SparkFun Forums

[kirash4]

Something else I'm not entirely understanding. I learned a long time ago that as voltage goes up, Amps goes down. So, if I'm using two single cell LiPos, 6,600mA each, in parallel, at the point where they are tied together (with diodes), technically I should have 3.7V @ 13,200mA. Yes?

And if I buck/boost that to 5V, is the amount of Amps equally affected, bringing it down to ... uh, I think about 8,712ma?

[Mee_n_Mac]

Something else I'm not entirely understanding. I learned a long time ago that as voltage goes up, Amps goes down. So, if I'm using two single cell LiPos, 6,600mA each, in parallel, at the point where they are tied together (with diodes), technically I should have 3.7V @ 13,200mA. Yes?

Almost yes. Two identical batteries in parallel will output the same voltage as a single battery but with (ideally) double the current capacity. If a single LiPo is 3.7V (nominally) then after a diode drop of about 0.65V the output voltage would be about 3V at 13,200 mA.
http://www.onsemi.com/pub_link/Collater ... 80CT-D.PDF

And if I buck/boost that to 5V, is the amount of Amps equally affected, bringing it down to ... uh, I think about 8,712ma?

If the converter were 100% efficient then the power output would equal the power input but since no device is 100% efficient, more power must go in then comes out.

Pout = Pin * efficiency

So with 3V and 13,200 mA input that's 39.6W input. Less use an 85% number for efficiency, making the max output power be 39.6 * 0.85 = 33.66W. At 5V that's 6.73A (6732 mA). Were you to have 3.7V input and depending on what you used for an efficiency #, your estimate would be correct. Mine is intended to be on the conservative side.

[kirash4]

Ok, question. The Atmel in the circuit will be running at 5V, however there's a USB controller that runs at 3.3V. So, I'm thinking after tying the LiPos together (I will use MCP73833 charging circuit) and then buck/boost it up to 5V and a regulator after that to get 3V3. But, before to go on an adventure and try to produce some blue smoke, would someone be so kind and look at the attached schematic and give their advice/comments/suggestion. I also need to figure out a buck/boost setup for this. Suggestions?

Thanks everyone who has helped so far!

[waltr]

LiPos, 6,600mA each

Are you drawing 6.6A from the battery or do you mean that you are using a battery that has a capacity of 6.6A-Hr?
Big difference.

[kirash4]

Capacity. At the most I'll be drawing 2A, and that's intermittent ... it'll be less for most of the time.

[kirash4]

One thing I have to keep reminding myself is that the external supply can be as much as 6V (which is the upper limit for the MCP73833). This means I do need a buck/boost circuit after the LiPos. If external power is supplied, I need to bring that down to 5V for the remaining circuit, and if the whole thing is running off of LiPos, then I need to boost that to 5V. And I need to maintain a steady 2A (minimum) out of that ...

[Mee_n_Mac]

What's the range of voltage that might come in at J1 ? What's the purpose of Q1 ?

I know the RC guys use a variety of LiPo configurations, both serial to get voltage and parallel (no diodes) to get amperage. How do they charge up their parallel cells ?

[kirash4]

J1's range will be 5V-6V. The upper limit for the MCP73833 is 6V, so anything above that will damage them. And Q1 is used as a switch between LiPo power or external power from J1. If external power isn't supplied, everything runs off of LiPos. If external power is supplied, Q1 will switch and cut off power FROM the LiPos. At least, that's the idea here.

[Mee_n_Mac]

I'm not sure Q1 will work the way you want it to (as shown) but the good thing is, given the 5V min @ J1, you don't even need Q1. D1 should steer the J1 voltage to the boost converter. The LiPo's will always present less voltage at that same point (after their diode drops) and so won't contribute any current when voltage is present @ J1. D2 and D3 block any "backfeeding" from J1 to the LiPo's.

Let me also add you should be aware of the discharge curve (voltage vs state-of-charge) for LiPo's. The voltage can vary a bit and to be safe you should have a cut-out circuit that stops any further discharge once the battery voltage gets to a certain (low) value.

(generic curve)

[kirash4]

Actually, you're right. I forgot what I was doing there. Q1 is to prevent reverse current from an incorrectly connected supply. It's taken from an old application report by TI: http://www.ti.com/lit/an/slva139/slva139.pdf

But now that I look at it, I'm not sure it will do much good since there are two batteries at play here, and the possibility of an external supply. I have to figure out how to take all of that into account. Someone could potentially connect one of the two batteries in reverse, or both, or the external supply ... or all three for that matter.

But I'm still trying to find an appropriate buck/boost circuit that can take the LiPo voltage up to 5V or if external power is higher than 5V, it needs to regulate it down. So far all the ICs I've found don't supply nearly as many amps as I would need out of them, so maybe I am looking in the wrong direction?

[kirash4]

Still hoping for some suggestions for a boost circuit or IC to bring the voltage from the LiPos (through the Schottkys) up to 5V. So far I've come up with this and I'm wondering if anyone has any advice on this:

http://cds.linear.com/docs/Datasheet/3872fb.pdf

Thanks all.

[viskr]

That looks like it would do it. I'd go back to the system design, you are talking about 5V and 2A, that seems like a lot of power and why does it have to be 5V. Most micro's run off 3.3V now and most sensors are available at 3.3 as well.

I second Mee_n_Mac about a voltage cutoff point. A while back I was doing destructive testing on battery charge circuits in cell phones, and you can over-discharge them. While it's not going to produce a fire, it will destroy the battery, what happens is the cell will expand when over dis-charged and it loses the capability to be re-charged.

[kirash4]

Unfortunately, there are other devices attached to the circuit (via a USB controller on board) that will need 5V from it. Hence the 2A requirement as well (so that everyone has enough power.) Trust me, I've been trying to figure out how to keep this thing at 3V3 but that's just no way. The BlueTooth device needs 5V, the RFID device needs 5V. The stupid thing is, the only component on this whole thing that needs 3V3, is the USB controller itself. Everything else runs at 5V. So yeah, 5V all around with a smaller 3V3 reg to run the USB controller.

As for the cut off, I'm looking at the DS2764 ... But I'm not understanding its schematic. On one side the terminals are labeled "pack+" and "pack-" and on the other side they're "bat+" and "bat-". I'm unclear how to tie this in with the MCP73833 that we're using to charge the LiPos (see schematic posted earlier.)

[viskr]

I think the DS2764 is overkill for what you need, all you need is a comparator against some minimum battery voltage, when below that disable the buck-boost supply, most of them have an enable pin. There use to be a lot of voltage monitor circuits for generating reset that could do it.

Back to the load sharing, you might want look into the ideal diode circuits from LT and Maxim, as if you really are running 2A, the diode drop is a a large percentage of the power. One candidate is the LTC4412, that my reading of the spec would be you could tie 2 together and the aux supply connected with a diode.

[kirash4]

Ok, trying to understand the LTC4412, specifically Figure 5 on page 10 of it's datasheet since I'll be using two LiPos in parallel. The STAT pin is a bit of a mystery still to me. Where is VCC coming from? Is that the same VCC that drives the MCU (in the grand scheme of the "load")? And are those pins also connected to the MCU for status sensing, or can they drive, say LEDs (to indicate which battery's providing power for example)?

And this is specifically for load sharing between the batteries. Which leaves me still with the MCP73833 for charging the same batteries, and I need a low voltage cut-off mechanism. I'll try to tie the LTCs in the charging circuit tonight and see what I come up with ... Schematic to follow later tonight.