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I'm designing my first circuit and I have to use the TEC3-2419 DC-DC converter (datasheet) to convert 28V to 9V.

Until not long ago I used to work as a hardware tester where I tested and troubleshooted circuits that use completely different buck converters, such as this one.

When I looked at the datasheet for these buck converters, I saw there are one-by-one steps on how to properly design the part of the circuit with the buck. For example, how to choose the right inductor value, choose input/output capacitor, etc...

The datasheet of the TEC3-2419 does not provide those kind of instructions and this leads me to confusion about how to properly design this, which is why I have these questions:

  1. Why do I have Vout(+) and Vout(-)? From what I know about buck converters there are multiple switching pins, okay, but why both positive and negative Vout pins?

  2. Do I need an inductor at the output? If so, how do I calculate the value?

  3. I suppose it would be a good idea to connect bypass capacitors at the input and at the output. How many do I need and what should be their values?

  4. Pin #3 is On/off. Does that mean EN pin? If so, should I put a resistor between this pin and the VCC pin (again, something I saw in previous buck converters). What value should the resistor be?

  5. What is the difference between single and dual as noted on page 4 of the datasheet? Or, is there a website you can direct me to? I have Googled it but couldn't find any information about this.

Thank you!

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Why do I have Vout(+) and Vout(-)? From what I know about buck converters there are multiple switching pins, okay, but why both positive and negative Vout pins?

It's an isolating DC to DC converter and therefore has two pins for +Vout and -Vout. Those pins are isolated from other pins. It isn't a buck converter.

Do I need an inductor at the output? If so, how do I calculate the value?

No, because it's an isolating DC to DC converter. It uses an internal flyback transformer for passing magnetic energy across an isolated internal physical gap.

I suppose it would be a good idea to connect bypass capacitors at the input and at the output. How many do I need and what should be their values?

It has inbuilt input and output capacitors. Addding more at the input may reduce conducted emissions to other equipment sharing the same input DC system. The data sheet tells you how much you can add to the output (see page 3 of DS).

Pin #3 is On/off. Does that mean EN pin? If so, should I put a resistor between this pin and the VCC pin (again, something I saw in previous buck converters). What value should the resistor be?

Page 3 of the DS tells you - open circuit means on and off is typically using a 1 kohm pull-down.

What is the difference between single and dual as noted on page 4 of the datasheet? Or, is there a website you can direct me to? I have Googled it but couldn't find any information about this

Some of the models in the product range have dual outputs outputs i.e. a mid-rail common 0 volts and positive and negative output voltages. They are referred to as "dual output" on page 3.

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  • \$\begingroup\$ Thank you very much! About the On/Off pin, if I want it always on, should I just leave it not connected/floating? \$\endgroup\$ – Eran Jun 4 '18 at 14:38
  • \$\begingroup\$ That's what the data sheet says and, from memory, that's what I have done previously. \$\endgroup\$ – Andy aka Jun 4 '18 at 14:39
  • \$\begingroup\$ Also, about my first question, does that mean that Vout(-) would output -9V? If that's the case, I doubt I even need it... \$\endgroup\$ – Eran Jun 4 '18 at 14:40
  • \$\begingroup\$ Another question - about the on/off, how do you know it's a pull-down? it's not specified. Also, the datasheet says that a 2-4 mA current needs to be applied. Do I need to worry about the current or does the converter draw the exact amount if I put a 1KOhm resistor? \$\endgroup\$ – Eran Jun 4 '18 at 14:54
  • \$\begingroup\$ It means the negative terminal of the isolated output pair. Between + and - will be the voltage you require. They are isolated from other connections. \$\endgroup\$ – Andy aka Jun 4 '18 at 14:54
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First, why do you "...have to use the TEC3-2419 DC-DC converter..."? There are lots of DC-DC converters that should be able to do that with far better datasheets.

Second, you should look at the listed EMI considerations datasheet here, it has some example schematics with cap and inductor values that should get you going. If you are confused though, you should contact the manufacturer or at least your sales rep that is sourcing the part to get you some help with designing and to answer any of your future questions.

You should also buy a few of them and play with them on a breadboard to see how well it even works before you put it in a schematic or on a board, it's not a good idea to just slap any IC on your board and pray it works.

  1. Vout(-) is just a reference to what will be the common for the output, it's not unusual to reference it that way.

  2. You shouldn't need an inductor at your output, unless your circuit needs it for some reason.

  3. Bypass on the output will also be dependent on what your circuit requires, but my rule of thumb is either a 0.1 or 10 \$\mu F\$ capacitor to start with.

  4. Probably, but the datasheet isn't clear on how to connect it. You only need a pull-up resistor if you are planning on grounding the enable pin to disable the supply. Otherwise you are fine to connect it directly to VCC (IF it's supposed to connected to VCC for an ON state.

  5. I'm not sure what they mean either, but you should contact them to find out.

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  • \$\begingroup\$ Thanks for your answer. I have to use that exact component because it's a circuit I'm designing at work and I've been asked to use this specific IC. \$\endgroup\$ – Eran Jun 4 '18 at 16:03
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The unit is isolated, as others have informed you. There is insufficient information in the datasheet to design the application circuitry, so you should refer to the relevant application note for appropriate EMI filtering components.

enter image description here

Note the LC \$\pi\$ filter at the input and C3 in particular.

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