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Hi Russ and Brian,
I remember that the LCBPS/LCDPS "performance" can be optimized by changing the value of the resistor in the C-L-R-C circuit when you know what current you will be drawing. I just cannot locate the "how to calculate".
I have some low current applications in mind for these regs and if any of you could point me in the right direction I would be very happy (and possibly order a few more!)
Cheers,
Nic
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I have had good success aiming for about 1-2V drop across the resistor. So if you know your load its pretty easy to calculate. Edited by user Thursday, February 17, 2011 5:50:27 PM(UTC)
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Since those resistors are in series, the voltage drop across them 2X(IxR) must be added to the required voltage headroom of the transformers. Suppose your current is 500 mA, the resistors are 10 ohm, therefore the voltage across them is 5V. Since they are two of them, you need to add 10V (DC) to the headroom of the transformers.
Exanple: you want 5V @ 500 mA at the output. The regulator takes 3V, the resistors at 10 ohm takes 2X5v or 10 volts and the bridge maybe 1V. Total DC headroom required: is 5V+3V+10V+1V=19V DC or 19/1.4 = 13.6V AC
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Yes good point always remeber to leave enough headroom for the regulator. Sometime I forget things that might be obvious to me may not be to others. Thanks glt.
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Actually slight correction in the latest version of LCBPS there is only one series resistor per rail.
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glt wrote:Exanple: you want 5V @ 500 mA at the output. The regulator takes 3V, the resistors at 10 ohm takes 2X5v or 10 volts and the bridge maybe 1V. Total DC headroom required: is 5V+3V+10V+1V=19V DC or 19/1.4 = 13.6V AC Thanks glt for the well explained example, Let me see if I get the calculations right (for one theoretical single resistor rail): I will be drawing 200 mA @ 15V => 2V drop over 10 ohm (which is fine according to Russ). Required DC: 15(load) +3(regulator) +2(resistor)+1(bridge) = 21V DC = 21/1.4 = 15V AC => i.e. minimum a 15V, 3VA transformer secondary. With low current loads (say 50 mA) on may consider increasing the resistor to 2V/0.05A = 40 ohm With high current loads (say 500 mA) on may consider decreasing the resistor to 2V/0.5A = 4 ohm Changing the resistor value according to the actual load current will maintain the requirement to transformer voltage constant and keep the noise at a minimum. Correct? Russ, What is the value of the resistor supplied with the latest version of LCBPS (the BOM lists 4 pcs of 2.7R which I assume refer to the "old" version)? Cheers, Nic
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I think its the same value.
Keep in mind the CRC is effective even at lower voltage drop. So I would always use the lowest practical value.
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Hi Nic,
The limitation of the CLRC filter is the power dissipation of the resistor.
In your example, you need .2A
P=VI; V=IR. P=I2R. If the resistance is 10 ohm, then P= .2 x .2 x 10= .4 Watts. If you are pulling .5A, then P=2.5Watts...
The calculated headroom is the minimum required for the transformer. If your transformer supplies higher voltages, then the "excess" will be dropped across the regulator which is designed for such task.
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.5A accross the 10R would likely cause a drop that would make the regulator sag, depending on the trafo. But you are correct glt (as usual) :), one must be sure to keep that resistor within its spec.
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