- Mon Jun 09, 2008 5:41 pm
#49944
Greetings Andy,
is eight times the average. For 100mA peak the total
resistance is (in words) "Supply minus LED drop, divided
by desired peak current".
(5 - 2.2)/0.1 = 28 ohms.
But, there is equivalent resistance in the FETs so the
next step is to find "28 ohms - Fet resistance".
The TPIC is 7 ohms, the other FET is 200mR and can be
ignored. So 21 ohms (use 22 ohms standard value).
I Spiced the circuit and looked at a 5% loss in supply
voltage and worst case tolerance on the other components,
that brought me to 15 Ohms. You can pick the mean
average and use 18 Ohms. 15 to 22 Ohms is not going to
make much difference, and is within Abs Max for all
components.
Comments Welcome!
arader wrote:Remember the LEDs are 8:1 MUX'd, so the peak currentbigglez wrote:(1) The LED ballast resistors (100 ohms) will only yieldThanks Peter! quick question though: how did you calculate these numbers? I would have thought at 100 ohms I'd have I = 5v / 100ohms = 50mA (still not enough however).
27mA peak for the LEDs. A value of 15 ohms will put
you at 106mA peak (desired).
is eight times the average. For 100mA peak the total
resistance is (in words) "Supply minus LED drop, divided
by desired peak current".
(5 - 2.2)/0.1 = 28 ohms.
But, there is equivalent resistance in the FETs so the
next step is to find "28 ohms - Fet resistance".
The TPIC is 7 ohms, the other FET is 200mR and can be
ignored. So 21 ohms (use 22 ohms standard value).
I Spiced the circuit and looked at a 5% loss in supply
voltage and worst case tolerance on the other components,
that brought me to 15 Ohms. You can pick the mean
average and use 18 Ohms. 15 to 22 Ohms is not going to
make much difference, and is within Abs Max for all
components.
Comments Welcome!
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