I only ever used fet drivers. Can someone give me the run down on buck drivers? How do increase/ decrease current?
Lagman put together this post to explain the differing driver types.
Thank you. That helped understand them. Still confused on how increase and decrease amps on these things though. Maybe tomorrow ill disect the entire post to see if theres anything about it.
Buck, and it’s sibling Boost, are constant power devices.
Power=Current*Voltage or more commonly stated as P=I*V
So, say if I have a 3V LED @ 1.5A, so:
So the LED needs 4.5Watt. To do the example, it is easier first to assume efficiency=100%, we will add that in later.
If you are bucking, say from 4.5Volt down to 3V for the LED, you need 4.5Watt output,
so you need to put in 4.5Watt.
4.5W = ?A * 4.5V
obvious the ? is a 1.
In other words, to get 4.5Watt out of 4.5Volt, you need merely 1Amp. So you can see with buck, you put in a higher voltage and you get more current, in this case. So 4.5V@1A buck-in for 3V@1.5A buck-out. Your current output is increased by 50%.
The math is, doubling the voltage in and you halve the current needed. (Assuming 100% efficient)
If you are boosting say from 1.5V to 3V for your LED, you need 4.5W output
so you need to put in 4.5Watt
4.5W = ?A * 1.5V,
obvious the ? is a 3.
So here, to get 4.5W out of 1.5V, you need 3Amp. So, 1.5V@3A boost-in for 3V@1.5A boost-out. With boost, you get more voltage but you need more current from the input.
Obviously, efficiency is never going to be 100% (Laws of Thermodynamic). So, to get 4.5W out, you need to put in more than 4.5W. So in the above buck example, you need more than 1A@4.5V to get 1.5A@3V. Typically, you can expect buck to be 85-95% efficient.
With boost, the more you are boosting, the worst the efficiency. 60% would be very good efficiency if you are boosting 1.5V to 3V at 3Amp.
Hope this helps
Back when I was learning electronics theory it was P = I x E
I think they use E for “Electronic Voltage”. Either way, it is the voltage.
There’s usually a small value sense resistor on the output side. The boost ic uses the voltage drop across this resistance to calculate/maintain a constant current. Changing the resistor can raise or lower the set current.