Searching for an Op-Amp

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** OK, but I was alerting the OP to a serious oversight.

..... Phil

Understood. They changed the voltage limits on the part(s) about a decade back. I was just pointing that out. Certainly the increase from 5.5V to 6V isn't going to help much at 18V.

Yeah, but he's opted for NiMH and those things are near-impossible by properly charge by watching voltage, especially batteries, cells are easier but still hard.

C/100 trickle charge seems reasonable here, but there's no reason to use more than a resistor and diode to achieve that.

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On which planet? All the dumb domestic battery chargers I have ever owned used a 14 hour C/10 charge rate - an order of magnitude higher than C/100. These days they are all fast chargers even the budget ones.

It was possible to kill batteries by leaving them on too long.

Where I live one power failure is quite often followed by another since loss of power is usually down to storm damage taking out power lines. Storms typically last 3 or 4 days so you want emergency lights to grab power back quickly when the grid returns. Sometimes it comes back and stays other times it drops repeatedly after an hour or so due to a residual fault. Not uncommon to have most of a day off power in bad conditions (and at the moment I am snowed in - but with mains for now).

An LDO three pin regulator will work with just a fraction of a volt headroom. See the TI paper on choice of LDOs:

Guessing at your spec from the choice of diodes the LM1085 ought to do. (see fig 1)

An opamp is with crazy parameters is really not the solution.

Or you could do it the good old fashioned bipolar discrete way that seems to have been largely forgotten these days. Power transistor (eg TIP31) as an emitter follower with a 10v6 zener base to ground and a 1k resistor collector to base. Apply raw voltage to the collector and get regulated DC at the emitter (max charge current limited by Hfe). eg.

Put a Schottky diode and 10R in series to the reserve battery and then connect the load to that - this way no glitch when mains fails either. If you allow the regulated voltage out to be at sufficient current.

Otherwise you could still use the diode or of the two voltage sources.

It would be so much easier if you said how much power is required at the output to the load and at what voltage.

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Regards, 
Martin Brown

:

put and + input

ng them all together in a single chip is another matter. I was thinking of something like

I would revise your basic topology to as shown below. Use a cheaper 6V NiMH , which doesn't need regulation, and bypass the converter. You can transpor t most of your existing subcircuits into this setup. Indefinitely long term trickle charging of NiMH is C/40, and C/10 is a no brainer trickle charge rate that doesn't require going to extremes, but must be cut back when term inal voltage is 1.2V, maybe with temp dependent threshold.

Please view in a fixed-width font such as Courier.

. . . -------- . | | 1N5400 . Vin>-+----------|5V DC-DC|----|>|---- . | | | | . | -------- | . | | . --------- | . | C/40 | | . | | | . | CHARGER | +---->VDD . --------- | . | | . | | . | SW CTL>- | . | | | . | | | . | -------- | . | | | 1N5400 | . +----------|VO14642 |----|>|---- . | | | . NiMH | -------- . 6V --- . 5xAA - . 2000 | . mAh --- . GND BATT BKUP . . .

. .

I wasn't saying C/100 charging has been the historical means of charging. I was merely saying trickle charging is a venerable amd acceptable method o f keeping a battery charged. Yes, C/100 is very conservative.

Absolutely. Depending on the chemistry, C/14 is a bit harsh. C/100 is far more sustainable.

It can happen, there is no question of that. This is why in my industry (t elecommunications) we always provide a minimum of 8 hours of battery backup plus a minimum of 12 hours of Diesel or natural gas generator backup for e very internal site.

That is one reason why I want to give the user a choice of how much backup time he will provide. The 7.6AH battery stack I am using for the prototype should last about 4.6 hours at full load or a little over a day under typi cal loads.

That's an idea. I'll dig more later.

Well, as you can see, that is not the current design. I am using a TL071.

From the charging circuit? As I said, from about 7V to about 16V at the ba ttery, depending on the chosen battery and ts state of charge, and I would expect C to be between 1.0AH and 20.0AH. I will probably be using 9.6V NiM H batteries in 3.8AH pairs. The terminal voltage range on these batteries will be about 8.0 to 10V, charging at about 76mA.

No, that won't work for several reasons, not the least of which being the R Pi won't work with less than 4.95V. Your layout only delivers 4.3 - 4.5V f rom the mains supply, and as the battery discharges, it also may deliver le ss than 4.95V.

erter.

I am not supplying the battery, the user will. Six volts is probably not a n option, or at least not a good one. Much of the battery capacity will re main unused. Not only that, but 6V batteries are only "cheaper" in the sen se they result in a lower gross capital outlay per ampere-hour. They are m ore expensive per watt-hour, which is the important figure.

The user might choose 2000mAH batteries. I never would, especially not at

6V. That is only 12 watt-hours of capacity, of which perhaps half could be used. He might not even get an hour of backup. Using 9.6V at 3800mAH x 2 , I will be getting around 73 watt-hours, of which close to 90% could be us ed. The user may choose even larger batteries. The prototype should get e asily a day of backup.

a 6V NiMH discharged to 1.1V per cell is 95% depleted (ballpark) so, with a series regulator that seems like about 75% efficient Also you can't safely discharge beyond 1.1V per cell without risking damage.

Whats this "x 2"? You can't parallel charge NiMH batteries - their charge curve is not monotonic. two batteries, two chargers, two diodes - I guess that could work.

given that you're losing 4% of that energy in the diode before it you want me to believe that the DC-DC converter you've chosen will be 96% efficient?

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A 6V NiMH battery will only produces 5.5V when discharged to 1.1V per ce ll. That is not enough to drive the DC-DC converter, but still too much to drive the RPi without further regulation.

One can when trickle charging. It does not matter (much) they are not mo notonic. As long as their float voltages are within a couple of percent, t hey will eventually be charged to essentially the same charge level.

The same with discharging. One set of cells will inevitably discharge a t a different instantaneous rate, depending on the exact terminal voltage a nd total internal resistance, but as one string discharges at a greater rat e, its terminal voltage drops faster than the other string, and the other s tring begins to take on a majority of the charge flux. Admittedly this wor ks much better with battery strings that have somewhat higher internal resi stances than NiMH cells, but it will work as long as the charge rate is not very high.

No. Where did I ever suggest a 92% efficiency? Or 100%? Even in my sta tement above (90% use), I am not saying I expect the converter to be 90% ef ficient. I am merely saying 90% or so of the energy rating of the battery can be extracted from the battery before the circuit must be shut down to p revent damage to the battery. Some will be used by the diode (about 7%, ac tually, not 4%). The converter is specified at 92%, so we are down to abou t 85% or so overall efficiency when running on battery.

Here is the latest iteration of the project, including both the charger and the switching hardware, but sans the monitoring hardware:

Note this should allow the following range of batteries with a 12.0V supply . Assuming a minimum of C/100 charge rate, any battery capacity from 500mAH to 10,000mAH should work, adjusting R10 accordingly. Anything more than 1

0,000mAH will suffer from a lower than C/100 rate. Batteries as large as 5 000mAH could be charged at C/50.

Nickel Chemistry:

7.2V (6 cells) will shut down at about 50-60% of capacity 8.4V (7 cells) 9.6V (8 cells) 10.8V (9 cells) will charge to about 90% capacity

Lead-Acid:

8.0V (4 cells) 10.0V (5 cells)

By increasing the mains power supply to 16V, the following could also be em ployed:

Nickel Chemistry:

10.8V (9 cells) 12.0V (10 cells) 13.2V (11 Cells) 14.4V (12 Cells) will charge to about 80% capacity

Lead-Acid:

12.0V (6 cells)

Thanks for sharing.

Is the anode of the 5.6V zener meant to go to +5V?

Have you thought about eliminating X2 and implementing that functionality in Q4/Q5?

piglet

Precisely what does it do? Gozinta/gozouta requirements? (*) Looks very Rube Goldberg to me.

(*) To date I have not seen a specification. ...Jim Thompson

--
| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    | 
| STV, Queen Creek, AZ 85142    Skype: skypeanalog |             | 
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  | 
| E-mail Icon at http://www.analog-innovations.com |    1962     | 
              
     It's what you learn, after you know it all, that counts.

Looks like a hairball's hairball, to me. Too much wrong to even get started. What's it supposed to do?

"What's it supposed to do?" is what I keep asking.

Rhorer seems to be a snarky POS who can't cope with being asked a criticizing question.

It's definitely a hairball... almost as bad as starting out with a rubber band, then asking, "what do I need to do to patch this into a charger?".

Sheeesh!

Dumbass at its finest. ...Jim Thompson

--
| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    | 
| STV, Queen Creek, AZ 85142    Skype: skypeanalog |             | 
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  | 
| E-mail Icon at http://www.analog-innovations.com |    1962     | 
              
     It's what you learn, after you know it all, that counts.

Reading from the start reveals it to be a super-duper world beating univeral Raspberry-pi backup power supply. You'd probably rustle up a nice solution from a TL431 and 3 BJTs but that would spoil the fun for the rest of us ;>

piglet

OK! "Problem" solved >:-} ...Jim Thompson

--
| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    | 
| STV, Queen Creek, AZ 85142    Skype: skypeanalog |             | 
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  | 
| E-mail Icon at http://www.analog-innovations.com |    1962     | 
              
     It's what you learn, after you know it all, that counts.

"What's it supposed to do?" is hardly a criticism. If he wants help, it's kinda important to know what he wants help with.

Easy! "Duct tape."

Now THAT'S critical.

rger and the switching hardware, but sans the monitoring hardware:

Oops. No. That slipped in when I was tidying up the schematic. It's fixe d, now. The page may need to be force-reloaded in your browser in order to update.

I did look at the fact I have essentially three relays here, but discarding one won't work. X2 disconnects the battery and only the battery when GPIO 19 is low. It is not possible to assert GPIO 19 until after the RPi is bo oted up, so whatever feeds the RPi from both power feeds must be Normally C losed, and not controlled by the RPi in any way. That constitutes 1 unique relay - Q4/Q5 in this case. Since it is a shutdown facility, it does nee d to be fed from both 1N54504 diodes. Meanwhile, the battery should always remain disconnected from the power train unless the RPi actively connects it during live operation. That constitutes a second, unique, Normally Open relay, X2. Finally, X1 is also Normally Open, because it isolates U1 from the battery whenever the mains fail, to prevent damage to U1. Whenever th e mains fail, X1 must be de-energized, but if the RPi is up and running, th en both X2 and Q4/Q5 must remain energized at all times. That makes three unique relays, with no way I can see for any one relay to take over the job of either of the other two. Here is a Boolean table:

Vin-yes Vin-No RPi-Up RPi-down X1 1 0 - - X2 - - 1 0 Q4 1 - 1 -

As you can see, the deterministic states for any relay cannot be reconciled with any other relay, and each relay has at least one non-deterministic st ate when the other two relays in turn have a deterministic state. Either of these facts would prevent one relay from doing the job of any other relay.

Now I could hypothetically replace Q4/Q5 with a relay, but the relays cost a bit more. Replacing X1 with a transistor pair wouldn't work as well as t he relay. I suppose I could replace X2 with a PNP transistor.