"rickman" <spamgoeshere4@yahoo.com> wrote in message news:40BFF1D7.51910938@yahoo.com...> Spehro Pefhany wrote: > > > > On Wed, 02 Jun 2004 19:07:44 -0400, the renowned rickman > > <spamgoeshere4@yahoo.com> wrote: > > > > > >What voltage do you measure on the LED with a current of 4 uA? > > > > 1.0 V @ 0.06uA > > 1.94V @ 1uA (visible in dim light) > > 2.23V @ 4uA > > 2.59V @ 100uA (quite visibly lit) > > 2.79V @ 1mA > > 3.08V @ 10mA (blindingly bright) > > What type of LED is this? I have never seen one that was "blindingly > bright" at 10 mA and I have not seen one with a 3 volt Vf. Can you test > a standard red LED that is designed to operate at 10 to 20 mA?Never seen a superbright blue led? They ARE blindingly bright at 10mA and almost hurt your eyes at 20mA. And yes, blue LEDs have a Vf of 3V Meindert
LED & Resistor befuddlement
Started by ●June 2, 2004
Reply by ●June 4, 20042004-06-04
Reply by ●June 4, 20042004-06-04
On Thu, 03 Jun 2004 23:51:51 -0400, the renowned rickman <spamgoeshere4@yahoo.com> wrote:>Spehro Pefhany wrote: >> >> On Wed, 02 Jun 2004 19:07:44 -0400, the renowned rickman >> <spamgoeshere4@yahoo.com> wrote: >> > >> >What voltage do you measure on the LED with a current of 4 uA? >> >> 1.0 V @ 0.06uA >> 1.94V @ 1uA (visible in dim light) >> 2.23V @ 4uA >> 2.59V @ 100uA (quite visibly lit) >> 2.79V @ 1mA >> 3.08V @ 10mA (blindingly bright) > >What type of LED is this?Super-bright green (525nm) 5mm dia. water-clear narrow angle lens. Common types of blue and white LEDs are probably in the same ballpark.>I have never seen one that was "blindingly >bright" at 10 mA and I have not seen one with a 3 volt Vf. Can you test >a standard red LED that is designed to operate at 10 to 20 mA?Yes, I can, but my purpose was to test one that was more unlikely to work, since Mike indicated he wanted to use a range of LEDs. In another post, I presented some typical numbers with a high quality GaAlAs super-bright red LED and indicated there was some margin in that particular case, even with the "measurement" at high current. However, it's LED parameter-sensitive and the manufacturer of the LEDs generally does not guarantee the drop at low current. Your circuit with the extra resistor would probably be acceptable for production if the LED type could be constrained to certain types of red, and the input could be constrained to TTL levels. An ultra-bright red LED that leaves spots in my eyes at 15-20mA drops just under 2V typically at that current. Of course the LED Vf has a temperature coefficient and a unit-to-unit variation. Ony a few manufacturers give you even typical values. <preach> As an EE I'm typically more concerned with ensuring a very low probability that it doesn't work rather than checking that it often does work. The latter is potentially a most dangerous situation, IMHO, because cursory (or even non-targeted in-depth) tests can allow such a system to be shipped, leading to real problems down the road. </preach> Best regards, Spehro Pefhany -- "it's the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com
Reply by ●June 4, 20042004-06-04
Peter Jakacki wrote:> Mike Turco wrote: > >> I'm not clear on the concept. Is this the circuit? >> >> +5 >> | >> | >> 1k ohm >> | >> | >> LED >> | >> ----------------- I/O >> >> >> If the LED is not in the circuit, given the capacitance of a CMOS >> output, I >> can see the output staying at whichever state it is driven for long >> enough >> to be measured (State 1 & 2). >> >> 1) No LED. Output 0 & measure back 0 >> >> 2) No LED. Output 1 & measure back 1 >> >> When you put the LED in the circuit, I guess I don't understand how it >> is to >> operate (State 3 & 4): >> >> 3) Has LED. Output 0 & measure back ? (Turns on LED) >> >> 4) Has LED. Output 1 & measure back ? >> >> Thanks, >> >> Mike >> >> > > This circuit fulfills your requirements and works just like you want it > to. Furthermore, you can use any value resistor practical. > > With LED in circuit > 1. Output logic low > Led draws current and output pin goes to within Vss > Parasitic capacitance of pin is now at this level > > 2. Release output (tristate) > Pin is now high impedance with parasitic input capacitance holding last > state. > > 3. very short delay > Resistor + led charges parasitic input capacitance up to within Vdd > > 4. Read input as logic high > > 5. Switch pin back to an output to whatever state you want. > > > With no LED in circuit > 1. Output logic low > Led draws current and output pin goes to within Vss > Parasitic capacitance of pin is now at this level > > 2. Release output (tristate) > Pin is now high impedance with parasitic input capacitance holding last > state. > > 3. very short delay > As there is nothing connected to input pin the parasitic input capacitor > will hold the last state (logic low). This may eventually float to some > indeterminate state if left in this condition. > > 4. Read input as logic low > > 5. Switch pin back to an output to whatever state you want. > > -- > Peter Jakacki >The above solution is a little flakey. If you like, add an external capacitor and a pull-down resistor. -- Joe Legris
Reply by ●June 4, 20042004-06-04
Meindert Sprang wrote:> > "rickman" <spamgoeshere4@yahoo.com> wrote in message > news:40BFF1D7.51910938@yahoo.com... > > Spehro Pefhany wrote: > > > > > > On Wed, 02 Jun 2004 19:07:44 -0400, the renowned rickman > > > <spamgoeshere4@yahoo.com> wrote: > > > > > > > >What voltage do you measure on the LED with a current of 4 uA? > > > > > > 1.0 V @ 0.06uA > > > 1.94V @ 1uA (visible in dim light) > > > 2.23V @ 4uA > > > 2.59V @ 100uA (quite visibly lit) > > > 2.79V @ 1mA > > > 3.08V @ 10mA (blindingly bright) > > > > What type of LED is this? I have never seen one that was "blindingly > > bright" at 10 mA and I have not seen one with a 3 volt Vf. Can you test > > a standard red LED that is designed to operate at 10 to 20 mA? > > Never seen a superbright blue led? They ARE blindingly bright at 10mA and > almost hurt your eyes at 20mA. > And yes, blue LEDs have a Vf of 3VOk, but I don't think the OP was working with superbright blue LEDs. Aren't they pretty pricey? -- Rick "rickman" Collins rick.collins@XYarius.com Ignore the reply address. To email me use the above address with the XY removed. Arius - A Signal Processing Solutions Company Specializing in DSP and FPGA design URL http://www.arius.com 4 King Ave 301-682-7772 Voice Frederick, MD 21701-3110 301-682-7666 FAX
Reply by ●June 4, 20042004-06-04
On Fri, 04 Jun 2004 11:56:50 -0400, the renowned rickman <spamgoeshere4@yahoo.com> wrote:>Meindert Sprang wrote: >> >> "rickman" <spamgoeshere4@yahoo.com> wrote in message >> news:40BFF1D7.51910938@yahoo.com... >> > Spehro Pefhany wrote: >> > > >> > > On Wed, 02 Jun 2004 19:07:44 -0400, the renowned rickman >> > > <spamgoeshere4@yahoo.com> wrote: >> > > > >> > > >What voltage do you measure on the LED with a current of 4 uA? >> > > >> > > 1.0 V @ 0.06uA >> > > 1.94V @ 1uA (visible in dim light) >> > > 2.23V @ 4uA >> > > 2.59V @ 100uA (quite visibly lit) >> > > 2.79V @ 1mA >> > > 3.08V @ 10mA (blindingly bright) >> > >> > What type of LED is this? I have never seen one that was "blindingly >> > bright" at 10 mA and I have not seen one with a 3 volt Vf. Can you test >> > a standard red LED that is designed to operate at 10 to 20 mA? >> >> Never seen a superbright blue led? They ARE blindingly bright at 10mA and >> almost hurt your eyes at 20mA. >> And yes, blue LEDs have a Vf of 3V > >Ok, but I don't think the OP was working with superbright blue LEDs. >Aren't they pretty pricey?<35 cents in small quantity for 3,000 mcd blue. Best regards, Spehro Pefhany -- "it's the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com
Reply by ●June 4, 20042004-06-04
Care to quantify flakey???? As I stated, this solution works, is repeatable, and is based on real-life component parametric behaviour. *Any fool knows that a rock can't burn Marco.* The original poster called for only two components, the led, and the resistor. You have just doubled his component count and left the solution open-ended by omitting details. Besides, where's the fun in doing things the easy way. Have fun in meeting the challenge and then afterwards the solution becomes the new "easy way", till next time... -- Peter Jakacki Joe Legris wrote:> Peter Jakacki wrote: > >> Mike Turco wrote: >> >>> I'm not clear on the concept. Is this the circuit? >>> >>> +5 >>> | >>> | >>> 1k ohm >>> | >>> | >>> LED >>> | >>> ----------------- I/O >>> >>> >>> If the LED is not in the circuit, given the capacitance of a CMOS >>> output, I >>> can see the output staying at whichever state it is driven for long >>> enough >>> to be measured (State 1 & 2). >>> >>> 1) No LED. Output 0 & measure back 0 >>> >>> 2) No LED. Output 1 & measure back 1 >>> >>> When you put the LED in the circuit, I guess I don't understand how >>> it is to >>> operate (State 3 & 4): >>> >>> 3) Has LED. Output 0 & measure back ? (Turns on LED) >>> >>> 4) Has LED. Output 1 & measure back ? >>> >>> Thanks, >>> >>> Mike >>> >>> >> >> This circuit fulfills your requirements and works just like you want >> it to. Furthermore, you can use any value resistor practical. >> >> With LED in circuit >> 1. Output logic low >> Led draws current and output pin goes to within Vss >> Parasitic capacitance of pin is now at this level >> >> 2. Release output (tristate) >> Pin is now high impedance with parasitic input capacitance holding >> last state. >> >> 3. very short delay >> Resistor + led charges parasitic input capacitance up to within Vdd >> >> 4. Read input as logic high >> >> 5. Switch pin back to an output to whatever state you want. >> >> >> With no LED in circuit >> 1. Output logic low >> Led draws current and output pin goes to within Vss >> Parasitic capacitance of pin is now at this level >> >> 2. Release output (tristate) >> Pin is now high impedance with parasitic input capacitance holding >> last state. >> >> 3. very short delay >> As there is nothing connected to input pin the parasitic input >> capacitor will hold the last state (logic low). This may eventually >> float to some indeterminate state if left in this condition. >> >> 4. Read input as logic low >> >> 5. Switch pin back to an output to whatever state you want. >> >> -- >> Peter Jakacki >> > > The above solution is a little flakey. > > If you like, add an external capacitor and a pull-down resistor. > > -- > Joe Legris >
Reply by ●June 5, 20042004-06-05
"Spehro Pefhany" <speffSNIP@interlogDOTyou.knowwhat> schreef in bericht news:beovb013ctqsrcr7pk7uj0qjd4g9cmlb6r@4ax.com...> On Wed, 02 Jun 2004 19:07:44 -0400, the renowned rickman > <spamgoeshere4@yahoo.com> wrote: > > >What voltage do you measure on the LED with a current of 4 uA? > > 1.0 V @ 0.06uA > 1.94V @ 1uA (visible in dim light) > 2.23V @ 4uA > 2.59V @ 100uA (quite visibly lit) > 2.79V @ 1mA > 3.08V @ 10mA (blindingly bright)How did you measure the 1.0V @ 0.06 uA ? -- Thanks, Frank. (remove 'x' and 'invalid' when replying by email)
Reply by ●June 5, 20042004-06-05
On Sat, 5 Jun 2004 14:04:26 +0200, the renowned "Frank Bemelman" <f.bemelmanx@planet.invalid.nl> wrote:>"Spehro Pefhany" <speffSNIP@interlogDOTyou.knowwhat> schreef in bericht >news:beovb013ctqsrcr7pk7uj0qjd4g9cmlb6r@4ax.com... >> On Wed, 02 Jun 2004 19:07:44 -0400, the renowned rickman >> <spamgoeshere4@yahoo.com> wrote: >> >> >What voltage do you measure on the LED with a current of 4 uA? >> >> 1.0 V @ 0.06uA >> 1.94V @ 1uA (visible in dim light) >> 2.23V @ 4uA >> 2.59V @ 100uA (quite visibly lit) >> 2.79V @ 1mA >> 3.08V @ 10mA (blindingly bright) > >How did you measure the 1.0V @ 0.06 uA ?1M resistor in series, HP34401A set to DCV & >10G ohm input impedance, measure voltage across LED and increase supply until 1.0V across LED, switch to measure voltage across 1M resistor. Best regards, Spehro Pefhany -- "it's the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com
Reply by ●June 6, 20042004-06-06
"Spehro Pefhany" <speffSNIP@interlogDOTyou.knowwhat> schreef in bericht news:ite3c0dhcs3clmfifacck9s7vho23oet1m@4ax.com...> On Sat, 5 Jun 2004 14:04:26 +0200, the renowned "Frank Bemelman" > <f.bemelmanx@planet.invalid.nl> wrote: > > >"Spehro Pefhany" <speffSNIP@interlogDOTyou.knowwhat> schreef in bericht > >news:beovb013ctqsrcr7pk7uj0qjd4g9cmlb6r@4ax.com... > >> On Wed, 02 Jun 2004 19:07:44 -0400, the renowned rickman > >> <spamgoeshere4@yahoo.com> wrote: > >> > >> >What voltage do you measure on the LED with a current of 4 uA? > >> > >> 1.0 V @ 0.06uA > >> 1.94V @ 1uA (visible in dim light) > >> 2.23V @ 4uA > >> 2.59V @ 100uA (quite visibly lit) > >> 2.79V @ 1mA > >> 3.08V @ 10mA (blindingly bright) > > > >How did you measure the 1.0V @ 0.06 uA ? > > 1M resistor in series, HP34401A set to DCV & >10G ohm input > impedance, measure voltage across LED and increase supply until 1.0V > across LED, switch to measure voltage across 1M resistor.I guess I should get one of those ;) -- Thanks, Frank. (remove 'x' and 'invalid' when replying by email)
Reply by ●June 6, 20042004-06-06
"Spehro Pefhany" <speffSNIP@interlogDOTyou.knowwhat> wrote in message news:msrrb0hev8iruegeb28ii4qqaca6dkludf@4ax.com...> On Wed, 02 Jun 2004 11:55:43 GMT, Spehro Pefhany > <speffSNIP@interlogDOTyou.knowwhat> wrote: > > P.S. one more idea, specific to the PIC with ADC and requiring ZERO > parts beyond the LED and the resistor to set the current and just ONE > port pin. > > You can set digital/analog port pins to be a digital output and still > measure the voltage on that pin with the ADC. >What a nifty idea!! Now I'm interested to check what other uCs can operate this way!
