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I'm building a portable 8x8 LED matrix powered off of two CR2450 coin cell batteries. Most of my components ( Attiny85 and LEDs ) can run easily off of the supply voltage range of the CR2450, but the shift registers aren't being as conducive...

There are three shift registers in my design: two DM7495N and one NTE74LS164. All three shift registers have a recommended voltage range of 4.75 to 5.25 in order to operate. If I chained the two batteries, they would supply a maximum voltage output of 6V and a minimum of maybe a little over 5.2V, before the other components become under-powered and the batteries would need to be replaced. That is not within the range associated with the shift registers..;(

Is there a really simple way to step down the voltage to fit the ratings of the shift registers without using a voltage divider? Like maybe with a couple of resistors or some other common electronics component? Something tells me that there is in this case, since I am using two separate batteries... If I have to, I could ignore the recommended ratings, as both registers have a maximum input voltage larger than 6V, but I would like to know if there is an efficient and easy way to step the voltage down in this case. Thanks!

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  • \$\begingroup\$ No one is answering :( \$\endgroup\$
    – Holden
    Jun 16, 2017 at 3:34
  • \$\begingroup\$ I'd avoid all the regulator hassle and go for devices that can handle the voltage. CD4035 or some HC device? \$\endgroup\$
    – jonk
    Jun 16, 2017 at 3:47
  • \$\begingroup\$ Yeah, I'm starting to think a regulator would be a little over-kill, too... I was mainly wondering if it was possible to easily adjust the voltage output of two chained batteries by chaining them in some special way or something... \$\endgroup\$
    – Holden
    Jun 16, 2017 at 4:01
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    \$\begingroup\$ Have you seen the datasheet for the cell you've selected? The "continuous standard load" is just 0.2 mA. How many LEDs are you hoping to drive with that? \$\endgroup\$
    – Dave Tweed
    Jun 16, 2017 at 4:11
  • \$\begingroup\$ Oh no.. I didn't see that. Do you think it would vary much depending on the manufacturer? The battery I'm using doesn't have that type of information on it. Could it handle 80mA or 40mA, maybe 90mA? Lol, I'm new to electronics.. \$\endgroup\$
    – Holden
    Jun 16, 2017 at 4:25

3 Answers 3

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I would recommend changing the parts in your design to use parts such as the 74HC195 and 74HC164, or change the design slightly to use other variants of CMOS technology.

The 74HC parts operate over the range of 2-6V and consume hardly any power, much less than those you are currently using - they could operate without any regulator. They can interface with the ATtiny the same as the TTL and LS parts and are pin compatible.

The static power consumption of CMOS devices is close to zero - for example the 74HC164 has a typical value of only 8uA at 25°C.

When the device is operating there will be additional current required to charge the internal capacitances that depend upon both the power supply voltage and the operating frequency.

The data sheet usually has a value called "Power Dissipation Capacitance" or similar - this is 40pF for the 74HC164. This can be used to calculate the dynamic power consumption. (See page 9, note 3 of the datasheet for the 74HC164).

If the device is operating at low frequencies this will be very low. For example if the 74HC164 is being clocked at 1MHz with a 5v supply and the input data is a constant zero or 1 the resulting supply current will be 1mA, much less than the 30mA of the 74LS164.

40*(5)^2 * 10^6 * 10^-12 = 10^-3 = 1mA.

I would expect the average frequency when used for driving a display to be in the kHz range with a correspondingly low current consumption.

Any current required to be sourced from the output pins would be added to this currents. Sinking current does not require any additional supply current.

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  • \$\begingroup\$ That'd be a good idea... I might do that in the next iteration of the matrix... For now I'm thinking about just supplying the 6V. That's still a volt under what the NTE register can handle and far under the max for the DM7495Ns can. For most of the batteries life, they would supply less than six volts anyways... \$\endgroup\$
    – Holden
    Jun 16, 2017 at 3:56
  • \$\begingroup\$ The CMOS ones are pin compatible with the ones you are using, so they only require substituting the new parts. \$\endgroup\$ Jun 16, 2017 at 3:58
  • \$\begingroup\$ Oh really... You know you're right. It may only take a couple of days for these components to ship from Amazon anyways... I'll try ordering something. Thanks! \$\endgroup\$
    – Holden
    Jun 16, 2017 at 4:06
  • \$\begingroup\$ How exactly is power consumption measured with these things? Like how do I know how much current I need to supply the shift register in order to achieve a certain current output from the pins? \$\endgroup\$
    – Holden
    Jun 16, 2017 at 19:54
  • \$\begingroup\$ I've added how to calculate the supply current. \$\endgroup\$ Jun 16, 2017 at 20:42
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This arrangement will not work at all, not even for a short time.

The CR2450 battery (per Sony, Panasonic datasheets) has a drop of ~100 mV with a load of 7.5k, or 0.4 mA. This means that the internal battery resistance is about 250 Ohms. Even if you try 4 LEDs at 2 mA each, the voltage will drop by 2-3 V. Your Attiny will eat another 1 mA even if you run it at 1MHz. The main killler will be DM7495 with typical consumption of 50 (!!!) mA. Two of them will take 100 mA, plus 17 mA for NTE74LS164. There is no way for these batteries to deliver 120mA, the short current (voltage drop to zero) will be ~12-15 mA.

So scrap your project and start over, with CMOS chips as Kevin suggested, and a different battery.

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  • \$\begingroup\$ The recommended max supply current for the DM7495N is 75 mA. I saw the typical 250mW dissapation, but how do you determine 50mA from that? That seems like a lot? Sorry if these are stupid questions. I'm new to electronics... \$\endgroup\$
    – Holden
    Jun 16, 2017 at 5:59
  • \$\begingroup\$ Also the LEDs are run in parallel. This thing is charleplexed. So where are you getting a 2-3 volt drop? \$\endgroup\$
    – Holden
    Jun 16, 2017 at 6:03
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    \$\begingroup\$ From Energizer datasheet, data.energizer.com/pdfs/cr2450.pdf they list 9mA at 2.7 V, making internal resistance about 33 Ohms. So the battery can deliver max 100mA at zero volts, for 2 seconds. Zero volts is not what you want. The 50 mA is "typical" for DM7495. I am really curious, where did you get these ancient parts? \$\endgroup\$ Jun 16, 2017 at 6:08
  • \$\begingroup\$ Lol, that's understandable.. ;) I just really wanted to make an led matrix so I used what I could scavenge at a local makerspace. \$\endgroup\$
    – Holden
    Jun 16, 2017 at 6:12
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    \$\begingroup\$ @Holden Reconsider using AA(A) batteries in combination with a boost converter. (or buck/boost depending on the amount of cells) \$\endgroup\$
    – Jeroen3
    Jun 16, 2017 at 6:29
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Try searching up voltage regulators. Linear voltage regulators can do the job of stepping down the voltage but the voltage difference between the input voltage and output voltage of the linear regulator gets dissipated as heat. If you want an efficient voltage regulator, try switching voltage regulators like the Buck Converter.

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  • \$\begingroup\$ Ok, I'll do that... \$\endgroup\$
    – Holden
    Jun 16, 2017 at 3:42

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