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I'm designing my first PCB and I'm not sure if the diagram I found on a datasheet is complete. For what I have read, there should be a resistor between R1 and the transistor Tr1. I have followed this answer but I'm having trouble knowing what Vin should be. The circuit is the following: Circuit

It's directly from the HT78XX datasheet. My intentions are to power this circuit with 2 18650 batteries so VIN should be approximately 7.4 V. For current output I'm aiming to get 1 A max. Using the R1 formula below the image I get 1.4 Ω (@IREG = 500 mA, Ic1=500 mA) which is enough to get the 0.7 V needed to activate the transistor. So, if I were to use Kirchhoff's voltage's law: VBC = VIN(HT78XX) - VOUT = 2.4 V aprox.

=> Rb = (2.4-0.7)/0.0024 = 708 Ω

Is this correct?

Also, if anyone has a recommendation for an alternative to this circuit, it is much appreciated. I need to get 5 V, 1 A with a low Iq voltage regulator.

I intend to use the HT7850 voltage regulator and FMMT549 transistor.

Any extra recommendations are welcome.

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    \$\begingroup\$ The battery will be at about 2.7V when empty, so the voltage to regulator will be 5.4V. The regulator needs 7V minimum. There is very little point using a 7805 with external pass transistor to waste power as heat and stop working earlier what batteries are capable of. \$\endgroup\$
    – Justme
    Commented Mar 22 at 21:15
  • \$\begingroup\$ Hello @Justme. The voltage regulator i'm thinking of using is the HT7850 which it seems it would work when batteries are near full discharge. Do you know of any other voltage regulator which could work in this case while maintaining higher efficiency? \$\endgroup\$
    – Waripolazo
    Commented Mar 25 at 12:24

3 Answers 3

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A simpler approach would be putting 2 of these devices in parallel. 3 would be even better from a reliability and thermal management standpoint. Simpler still, would be to use a 1.5A 3-term regulator with a nice heatsink. These approaches will give you low Iq @ no load, but will have poor efficiency during operation. This will matter considering you are battery operated.

You should consider a switching converter (buck). That's how we do it these days for appreciable efficiency increase and a lot less heating. They make 3-terminal versions of what I'm talking about in a slightly larger size than the linear equivelent. Super simple to use and probably what you really want. Admittingly, Iq may be a little higher for the simpler devices, but I don't know - I haven't looked up the spec's on these modules in a while nor do I really care about the Iq in my applications.

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  • \$\begingroup\$ Thank you for your elaborate response. I will look into a switching converter. Since the device will be battery operated i'm looking for maximum efficiency. \$\endgroup\$
    – Waripolazo
    Commented Mar 25 at 12:00
  • \$\begingroup\$ @Waripolazo Check this out: digikey.com/en/products/detail/recom-power/R-78E5-0-0-5/… 1.5mA standby current. Or this one: cui.com/product/resource/v78xx-1000.pdf \$\endgroup\$
    – MOSFET
    Commented Mar 27 at 13:25
  • \$\begingroup\$ The second one is just what i needed. Thank you very much. May i ask you what do you use for searching for electronic parts? I've been using Mouser.com and digikey. \$\endgroup\$
    – Waripolazo
    Commented Mar 27 at 15:09
  • \$\begingroup\$ @Waripolazo Mouser, Digi-key, and Arrow for finding parts. Octopart and Findchips when buying parts. \$\endgroup\$
    – MOSFET
    Commented Mar 27 at 15:18
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You say you want high efficiency, and 5V output at 1A, with 2S lithium pack.

These are not going to happen with a linear regulator like HT7850 LDO.

The lithium pack will be at 8.4V when fully charged, and maybe somewhat below 6.0V when empty.

With a full pack, you need to drop 3.4V at 1A, which means, to get 5W out at 5V and 1A, you are feeding in 8.4W at 8.4V and 1A. You are thus wasting 3.4W as heat, which is a lot. And the efficiency is terrible 59%.

With a switch mode regulator, the excess voltage is not wasted into heat and you can expect 90% or better efficiency.

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  • \$\begingroup\$ I think a switch converter might be the way to go. Do you know of any model which may be a good fit for this application? I did a quick google search and most of them look much more complicated than a 3 pin voltage regulator. Are these hard to use? \$\endgroup\$
    – Waripolazo
    Commented Mar 25 at 13:03
  • \$\begingroup\$ Use a DC/DC converter, so you don't need to implement the discrete transistors. Unfortunately, it's more complex than linear regulators, but it's a good practice to hone your electronics skills. \$\endgroup\$
    – Never
    Commented Mar 25 at 13:15
  • \$\begingroup\$ What makes it difficult for me is that I only have a soldering iron to work with and most of the buck converters I've seen are the ones that have to be installed with a heated bed or heat gun. And I'm not even mentioning the tiny inductors, resistors and capacitors too. \$\endgroup\$
    – Waripolazo
    Commented Mar 25 at 18:45
  • \$\begingroup\$ @Waripolazo Buy a switch mode module then. \$\endgroup\$
    – Justme
    Commented Mar 25 at 18:47
  • \$\begingroup\$ Any recommendations on a model/make? Or where to look for? I've looked in Texas Instruments page but all of them have to be soldered with a heat bed. \$\endgroup\$
    – Waripolazo
    Commented Mar 25 at 19:03
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I don't believe you need a resistor between the base of Tr1 and R1. In the question you linked, a resistor was needed to protect the relatively fragile GPIO pins of a microprocessor. You have nothing equivalent in your circuit to protect. The circuit shown in the datasheet should work fine as is.

One change I would make however, is to reduce the Ireg to 250 or 330 mA. The HT7850's maximum current is 500 mA, and I would not want to run it at it's maximum. This would mean changing R1 to 0.7/.25 = 2.8 \$\Omega\$ or thereabouts. Since this would leave 750 mA to be handled by Tr1, I would size it to handle at least 1.5 A. When running at 750 mA, and 2.4 V across it, it will dissipate 1.8 W. I would size it to handle 3 W, but even so, it will get hot, so use a heat sink, or if you are using surface mount technology, use a large copper pad for heat-sinking.

EDIT: When I originally answered this question, I focused on the question at hand. However, I notice that the linear regulator is an HT7805, which requires a minimum input voltage of 7 V. The sense resistor requires a voltage drop of 0.7 V, making the total minimum voltage of the circuit 7.7 V. A 7.4 V input will not work. If you go with a linear regulator, it needs to be a low drop out type (i.e. LDO). An LF50 is an example of a 5 V LDO. It has a maximum drop-out voltage of 1.3 V. Together with the 0.7 V across the sense resistor, the circuit will operate down to 7 V.

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  • \$\begingroup\$ Thank you for your answer. This device will be in a sealed box. Do you think overheating will be an issue if i follow your recommendation? \$\endgroup\$
    – Waripolazo
    Commented Mar 25 at 12:18
  • \$\begingroup\$ Your total power dissipation will be 7.4 V x 1 A = 7.4 W. If the sealed box is small and made of plastic, heat may be a problem. If you used a switching power supply you could reduce the power dissipation to near 5 V x 1 A = 5 W, which would be some savings, but heat might still be a problem. Why do you need 1 A? \$\endgroup\$ Commented Mar 25 at 12:43
  • \$\begingroup\$ The device i'm designing has antennas and modules for them. Each of them can peak to around 300 to 400mA. With a microcontroller, the maximum peak is about 900mA. \$\endgroup\$
    – Waripolazo
    Commented Mar 25 at 12:54
  • \$\begingroup\$ what are the dimensions of your sealed box? If plastic, can you attach an aluminum plate directly to an inside surface of the box? \$\endgroup\$ Commented Mar 25 at 12:59
  • \$\begingroup\$ The dimensions are about 40x160x120mm aprox. I'm considering using a plastic box and eventualy i could attach an aluminum plate but not ideal. \$\endgroup\$
    – Waripolazo
    Commented Mar 25 at 13:07

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