL2575HV-05 voltage regulator, so I made a new circuit diagram to show how I thought it could work with the coil before the regulator (as my flyback circuit simulation above), but I'm not sure if its possible for the regulator to operate this way.
You have picked up a lot in one day. Now another simulator may help, one that contains specs which model such a regulator.

Finally, the TL2575HV-05 switching frequency is 52kHz, but the simulator doesn't allow anything over 25. When I set the frequency to 25kHz I was unable to find a combination of parameters for the transformer that would produce a reasonable output.
I have never seen Falstad's allow more than 25 kHz. It appears there is an internal limit to how short a time-increment can be, per each interation.

This makes it difficult to simulate coils which have very small henry values, as are typically used in high-speed switched-coil converters.

It seems to me then, that the transformer choice has a huge impact on the output. How can i find out how to choose the right transformer?
This will be of prime concern. It needs a certain ratio of primary-to-secondary, although you can adapt the duty cycle to suit.

Its wires must be thick enough to handle the expected current.

There are formulas to aid the selection.

A poster named Goldsmith has a thread about it, on inductor design methodology. A search on 'methodology' might turn it up.

thepaan

### thepaan

points: 2

##### Super Moderator
Staff member
Re: Buck circuit questions.

'Inductor and transformer design methodology!'

Rather than wind your own, you'll probably find it easier to go with a store-bought transformer.

The primary acts as an inductor during switch-On.
Your switching device datasheet may suggest a value. Say 50 uH or 100 uH.

You are operating at about 50 percent duty cycle (give or take although it might become 60 or 70).

The primary will have current flow resembling a triangle wave ramping up and down, between zero and 350 mA max. Average 150 mA or so.

You want 5 x the average current. This is feasible because you are stepping down 43V at 150 mA, to 5V at 700 mA.
The turns ratio is flexible. 5 is reasonable.

If you use regulation, the switching device will alter frequency and duty cycle, to yield your desired output.

#### thepaan

##### Newbie level 6
Re: Buck circuit questions.

I really was toying with the idea of winding my own transformer but, after reading a bit about it, it seems a more monumental undertaking than I'd like. Interestingly, most transformer manufacurers make ones specifically for my application, which I didn't know because very few of them (and none of the inexpensive ones) are listed on Digikey. Last night I think I saw one manufactured by Wurth Electronics on Farnell (which I think is cheap because both Wurth and Farnell are out of Europe?) for ~$4, but I think my best bet will be to get one directly from Coilcraft for which they quoted me under$3 each. Cooper Bussman also had some good ones listed, but the cheapest one I could find for purchase was ~$15. When looking at the spec sheets, most of the ones designed for 5V output had a primary inductance of 120-200uH, winding ratios of 1:0.125 to 1:0.25, and were meant to be operated at 250kHz. I think the 52kHz of the IC I was looking at is too far from the target to use it with a premade transformer. So, I started looking at ICs and saw one from Maxim that was under$2 (or under \$1 if you buy 100!) that had a built-in under-voltage lockout and can be run at 250kHz. I think if I eliminate the UVLO segment of my current topology and use the cheaper IC, then I can incorporate the transormer without driving up the cost. Also, I won't have to do anything weird with a regulator designed for a buck.

#### thepaan

##### Newbie level 6
Re: Buck circuit questions.

Ok!
I decided on the SG3524N which seems a bit old, but it's really cheap. The first major question is: Can I power the controller from the bias winding of the transformer - does it present a chicken-and-the-egg problem? Assuming I can, I went ahead and drew out the schematic and found what I think are appropriate parts.

The first part, according to the data sheet, is calculating the oscillator frequency controlled by the RT and CT pins. The equation the data sheet gave yielded .052kohms for RT, but that was below the minimum also listed in the data sheet. Using the equation for minimum and maximum recommended values, it appears they are off by two decimal places which means I think I really need 5.2kohm for RT while using the maximum value for CT to get 250kHz. I'm using a 5.1kohm resistor instead though for cost reasons. Calculating it back out gives me 255kHz for my oscillator.

The second part was calculating the resistor values to compare to the 5V reference. Since my output will be 5V, I just used the same value all around. The data sheet said 5kohm, but I opted for a cheaper 5.1kohm. I thought about using 51kohm resistors because of this write-up, but they converge at 1A, so I don't expect it to make much difference.

From there I just mimicked the configurations from the data sheet, but I still had some leftover questions.
In the data sheet it says the two transistors are alternately turned on, but the flyback design has them connected in parallel. Wouldn't this cause the circuit to be perpetually on instead of switching?
I'm not really worried about low frequency noise. Do I really need to connect the COMP pin?
I'd like to use the SHUTDOWN pin as an under-voltage lockout, but it shuts down when there is voltage, and not when there isn't. Is there an easy way to replace my Z1, R2, R3, and Q1 with something that works with the SHUTDOWN pin?

Does this design look like it will work?

Edit: One last thing, I was thinking I could use some tweezers and dip some SMT capacitors in some already-melted solder then attach a wire to them, but I'm afraid that may be unrealistic. How likely am I to be successful with that idea?

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##### Super Moderator
Staff member
Re: Buck circuit questions.

I decided on the SG3524N which seems a bit old, but it's really cheap.
Looking at the datasheet, it takes up to 40V max supply. The PoE is higher than that. Up to 57V as you state. So you would need to drop/regulate the incoming PoE supply.

If 57V is the peak expected PoE, you must drop about 20V. The datasheet mentions a reference output current of 50mA. So if the IC uses 50 mA, then you must dissipate 1W.

This complicates things. It will help if you can reduce current draw. (Perhaps you can avoid drawing 50 mA by using the default setting of 5V).

The first major question is: Can I power the controller from the bias winding of the transformer - does it present a chicken-and-the-egg problem?
I don't believe we can expect the output stage to supply the same device which controls the output stage, no.

Only by fortunate happenstance could the mosfet turn on at the moment of power-up, letting current flow in the transformer, and thus send power to the IC, so it can continue switching the mosfet.

The first part, according to the data sheet, is calculating the oscillator frequency controlled by the RT and CT pins. The equation the data sheet gave yielded .052kohms for RT, but that was below the minimum also listed in the data sheet. Using the equation for minimum and maximum recommended values, it appears they are off by two decimal places which means I think I really need 5.2kohm for RT while using the maximum value for CT to get 250kHz. I'm using a 5.1kohm resistor instead though for cost reasons. Calculating it back out gives me 255kHz for my oscillator.

The second part was calculating the resistor values to compare to the 5V reference. Since my output will be 5V, I just used the same value all around. The data sheet said 5kohm, but I opted for a cheaper 5.1kohm. I thought about using 51kohm resistors because of this write-up, but they converge at 1A, so I don't expect it to make much difference.
Yes, you want enough current in the resistors but not too much. These are details which can be worked out in the troubleshooting phase of development.

There is only so much you can figure out prior to buying parts.

From there I just mimicked the configurations from the data sheet, but I still had some leftover questions.
In the data sheet it says the two transistors are alternately turned on, but the flyback design has them connected in parallel. Wouldn't this cause the circuit to be perpetually on instead of switching?
I believe you'll only need one transistor/mosfet, as long as it can conduct enough current.

A possible reason for showing 2 transistors in parallel, is to illustrate a way to get more current through the coil, in case a single transistor would have too much On-resistance.

I'm not really worried about low frequency noise. Do I really need to connect the COMP pin?
I'd like to use the SHUTDOWN pin as an under-voltage lockout, but it shuts down when there is voltage, and not when there isn't. Is there an easy way to replace my Z1, R2, R3, and Q1 with something that works with the SHUTDOWN pin?
Sorry, don't know. It's things like this that will make it necessary for you to experiment and discover how the IC functions.

If it were me, I'd hook up a square-wave oscillator directly to the mosfet. I'd play with the frequency and duty cycle, and watch how it affects performance.

However since you say you might make 10 or 20, it could be worth whatever effort you put in, to create a project that uses the IC, since it has safeguards built in.

Does this design look like it will work?
You're getting there. This is rapid progress. Fortunately we realized a buck converter would not do the job before pursuing it and thus we save time and effort.

You have R4 (at the top of the circuit) alone going to the output stage. It will need a return wire from the negative side to the IC. I don't know for sure if it will work to connect the return wire to IC ground. There may be another method of obtaining feedback.

Several resistors are named R4.

You show a grasp of the principles of operation. You already recognize a few spots where trouble might pop up.

Any new project has its quirks. You will be able to handle them.

Edit: One last thing, I was thinking I could use some tweezers and dip some SMT capacitors in some already-melted solder then attach a wire to them, but I'm afraid that may be unrealistic. How likely am I to be successful with that idea?
The wire must be long enough so that when you solder the loose end of the wire, you do not melt solder at the component end as well.

I have not worked with surface mount components, but I understand they can only take a few reheats and then their operating characteristics are never the same.

As you develop this project you can expect to do some soldering and de-soldering. Better to use parts that are easy to work with. The kind with leads.

You'll gravitate toward a suitable layout. Add a component here or there. Etc.

Then if you have to miniaturize, you can move to SMD, and you won't have to move them around a lot.

thepaan

### thepaan

points: 2

#### thepaan

##### Newbie level 6
Re: Buck circuit questions.

This is great stuff - thanks a lot.

About the R4 (now R5) at the top of the schematic: This is for the voltage comparator. As far as I understand it, the controller attempts to make pins 1 and 2 the same voltage. Since pin 2 is using the 5V reference and equal resistors, it should be about 2.5V. Using the output voltage, R5 and R6 should put pin 1 at 2.5V.

I spent some time this morning pondering how to power this IC (as opposed to looking for another one) when suddenly I remembered reading somewhere that you can build a voltage divider instead of using the 19-26kohm PoE sensing resistor. I've attached my new schematic to show how I thought that would work. I also renumbered the components. One part I am worried about with this configuration is that the controller's transistors max current rating is only 100mA. But later in the datasheet it says "Each transistor has antisaturation circuitry that limits the current through that transistor to a maximum of 100 mA for fast response." So, I'm not sure if I'm allowed to exceed that or not. I used the voltage divider calculator at hyperphysics to determine the power available at the IC and divided by the voltage to get the current which seems to be at acceptable levels.

I guess I'll stick to components with leads for now.

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