NewBie
*Retired*
Re: Zetex 310 based boost converter with 1 amp out
[ QUOTE ]
greenLED said:
[ QUOTE ]
dat2zip said:
In fact I have a derivative of the Nexgen that will do more than 1A from a single 123 cell and it is fully regulated constant current design.
[/ QUOTE ]
/highjack alert!/
Also, am I understanding correctly that driving a circuit at 3v with 1x123 is "better" (whatever that means electronically) than using a 2xAA config? Can someone please explain why this is so, or point to a source I could check? I'm trying to educat myself on the electronix world.
Thanks!
/Highjack off/
[/ QUOTE ]
greenLED
Lets take an imaginary converter.
Consider the sag of the NiMH under high loads and also the 123 cell. Think about the internal resistance of the battery. Then think about how the cell voltage sags over time under load. Consider the starting voltage. Remember power in is voltage * current = power. You want power.
But remember there is resistance in the MOSFET/Transistor to ground. Higher voltage means you need less current for the same power. The majority of loss in the MOSFET is due to Power = Current^2 * resistance. So, a small decrease in input voltage means you need more current, but the losses in the mosfet are due to Current being squared, so it takes it's toll quickly.
The same type of thing happens as the battery sags.
In every battery, you will find some internal resistance. This causes the cell voltage to sag such as when you put a 1 Amp load on it. A NiMH that I have here charged two days ago. When it isn't under load, the voltage across it reads 1.287V. When I put a 1 Amp load on it, it's output voltage drops to 1.190V. If I was utilizing two of these, the voltage would be 2.574 Volts, and under a 1 Amp load, the voltage would drop to 2.380 Volts, and they would sag from this over time as they were drained.
But, if I set everything up, such that I needed 2.574V on the input at 1A for 2.54W going in, and then the voltage dropped to 2.38V, then I'd need to draw 1.07A to maintain the same power going in. As the cell is used, the battery voltage continues to drop. Lets look at the point As the cells drain down to their 1.8V ( 0.9V each) cut-off point, I'd need to draw 1.41 Amps to maintain a flat light output level down to the battery cut-off point.
Now, let say your input MOSFET had 0.1 ohms. With 2.54V off the battery and 1A flowing into the input, there would be Power = Current^2 * resistance loss, or 0.1 Watt of loss. Keep in mind we are not including the output MOSFET/schotty diode loss or the power needed to drive the MOSFET gate, nor the power needed to make the control circuits function. So then we take the power lost and divide it by our input power. 0.1W/2.54W= 0.04 or 4% loss due to the resistance just in the input MOSFET.
Now, when we drop on down to 1.8V at 1.41A, things change.
Power = Current^2 * resistance. The MOSFET had 0.1 ohms of resistance. So 1.41^2 * .1 = 0.20 Watts of loss. We have 2.54W input, so the power loss in the MOSFET due resistance/input power, or 0.20W/2.54W = 0.0787 or 7.87 % loss.
Then we go on to the schottky diode, such as used in the BB and these Zetex converters, on the output side of things. There are some super low loss schottky diodes that will have a Vf of only 0.3V when 1A flows through them. Well, power also equals Power = Voltage * Current. Since we are always putting out 1A in a constant currrent converter, this loss is easy to figure. 0.3V * 1A = 0.3W So, we had 2.54W, so we take this 0.3W/2.54W = 0.031 or 3.1% loss.
So with a new battery, only accounting for just two of the loss factors (there are many more), we'd have 7.1% losses with a fresh cell, and 10.97% losses near the cut-out.
You still need to add in losses due to the inductor DC resistance, eddy current losses in the inductor, B-H losses in the inductor, power lost driving the gate charge in the MOSFET, power lost in the circuit that drives the gate, losses due to charging and discharging the MOSFET body diode (internal to the MOSFET), losses from driving the capacitance in the schottky diode, losses from operating the control circuitry in the chip, losses due to the ripple current in the capacitors, and several other minor loss factors. But for know, we will keep it basic and simple.
Remember how we wanted 2.54W on the input? Well if I start off with a 123 cell that has 3.2V, remembering that Power = Voltage * Current, swapping things around Current = Power/Voltage, so the input current needed is 2.54W/3.2V= 0.79 Amps. Guess what, you have less amps flowing in the input MOSFET, and the losses here drop down to only 2.5%. If the batteries had the same internal resistance, this lower input current necessary, would also mean you'd have less input voltage drop due pulling less current out of the battery. Then the 123 Lithium battery typically has a rather flat discharge curve, so the current drawn towards the end wouldn't go as high, both from the less internal loss and from the higher voltage at the end of battery life.
It's late, and I hope you were able to follow things through all of this.
Now, lets look at things further. Most Luxeon LEDs are around 3.9V when you have 1 Amp flowing in them. P=I*V (I is Current), so in reality, we need 3.9 Watts.
For a moment, imagine you had an imaginary unrealistic absolutely losses converter.
Under load, my two NiMH cell voltages added up to 2.380 Volts. Okay, so how much input current do we need to produce 3.9W? Power/voltage= current 3.9W / 2.38V = 1.64 Amps. Humm, I go back and measure my NiMH voltage at 1.64 Amps and find that the Output voltage of the two NiMH cells drops to 2.28 Volts. Ouch. So I go back and take 3.9W/2.28V= 1.7A I'll need to pull from the batteries. So I again go back and measure the cells, this time at 1.7A, and I find the output voltage of the cells drops to 2.268 Volts. Going back and doing back and repeating again, 3.9W/2.268 Volts = 1.72A, and we get to 2.2648V. Okay, lets stop here.
The next major loss in the circuit is most likely the inductor, but we won't go into things that deep right now.
Hopefully you get the basic idea...when you start with a higher voltage, you need less current on the input, and the losses in the converter (for a boost) will be less. Basically a win-win type thing.
[ QUOTE ]
greenLED said:
[ QUOTE ]
dat2zip said:
In fact I have a derivative of the Nexgen that will do more than 1A from a single 123 cell and it is fully regulated constant current design.
[/ QUOTE ]
/highjack alert!/
Also, am I understanding correctly that driving a circuit at 3v with 1x123 is "better" (whatever that means electronically) than using a 2xAA config? Can someone please explain why this is so, or point to a source I could check? I'm trying to educat myself on the electronix world.
Thanks!
/Highjack off/
[/ QUOTE ]
greenLED
Lets take an imaginary converter.
Consider the sag of the NiMH under high loads and also the 123 cell. Think about the internal resistance of the battery. Then think about how the cell voltage sags over time under load. Consider the starting voltage. Remember power in is voltage * current = power. You want power.
But remember there is resistance in the MOSFET/Transistor to ground. Higher voltage means you need less current for the same power. The majority of loss in the MOSFET is due to Power = Current^2 * resistance. So, a small decrease in input voltage means you need more current, but the losses in the mosfet are due to Current being squared, so it takes it's toll quickly.
The same type of thing happens as the battery sags.
In every battery, you will find some internal resistance. This causes the cell voltage to sag such as when you put a 1 Amp load on it. A NiMH that I have here charged two days ago. When it isn't under load, the voltage across it reads 1.287V. When I put a 1 Amp load on it, it's output voltage drops to 1.190V. If I was utilizing two of these, the voltage would be 2.574 Volts, and under a 1 Amp load, the voltage would drop to 2.380 Volts, and they would sag from this over time as they were drained.
But, if I set everything up, such that I needed 2.574V on the input at 1A for 2.54W going in, and then the voltage dropped to 2.38V, then I'd need to draw 1.07A to maintain the same power going in. As the cell is used, the battery voltage continues to drop. Lets look at the point As the cells drain down to their 1.8V ( 0.9V each) cut-off point, I'd need to draw 1.41 Amps to maintain a flat light output level down to the battery cut-off point.
Now, let say your input MOSFET had 0.1 ohms. With 2.54V off the battery and 1A flowing into the input, there would be Power = Current^2 * resistance loss, or 0.1 Watt of loss. Keep in mind we are not including the output MOSFET/schotty diode loss or the power needed to drive the MOSFET gate, nor the power needed to make the control circuits function. So then we take the power lost and divide it by our input power. 0.1W/2.54W= 0.04 or 4% loss due to the resistance just in the input MOSFET.
Now, when we drop on down to 1.8V at 1.41A, things change.
Power = Current^2 * resistance. The MOSFET had 0.1 ohms of resistance. So 1.41^2 * .1 = 0.20 Watts of loss. We have 2.54W input, so the power loss in the MOSFET due resistance/input power, or 0.20W/2.54W = 0.0787 or 7.87 % loss.
Then we go on to the schottky diode, such as used in the BB and these Zetex converters, on the output side of things. There are some super low loss schottky diodes that will have a Vf of only 0.3V when 1A flows through them. Well, power also equals Power = Voltage * Current. Since we are always putting out 1A in a constant currrent converter, this loss is easy to figure. 0.3V * 1A = 0.3W So, we had 2.54W, so we take this 0.3W/2.54W = 0.031 or 3.1% loss.
So with a new battery, only accounting for just two of the loss factors (there are many more), we'd have 7.1% losses with a fresh cell, and 10.97% losses near the cut-out.
You still need to add in losses due to the inductor DC resistance, eddy current losses in the inductor, B-H losses in the inductor, power lost driving the gate charge in the MOSFET, power lost in the circuit that drives the gate, losses due to charging and discharging the MOSFET body diode (internal to the MOSFET), losses from driving the capacitance in the schottky diode, losses from operating the control circuitry in the chip, losses due to the ripple current in the capacitors, and several other minor loss factors. But for know, we will keep it basic and simple.
Remember how we wanted 2.54W on the input? Well if I start off with a 123 cell that has 3.2V, remembering that Power = Voltage * Current, swapping things around Current = Power/Voltage, so the input current needed is 2.54W/3.2V= 0.79 Amps. Guess what, you have less amps flowing in the input MOSFET, and the losses here drop down to only 2.5%. If the batteries had the same internal resistance, this lower input current necessary, would also mean you'd have less input voltage drop due pulling less current out of the battery. Then the 123 Lithium battery typically has a rather flat discharge curve, so the current drawn towards the end wouldn't go as high, both from the less internal loss and from the higher voltage at the end of battery life.
It's late, and I hope you were able to follow things through all of this.
Now, lets look at things further. Most Luxeon LEDs are around 3.9V when you have 1 Amp flowing in them. P=I*V (I is Current), so in reality, we need 3.9 Watts.
For a moment, imagine you had an imaginary unrealistic absolutely losses converter.
Under load, my two NiMH cell voltages added up to 2.380 Volts. Okay, so how much input current do we need to produce 3.9W? Power/voltage= current 3.9W / 2.38V = 1.64 Amps. Humm, I go back and measure my NiMH voltage at 1.64 Amps and find that the Output voltage of the two NiMH cells drops to 2.28 Volts. Ouch. So I go back and take 3.9W/2.28V= 1.7A I'll need to pull from the batteries. So I again go back and measure the cells, this time at 1.7A, and I find the output voltage of the cells drops to 2.268 Volts. Going back and doing back and repeating again, 3.9W/2.268 Volts = 1.72A, and we get to 2.2648V. Okay, lets stop here.
The next major loss in the circuit is most likely the inductor, but we won't go into things that deep right now.
Hopefully you get the basic idea...when you start with a higher voltage, you need less current on the input, and the losses in the converter (for a boost) will be less. Basically a win-win type thing.
