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SuperCap on Horizon?


acadiel

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Anyone try putting a 1F or larger SuperCap in place of the batteries on the Horizon?  Anyone know the current draw in standby and the math to figure out how long it should last with one in?

 

IMHO, it’ll recharge instantly and doesn’t need any kind of resistor, diode, etc.  

 

Something like this:  https://www.mouser.com/ProductDetail/Cornell-Dubilier-CDE/EDC155Z5R5C?qs=sGAEpiMZZMv0NwlthflBi9Hx2wYSwzGt7LwA30QFefM%3D&countrycode=US&currencycode=USD

 

From the TI Tech Page:  “To save power, only part of the Ramdisk is connected to the battery-backed power line: the power lines for the SRAMs, the 6264 RAM and the two 74HC154 decoders (that's why these are HC, not LS: we need CMOS chips, not TTL).”

 

Adding the 62256 SRAMs on my 3000, it looks like there’s 24.  Plus I’m guessing some of the decoders.

 

My green fully populated Horizon (not sure which model) have 6264s on them - 37, looks like.  The last two have just 32.  Plus the stack of 3 74HC154s. 
 

Yeah, I somehow wound up with four of them.  All need a better battery solution.

 

Two were the club Horizons from DTIHCG and two were from Louis/John Guion.  

 

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16 hours ago, Tursi said:

I'd be curious too! I've used a couple of super caps on Minimemory cards in the past, but I never thought of the ramdisk.

Either way, it will be a shorter duration charge to batteries, though. But for a daily use machine...??

 

I need to find either an ammeter (or someone with an ammeter) to see what current is being drawn from the 3.6v NiCD battery cells.  If we can get a baseline with the various chip count Horizons (or figure an average per SRAM figure), I'll see if I can track down how long certain supercaps can support that draw for for a certain farad supercap.  

 

Anyone want to volunteer to measure the battery draw current on their Horizon with the power to the PEB off?  Include the model card and how many SRAM chips you have on your board.

 

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I don't know if I did this right - someone please check my math.  I took the 21uA Grand RAM and plopped it into this calculator and got this.  This was for the aforementioned 1.5F SuperCap that I linked to.

 

So 12 hours for the Grand RAM?  Or over a day for the first one with the 9uA draw.  Or an hour and a half for the energy hogging HRD3000.

 

Edit: If you got this 4.7F bad boy - the calculator says you can get a whopping 37 hours before you got down to below 3V on the 21uA Horizon.

 

If the math is right - these might be a great thing for someone who needs to turn the system off every once in a while for maintenance but who might keep it going otherwise.  Or for power blips. 

 

supercap.png

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It looks like the capacitor will  feed that buss trace if that gray line is the cathode. So it's correct for that.

It's not clear to me how the capacitor gets charged. ?? Is there a path that the batteries used before?

It cannot get charged via the diode because no current can flow from the trace to the capacitor with the diode in place.

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46 minutes ago, TheBF said:

It looks like the capacitor will  feed that buss trace if that gray line is the cathode. So it's correct for that.

It's not clear to me how the capacitor gets charged. ?? Is there a path that the batteries used before?

It cannot get charged via the diode because no current can flow from the trace to the capacitor with the diode in place.

Oh, no - the SuperCap is something I’m doing as an experiment on another board.  The lithium battery is just a replacement battery period.  No SuperCap.  

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3 minutes ago, acadiel said:

Oh, no - the SuperCap is something I’m doing as an experiment on another board.  The lithium battery is just a replacement battery period.  No SuperCap.  

Ah.  So same question. The lithium battery needs a source of current to stay charged. It cannot get it via the diode.

And you cannot connect that lithium cell directly to the 5 volt rail.

 

I think you now need a path from 5V through some kind of 3.6..3.8 volt regulator that feeds the battery directly.

A resistor and Zener diode might be sufficient.

 

So what you have will work until the battery dies. :( (If I am seeing the circuit correcly) 

 

 

 

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I have an idea.  

Each silicon diode has a fixed voltage drop of .6 volts.

If you feed the battery directly with two silicon diodes in series the battery will only ever get 3.8 volts max.

 

So the circuit is like 

                                             3.8V                  this diode feeds buss when power is off and buss goes < 3.8 V 

   Earth: -- /\/\/\-----[ battery ] +  ------------------>|----------:  +5 Vcc

                                                    \

                                                      \ ---|<-- |<----------------: +5 Vcc 

                                                          2 diodes charge battery when buss is 5V

 

I think this will work... 

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Ah.  So same question. The lithium battery needs a source of current to stay charged. It cannot get it via the diode.
And you cannot connect that lithium cell directly to the 5 volt rail.
 
I think you now need a path from 5V through some kind of 3.6..3.8 volt regulator that feeds the battery directly.
A resistor and Zener diode might be sufficient.
 
So what you have will work until the battery dies. [emoji20] (If I am seeing the circuit correcly) 
 
 
 

It’s a disposable Lithium battery. Should last years backing this ramdisk up without power :-). Not meant to be charged. The diode is there to keep it from being charged per the TI Tech Guide.

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I've always used germanium diodes because the voltage drop is less than that of a silicon diode.  1N34As I believe?  It's getting hard to find good ones; the ones from China are all over the board on a curve tracer.  There are some good diodes that come out of Russia you can sometimes find on EBay.

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I've always used germanium diodes because the voltage drop is less than that of a silicon diode.  1N34As I believe?  It's getting hard to find good ones; the ones from China are all over the board on a curve tracer.  There are some good diodes that come out of Russia you can sometimes find on EBay.

1N914? I looked at those and the specs didn’t look too good on them, so I decided to use the 1N4001 instead. The voltage drop shouldn’t be an issue because it can go down to 2.2 before it gets to be an issue.
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Ok, you asked for it![emoji3]
 
But seriously, I'm not sure about the ranging on your meter. It appears to read 21mA(Milliamps), not 21MA(Microamps/uA).
 
That seems a bit high to me, so maybe I'm wrong.:twisted:

To be fair, I used Google. My brain is lazy tonight.

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