I have a circuit that runs 5V. I need to integrate a CO2 sensor that requires 6V. What would be the simplest, most cost effective way to do this? I need about 200ma for the sensor
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2\$\begingroup\$ Stock DC-DC boost regulator. 200mA wouldn't need very large discretes. \$\endgroup\$– Ignacio Vazquez-AbramsCommented Aug 30, 2012 at 3:55
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3 Answers
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boardbite's TPS61040 is just one of many step-up converters which will do the job. But it costs 1.88 dollar in 1s at Digikey, while the cheapest I found is only 70 cent. That's the Semtech SC4503.
This is the typical application schematic from the datasheet. Again, this will be similar for many other step-up switchers. The SC4503's reference voltage is typically 1.25 V, so for 6 V out you have to set R1 to 190 kΩ. At 12 V out and 200 mA the SC4503 will have a 90 % efficiency, for 6 V you can expect a slightly higher value. It can supply more than 1 A of output current, so you'll have lots of headroom. Comes in a SOT23-5 package, and the 1.3 MHz switching frequency means you only need a small inductor.
answered Aug 30, 2012 at 11:15
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\$\begingroup\$ Nearly a third the cost, nice! May I ask which search/catalog you generally use for finding parts? Or is this Semtech part just from experience \$\endgroup\$ Commented Aug 30, 2012 at 12:11
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1\$\begingroup\$ For me always Digikey. I haven't purchased much from them (I use them for information!), but their on-line selection tools are excellent; I use them all the time. Direct links to manufacturer's site for datasheets on all parts as well. Just throw in the right parameters, and sort by price. IIRC I have used Semtech parts in the past, but the advantage of a site like Digikey ensures availability (they have 10 000 in stock), however exotic the brand may be. \$\endgroup\$– stevenvhCommented Aug 30, 2012 at 12:23
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The TPS61040 (from TI), among other adjustable-output step-up regulators, can accommodate 200 mA or more current draw at output voltage of 6V, from input voltage of 5V.
It is available in SOT23-5 package as well as 6-SON (similar to QFN), so you can start with a larger prototyping-friendly footprint, then move on to a smaller one for a final product.
Also, it's a fairly popular and inexpensive part, so you should have no trouble obtaining it. I think it costs a $1-2 in single-unit quantities.
I have used it before (for an LCD backlight application), and had good results with little layout effort (minimal number of peripheral parts, including a 10uH inductor).
Here is its datasheet.
If you can't get that one, frankly there are dozens of options out there; just look under stepup/boost regulators in Digikey or Mouser. I just find that particular one to be convenient and reliable.
answered Aug 30, 2012 at 4:47
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Check the acceptable voltage range for all your components. There's a good chance you can run both all of your logic AND the CO2 sensor at 5.5V, so you'd just need to adjust your primary supply.
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\$\begingroup\$ If a logic IC's datasheet mentions 5.5 V as maximum for Recommended Operation Conditions, that's already 10 % margin included on a nominal 5.0 V. If you operate at 5.5 V and have a 10 % tolerance on that you may meet/exceed Absolute Maximum Ratings. \$\endgroup\$– stevenvhCommented Aug 30, 2012 at 18:47
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\$\begingroup\$ @stevenvh Sure, but plenty of 5V devices list 4.5-6V as the allowable range, and plenty of 6V devices allow 5-7.5V, so 5.5V makes everyone happy even with 10% tolerance. \$\endgroup\$– SparrCommented Aug 30, 2012 at 18:56
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\$\begingroup\$ In my experience that 6 V is often Absolute Maximum Ratings. Many users think that's maximum for normal operation, but it isn't. You should stay away from AMR. \$\endgroup\$– stevenvhCommented Aug 30, 2012 at 18:59
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2\$\begingroup\$ also CO2 sensor requires 6v +/- .1v \$\endgroup\$ Commented Sep 2, 2012 at 17:09