Absolute Minimum Voltage
Electronegativity
Posts: 311
Hello, I am thinking about a tiny coin battery powered project based on the SX20.
The data sheets mention 2.7V, but the traces of power consumption versus voltage seem to terminate around 2.2V.
How low can you go?
Also, how important is a voltage regulator?
Would it cause a problem if I ran an SX20 directly off a 3V coin battery with no intervening regulator?
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I wonder if this wire is hot...
The data sheets mention 2.7V, but the traces of power consumption versus voltage seem to terminate around 2.2V.
How low can you go?
Also, how important is a voltage regulator?
Would it cause a problem if I ran an SX20 directly off a 3V coin battery with no intervening regulator?
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I wonder if this wire is hot...
Comments
I think 2.7V is the minimum for all conditions (temperature ect..).
Bean.
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"SX-Video·Module" Now available from Parallax for only $28.95
http://www.parallax.com/detail.asp?product_id=30012
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"I'm a man, but I can change, if I have to, I guess"
Red Green
·
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I wonder if this wire is hot...
At 3mA regulator leakage for the 78L05 ( small ) regulator, the coin battery will finish in less than 24 hours.
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·1+1=10
Just curious, what are you trying to do at sub-3V and presumably low current?
How much current are you expecting to deliver out of the system, and for how long?
Perhaps this is a question of "Absolute Minimum Current" conservation rather than voltage.
I once designed a data logger around a 16F84 that·was configured in·RC oscillator
mode running at about 10kHz ... essentially a "functioning sleep mode" (FSM)... to
conserve power.· I don't see why something similar couldn't be applied to an SX.
When an event occurred one of the I/O pins would switch in a parallel resistor
to the RC oscillator increasing the speed to about 4MHz.· At this point other devices
were powered up... ·RTC (Real Time Clock) and Memory to "log" the event.· Once
a time stamp and·event status was logged into memory, everything would shut down
and go back to the 10kHz FSM mode.· Battery life from a 3V lithium (CR2032) was about
4 months.
In order for something like a data logger to work under these conditions, we need to
know more information about what you are trying to accomplish.
·
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Beau Schwabe
IC Layout Engineer
Parallax, Inc.
I am thinking of building a "Micro Simon Sez".
It will be an SX20 mounted on one of those Sparkfun boards with blue, red, green, and yellow LEDs around it.
The processor will generate a table of random numbers from 1 to 4 and flash the digits in progressively longer sequences.
The user will press colored tactile switches according to their memory of the pattern.
I want to make it as small and light as possible, which makes a 3V coin battery ideal.
The blue and green LEDs can run off 3V, and I am going to take advantage of the voltage drop across diodes to step the voltage down for the red and yellow.
The SX will trigger transistors to light the diodes so as to aviod sourcing too much current since each diode can draw 20 to 30 mA and I want to put 2 of each color per side.
William: Thanks for the advice, but since it is only a toy, the user (me) will shut it off when not in use so the processor will never actually be in sleep mode.
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I wonder if this wire is hot...
Post Edited (Electronegativity) : 9/28/2005 3:45:22 PM GMT
Bean.
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"SX-Video·Module" Now available from Parallax for only $28.95
http://www.parallax.com/detail.asp?product_id=30012
Product web site: www.sxvm.com
Available now... SX-Video OSD module $59.95 www.sxvm.com
"I'm a man, but I can change, if I have to, I guess"
Red Green
·
With all diodes continuously lit the battery would drain in an hour but it will be flashing them one at a time, so that's 8 hours.
I get a little more time because nothing is lit while it's waiting for user input.
there is also room to sacrifice some brightness for longer battery life since these are very bright LEDs (between 800 to 1500 mcd).
I don't really know what mcd stands for, but its probably something like millicandlepower.
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I wonder if this wire is hot...
Here is an excerpt from http://www.electro-optical.com/whitepapers/candela.htm
Definition:
The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency 540
I remembered the steradian part but couldn't dredge up the 1/683 Watts.
The inclusion of 555 nm in the definition is a bit strange though.
It works fine for black body radiation, but technically a 632 nm HENE laser would always have zero candlepower no matter how bright the beam is!
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I wonder if this wire is hot...
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·1+1=10
Do you really want the LED's to be that bright?· There are some very impressive 2mA LED's that put out lots of light.
Anyway, there are some other ways to reduce the amount of current without any external parts, by making use of
the internal RC oscillator settings of the SX you can have dramatic results.
I did not have a SX20, but I did have a SX28.· The datasheet indicates that they should be the same.
Here is the DATA I obtained by placing a 10 Ohm resistor in series with the power to the SX and measuring the voltage
across the resistor.
The program was just a simple program to toggle one of the I/O pins at 1/10th the clock frequency.
·
Tris_A = %00000000 'Make ALL pins on PortA output This is because we just don't want any pins to float. Tris_B = %00000000 'Make ALL pins on PortB output Floating pins EAT current. Tris_C = %00000000 'Make ALL pins on PortC output Asm Main: MOV RB.0,#1 '2 Clocks NOP '1 Clock NOP '1 Clock NOP '1 Clock (5) Clocks "ON" MOV RB.0,#0 '2 Clocks JMP Main '3 Clocks (5) Clocks "OFF" EndAsm▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
Beau Schwabe
IC Layout Engineer
Parallax, Inc.
It will be especially useful for breadboard prototypes where the 10 ohm resistor can be easily swapped in and out.
I'm a little confused by your example though.
You say it will toggle one of the pins at 1/10 the clock frequency, but it looks from the code like it would actually toggle at 1/2 the clock frequency.
Also, which 2mA LED's are you referring to?
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I wonder if this wire is hot...
Post Edited (Electronegativity) : 9/28/2005 11:59:42 PM GMT
The code is arbitrary, I just wanted something going on inside the SX.
But it does come in handy for measuring the frequency without tapping
off of the OSC pins. The code divides the input clock pulse by 10, because
it takes a total of 10 clock pulses to get through the code. 5 of the clock
pulses are spend with RB0 High and the other 5 clock pulses are spent with
RB0 in the Low state.
As far as the 2mA LED's let me look into that. Digikey has some, but I don't
have any hands on experience.
http://www.digikey.com/
Key words "LED 2mA"
I will try to find the manufacturer of some 2mA LED that I do have.
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Beau Schwabe
IC Layout Engineer
Parallax, Inc.
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Tracy Allen
www.emesystems.com
Last night it occurred to me... The 2mA LEDs that I have are about normal brightness (what I would expect on an appliance) ...
...BUT If I use a super bright LED and supply a minimal amount of current, it it's still going to be VERY bright. Doing just that on a
red LED with a forward voltage of the LED measuring 1.65V (<-- Diode check on a DMM) and a fresh 9V battery measuring 9.23V ,
I used a current limiting resistance of 2K (two 1K's in series). With only 3.79mA of current to the LED and lights out, I could see
the projection of the LED light on the ceiling. The LED was aiming straight up from my work bench.
I_to_LED = (Vsupply - LEDforward) / CurrentLimitResistor
I_to_LED = (9.23V - 1.65V) / 2000Ohms
I_to_LED = 7.58V / 2000Ohms
I_to_LED = 3.79mA
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Beau Schwabe
IC Layout Engineer
Parallax, Inc.
And if you try to do it with any bit other than 0 you will get an error. "MOV RB.1,#1" gives an error.
Bean.
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"SX-Video·Module" Now available from Parallax for only $28.95
http://www.parallax.com/detail.asp?product_id=30012
Product web site: www.sxvm.com
Available now... SX-Video OSD module $59.95 www.sxvm.com
"I'm a man, but I can change, if I have to, I guess"
Red Green
Post Edited (Bean (Hitt Consulting)) : 9/29/2005 3:20:14 PM GMT
What if you used "SETB RB.0" instead?
Of course you would have to add a NOP behind it in Beau's example to keep the number of clock cycles the same.
Beau, I think I will end up doing something like that if only to equalize the (perceived) intensity of the different colored diodes.
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I wonder if this wire is hot...
Main
setb RB.0 ; 1 LED is on after executing this instruction (total off time: 5 cycles)
nop ; 1
nop ; 1
nop ; 1
nop ; 1
clrb RB.0 ; 1 LED is off after executing this instruction (total on time: 5 cycles)
nop ; 1
jmp Main ; 3
As power consumption is an issue, I assume that you clock the SX at a low rate. But even when you use the 32 kHz internal clock, an on/off cycle will be as short as 312.9 µs, i.e. the LED will flash at a rate of 3.2 kHz which is way too fast. Well, yes I know, Beau provided this sample code to measure the power consumption at different clock rates - it was not intended to blink LEDs.
Nevertheless, after seeing this example an idea came up to my mind: Reducing the LED current by applying a PWM signal.
Beau has calculated his results based upon the voltage of a fresh 9V battery. Let me re-do it just for this example based upon a 5V supply, and an LED current of 3mA:
R = (5V - 1.65V) / 3mA = 1.12 kOhm
Now, let's calculate the power balance:
Total power consumption is
5V * 3mA = 15mW
The LED converts
Pled = 1.65V * 3mA = 4.95mW into light and some heat, but the majority of
Pres = (5V - 1.65V) * 3mA = 10.05mW (or 15mW - 4.95mW) is dissipated into heat by the resistor.
Now, let's assume, you generate a square wave with a 50% duty cycle on the LED-driving output pin as long as the LED shall be on. To make it simple, you can assume that the 50% duty cycle has the same effect as reducing the supply voltage by 50%, i.e. down to 2.5V.
The results are now as follows:
R = (2.5V - 1.65V) / 3mA = 283.3 Ohm
Pled = 1.65V * 3mA = 4.95mW (same as before)
Pres = (2.5V - 1.65V) * 3mA = 2.55mW
As you can see, this is a much better energy balance. You could even optimize it by selecting a duty-cycle of less than 50% for the output signal for lower resistor values. Nevertheless, you should keep in mind that reducing the resistor value means higher peak currents to be sourced by the output pin, which is limited to an absolute maximum of 45mA according to the SX28 datasheet.
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Greetings from Germany,
Günther
I totally overlooked the energy dissapated by the resistor.
I was going to incorporate 4 switches that use a 10K resistor to an input pin and have the switch draw the trickle current to ground when pressed. It seems from what you wrote that this would continuously drain power by heating the resistors.
I must admit though that I don't completely understand the second part of your post.
If you generate a 50% duty cycle then that is the same as putting 5V through the resistor for half as much time, so why isn't the power dissapated 1/2 as much (5.025 mW)?
Where did the extra factor of 2 come from?
I could see if you used capacitors to smooth out the square wave that you might be able to generate an actual 2.5V supply, but it seems to me that this is a 5V supply that is on 50% of the time.
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I wonder if this wire is hot...
The same current through half the resistance is half the power.
Bean.
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"SX-Video·Module" Now available from Parallax for only $28.95
http://www.parallax.com/detail.asp?product_id=30012
Product web site: www.sxvm.com
Available now... SX-Video OSD module $59.95 www.sxvm.com
"I'm a man, but I can change, if I have to, I guess"
Red Green
Post Edited (Bean (Hitt Consulting)) : 9/29/2005 5:48:08 PM GMT
RB.0 = 1 and RB.1 = 0 will generate SETB and CLRB in SX/B.
Just make sure the pin is setup as an output first.
Bean.
▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
"SX-Video·Module" Now available from Parallax for only $28.95
http://www.parallax.com/detail.asp?product_id=30012
Product web site: www.sxvm.com
Available now... SX-Video OSD module $59.95 www.sxvm.com
"I'm a man, but I can change, if I have to, I guess"
Red Green
·
In the first part of Guenther's example the resistor os 1.12 kOhms, and in the second part it is 283.3 Ohms.
283.3/1120 = around 1/4 the value. If you also include the fact that the current is on for 1/2 as long then the power dissapation should be 1/8 as much.
?????
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I wonder if this wire is hot...
I'm not sure that you can relate the two numbers 283.3 and 1120 the way that you are referring to them.
( I could be having a dense moment though)
This ratio you see is originating from the voltage across the current limiting resistor in series with the LED.
Across one scenario, the voltage is 3.35V while in the other circumstance the voltage is 0.85V.
In BOTH cases the goal is to maintain 3mA across the LED.
Let me try to put it another way...
If there is a 100% Duty cycle we would assume 5V 100% of the time.
If there is a 50% Duty cycle we would expect an average of 2.5V 100% of the time.
If there is a 40% Duty cycle we would expect an average of 2V 100% of the time.
The normal formula for determining the Current limiting resistor for a LED is:
Rled = (Vsupply - Vled) / Iled
If we want to include Duty cycle, the formula becomes:
Rled = ( ( Vsupply x DutyCycle ) - Vled) / Iled
Basically what you are comparing between 283.3 and 1120 is the Duty Cycle and the resistor
required to maintain 3mA across the LED. See table below:
Note: -50 Ohms at 30% indicates a voltage less than the forward LED voltage ( LED is OFF)
Power in all cases will be 4.95mW across the LED since we are forcing the current to be 3mA.
Power across the Current Limit resistor will vary between 10.05mW at 100% and 0.299mW at 35%
P = I * V
or
P = I^2 * R
For 100%...
P = 3mA^2 * 1116.7Ohms
P = 0.000009 * 1116.7Ohms
P = 10.05mW
For 35%
P = 3mA^2 * 33.3Ohms
P = 0.000009 * 33.3Ohms
P = 0.299mW
The goal here is to get as little current as possible across the resistor. Ideally, you would not even
have a resistor, and just generate a Duty Cycle that is equal to the forward voltage drop of your LED.
However in the real world TEMPERATURE is your enemy, and can cause the forward voltage of your
LED to slightly increase or decrease. Having a resistor here for a little "padding" can be your friend
in this case.
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Beau Schwabe
IC Layout Engineer
Parallax, Inc.
Post Edited (Beau Schwabe (Parallax)) : 9/30/2005 4:28:52 AM GMT
To calculate the resistor for the assumed 2.5V, I used this formula:
R = (2.5V - 1.65V) / 3mA = 283.3 Ohm
At first glance, you might think, reducing the voltage by 50% would also require reducing the resistor by 50%. But this is not the case because the LED's forward voltage is (almost) a constant of 1.65V. Thus, the voltage across the resistor is now 2.5 - 1.65V = 0.85V which is much lower than at 5V supply: 5 - 1.65V = 3.35V. Therefore, the resulting value is less than 50%.
Well, what if we would reduce the duty cycle to achieve an effective voltage of 1.65V? According to my formula, the resistor would be 0 Ohms in this case, i.e. no loss at all. But you should keep in mind that this would cause high peak currents to be sourced from the output, so you better use a resistor of no less than 120 Ohms. This limits the peak current to some value below 30mA which is a safe value for the SX.
If the LEDs are still bright enough, you can try to further reduce the duty cycle down to a value that still results in an acceptable brightness (w/o further reducing the resitor value).
Yes, you are right, in case you have connected 10kOhm pull-up resistors between the inputs and +5V, each of the resistors will dissipate 2.5 mW as long as the switches pull the inputs low. You might consider using pushbuttons instead, and handle the toggle functions by software. In this case, power is dissipated only as long as the buttons are pressed. Don't forget to de-bounce the pushbuttons. You can do this by software. For example, when you poll the button states -say - every 20ms, you accept a state change ony, when the state is still the same after the next poll.
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Greetings from Germany,
G