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Old 11-03-2008, 05:31 AM #1
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Default Boost Driver

So I was looking into making the boost driver posted by woop which is able to run off of a 0.3v input [a single AA]. I have a few questions about it. Thread Here

Woop 1.png-

Rsense- do these have to be current sensing resistors? I saw something saying that they should be low value, but isn't this where you would set the current for the LD?

Then there was another board that woop had posted which allows for the use of a case negative diode (vs the LD - being at 0.5v compared to ground) [Woop 2.png]

Woop 2.png-

Would it be possible to throw a pot in there somewhere? If so, where?


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Old 11-05-2008, 03:13 AM #2
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Default Re: Boost Driver

.................................................. .................................................. .................................................. .................................................. .................................................. .................................................. .................................................. .................................................. .................................................. ..............................

Any ideas... current sensing resistors or not for both boards & where a pot would go on board 2...? :-?
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Old 11-05-2008, 04:19 AM #3
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Default Re: Boost Driver

I found in the datasheet where it talks about the Rsense resistor (page 7) but i'm having some difficulty understanding it. Maybe someone could help me with that..
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Old 11-05-2008, 05:44 AM #4
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Default Re: Boost Driver

What I found is that a current sensing resistor is a high precision resistor (~1% variance) as compared to a normal resistor (up to 10% variance), and has very low inductance values. *You use a low value to minimize the effect on the rest of the circuit.
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Old 11-05-2008, 12:53 PM #5
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Default Re: Boost Driver

If it's supposed to be a small value (i think i saw 0.02 ohms mentioned somewhere in the datasheet), then how would you go about setting the current in the circuit?

why can't woop just log in :-/
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Old 11-05-2008, 04:07 PM #6
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Default Re: Boost Driver

The resistor is used as a feedback path. Look at this datasheet.

The TPS61200 voltage regulator attempts to set the output voltage based on the voltage value sensed at the current sensing resistors. If the feedback voltage is greater than 0.5 V, the regulator reduces the output voltage, and if it is less, it increases the output voltage.

By resolving Ohm's law when Vfb is fixed, and the current is determined by us, you get R = 0.5 / I. So, for a 200 mA circuit, you would use a 2.5 ohm resistor, and for a 125 mA circuit you would us a 4 ohm resistor.

---

This technique reduces the efficiency in the circuit by about 10%. If you could find the identical chip with a feedback voltage of 0.25 V, you only lose about 5%.

Overall, the efficiency looks to be around 65% at 1.2 V input, and 80% at 3 V input and higher.
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Old 11-05-2008, 06:50 PM #7
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Default Re: Boost Driver

So I get that part that you're talking about.

I saw in the datasheet for LT6106 that it says to get the "input sense voltage full scale" use a 500 ohm resistor for Rin. And for Rout, it never specifically says which to use but in most of their tables it lists 1k, 5k, and 10k ohm as the value. How do these two resistors effect the circuit and what would be the best choice for Rin and Rout?

And with the sense resistor, they don't have to be current sensing, just a tight tolerance?


Thanks for the help
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Old 11-05-2008, 08:48 PM #8
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Default Re: Boost Driver

Quote:
Originally Posted by treb76
I saw in the datasheet for LT6106 that it says to get the "input sense voltage full scale" use a 500 ohm resistor for Rin.
That's just the maximum Vsense it can handle reliably. You don't really want to run the current sensor near that range.

Quote:
For Rout, it never specifically says which to use but in most of their tables it lists 1k, 5k, and 10k ohm as the value. How do these two resistors effect the circuit and what would be the best choice for Rin and Rout?
The datasheet lists its values based on Rsense being a 0.02 ohm test resistance. From looking at the graphs, if you use a 0.02 ohm Rsense, you want Rin = I / 0.05, or a 2k resistor for 100 mA of current to output 0.5 V. Rout looks like a subjective value to me - I would probably go with a 10k ohm, and it doesn't look like it needs to be current sensing. It mainly seems like a discharge path when power is disconnected.

Quote:
With the sense resistor, they don't have to be current sensing, just a tight tolerance?
That appears to be the purpose of current sensing resistors. They're low resistance but high precision (and seem to be heat tolerant I think). You're likely to get a large current swing if you use imprecise resistors.


Overall, I'd say the second board isn't worth the increase in complexity for the higher efficiency. You gain no more than 10% efficiency with the second design, at a cost of extra components, time, and area. Do you need that extra 10%?
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Old 11-06-2008, 12:00 AM #9
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Default Re: Boost Driver

The main reason for the second board was to be able to use a case ground diode in a case neg host- so it would be nice to be able to use the second board, especially for reds.

But couldn't you just get regular SMD resistors (thick film is what mouser lists it as) with 1% tolerances? I thought current sensing resistors converted the current to some voltage drop. Just want to make sure I get the right parts- at least for board 1.
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Old 11-06-2008, 01:15 AM #10
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Default Re: Boost Driver

Okay...I really should read through that entire thread...especially cause I'm thinking about producing some of these boards...

Anytime you measure the voltage drop across a resistor, you are determining the current in a circuit. *I thought I saw that current sensing resistors were supposed to be more temperature invariant than normal resistors, but I really can't find much of anything on current sensing resistors right now.

Right now I'm trying to figure out how to convert design #1 to the TPS61020 - it's a little bit cheaper than the TPS61200, and it looks like the only compromise is a minimum working voltage of 0.9 V and not 0.3 V. *The chip seems a tad more efficient anyways.
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