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From the datasheet on the EP4CE6 under Power Supply Guidelines, it looks like it requires a switching regulator to produce the +1.2V? Can I use a linear regulator to do it, or why is it not recommended?
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You can use a linear regulator, but you will need to check the power dissipation of that regulator.
For example, the DE0-nano uses a linear regulator to create 1.2V from 5V (the USB supply). The attached plot shows the current drawn by the board for designs containing toggle registers (1-bit counters), 4-, 8-, and 16-bit counters. The worst-case current is a little under 500mA.
The power dissipated by a linear regulator with 5V input, 1.2V output, and 500mA load current is;
Pdissipated = (5V - 1.2V) x 500mA = 1.9W
The DE0-nano uses LP38500 linear regulators.
http://www.ti.com/lit/ds/symlink/lp38500-adj.pdf The DE0-nano uses LLP-8 packaged devices which have a thermal resistance of 80C/W.
For 1.9W power disspation, the linear regulator die temperature would rise 1.9 x 80 = 152C, and would shut-down.
To avoid thermal shutdown, the DE0-nano design uses two stages of linear regulators for the 1.2V; 5V to 3.3V (0.85W dissipation) and 3.3V to 1.2V (1.05W dissipation). So the linear regulator die temperature rise by about 80C. This is about as high as you'd ever want to go.
What is the efficiency of such a design;
Pin = 5V x 500mA = 2.5W
Pout = 1.2V x 500mA = 600mW
Efficiency = Pout/Pin = 24%
Pdissipated = 2.5W - 600mW = 1.9W
Now, how about a switching supply?
I've attached an LTspice design for the LTC3600 switch-mode controller for converting 5V to 1.2V at up to 1.5A. The efficiency is around 88%, i.e., for an output power of 600mW, the input power is 600mW/0.88 = 681mW, so the power dissipated by the regulator is 81mW.
Quite a difference isn't it?
Cheers,
Dave
Honored Contributor
