rhd
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I want to point out an issue that I've just come up against -
Lower IC Dropout Voltages Aren't Necessarily Useful
The reality is that in a 2X lithium-ion setup, a 1085 IC with a current setting resistor dropping 1.25V is still going to let you hit your desired diode current without the IC's dropout voltage ever becoming the limiting factor. Why? Because the IC is always going to be dropping more voltage than it's actual minimum dropout voltage anyway.
If you supply 8V in, and the diode needs 4.75V, then 3.25V still NEEDS to get dropped somewhere. In a DDL setup, you can drop 1.25V on the resistors, and the other ~2V on the IC. That's a fairly equal spread of heat on both sides of a driver.
Now, if you instead use one of these ICs that doesn't feature a resistor dropping 1.25V, what happens? Do you get a more efficient system? No. You just get the IC dropping more of the voltage itself. In this case, the IC would be dropping 3.25V, instead of just 2V. In other words, no matter how low your dropout voltage gets, you're STILL going to have to drop the same voltage somewhere, other than in the diode.
I just fairly extensively tested a driver that was intended to feature in the V3 of my Mosquito modules. The driver is a 4x or 5x AMC7135 setup. In testing, it runs FANTASTICALLY with 5V of input. The dropout is so low (a few hundred mV) that the full 1,400 or 1,750 mA of output is possible with just 5V in to the driver. Almost no heat is generated.
What happens if you power it by 2x lithium ions? Without current setting resistors dissipating heat, it ALL falls to the ICs, and they get scorching hot. Not a good design feature.
So I guess my message is - for Blues, we're at the limit of what we can do with linear drivers. The drop-outs are already below what we need them to be for 2x lithium ion setups. 1x lithium ion setups will never be possible from a linear setup because the Vf of a 445 is above a single cell's voltage. So - we're done
Lower IC Dropout Voltages Aren't Necessarily Useful
The reality is that in a 2X lithium-ion setup, a 1085 IC with a current setting resistor dropping 1.25V is still going to let you hit your desired diode current without the IC's dropout voltage ever becoming the limiting factor. Why? Because the IC is always going to be dropping more voltage than it's actual minimum dropout voltage anyway.
If you supply 8V in, and the diode needs 4.75V, then 3.25V still NEEDS to get dropped somewhere. In a DDL setup, you can drop 1.25V on the resistors, and the other ~2V on the IC. That's a fairly equal spread of heat on both sides of a driver.
Now, if you instead use one of these ICs that doesn't feature a resistor dropping 1.25V, what happens? Do you get a more efficient system? No. You just get the IC dropping more of the voltage itself. In this case, the IC would be dropping 3.25V, instead of just 2V. In other words, no matter how low your dropout voltage gets, you're STILL going to have to drop the same voltage somewhere, other than in the diode.
I just fairly extensively tested a driver that was intended to feature in the V3 of my Mosquito modules. The driver is a 4x or 5x AMC7135 setup. In testing, it runs FANTASTICALLY with 5V of input. The dropout is so low (a few hundred mV) that the full 1,400 or 1,750 mA of output is possible with just 5V in to the driver. Almost no heat is generated.
What happens if you power it by 2x lithium ions? Without current setting resistors dissipating heat, it ALL falls to the ICs, and they get scorching hot. Not a good design feature.
So I guess my message is - for Blues, we're at the limit of what we can do with linear drivers. The drop-outs are already below what we need them to be for 2x lithium ion setups. 1x lithium ion setups will never be possible from a linear setup because the Vf of a 445 is above a single cell's voltage. So - we're done
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