Pulsed LED driver vs. large resistor value, efficiency

Discussion in 'General Electronics Chat' started by hrs, Aug 16, 2014.

  1. hrs

    Thread Starter Member

    Jun 13, 2014

    I've modified a 555 circuit[1] I found online to drive a LED with pulses in
    order to reduce the brightness. I expected this to be more efficient than using
    a larger resistor value. An attempt to verify this produces confusing results.

    The 555 circuit is compared with just a LED and a resistor such that the LEDs
    in both circuits appear equally bright. Then I did some measurements of the
    current at the 9V battery with my digital multimeter and made a SPICE model of
    both. I don't think my DMM is a true-RMS kind so probably useless for this but
    I did it anyway.

    The SPICE results[2] show a pulse width of 0.0388 ms at 10.75 mA + a 3mA offset
    and a period of 1.4225 ms so I reccon the RMS current is 10.75 mA * 0.0388 ms /
    1.4225 ms = 0.29 mA not counting the offset. So that's where the numbers below
    come from.

    The results are as follows for the SPICE model, digital multimeter at direct
    current and at alternating current setting respectively.

    Code ( (Unknown Language)):
    1.                 SPICE                   DMM DC      DMM AC
    2. NE555           3 mA DC + 0.29 mA AC    4.65 mA     0.063 mA
    3. 33k resistor    0.25 mA                 0.2 mA      -
    1) Is there anything I can do with the 555 multimeter results or are they
    useless for this? I have no clue what to make of them.
    2) Why is the offset current (3mA) in the SPICE model so high? If an NE555 has
    a voltage divider made of 3 * 5k resistors wouldn't I get a leakage current of 9V
    / 15k ohm = 0.6mA?
    3) Is my interpretation correct that just a LED with a 33k resistor is more
    efficient than the 555 circuit?

    Ps. On my breadboard I put a 100n capacitor and a 2.2u polarised capacitor
    across Vcc and gnd of the 555 that are not present in the SPICE model as I
    thought it wouldn't matter for the idealised 555 that LTSpice has.

    [1] http://postimg.org/image/wem3qmll5/
    [2] http://postimg.org/image/bvf4t1zel/

    Edit: The original circuit can be found here:
    "Fig. 4.4.8 Improved Duty Cycle Control" at the bottom of the page.

    The SPICE model for the 555 circuit:
    Code ( (Unknown Language)):
    1. "ExpressPCB Netlist"
    2. "LTspice IV Version 4.21o"
    3. 1
    4. 0
    5. 0
    6. ""
    7. ""
    8. ""
    9. "Part IDs Table"
    10. "R1" "4.7k" ""
    11. "R2" "180k" ""
    12. "D1" "1N4148" ""
    13. "D2" "1N4148" ""
    14. "C1" "100n" ""
    15. "D3" "D" ""
    16. "U1" "NE555" ""
    17. "V1" "9" ""
    18. "R3" "820" ""
    19. "C2" "10n" ""
    21. "Net Names Table"
    22. "A" 1
    23. "N003" 5
    24. "N004" 7
    25. "C" 9
    26. "N002" 14
    27. "0" 16
    28. "B" 21
    29. "N001" 23
    30. "NC_01" 26
    32. "Net Connections Table"
    33. 1 1 1 2
    34. 1 2 1 3
    35. 1 6 1 4
    36. 1 7 3 0
    37. 2 1 2 6
    38. 2 3 1 0
    39. 3 2 2 8
    40. 3 4 2 0
    41. 4 3 2 10
    42. 4 4 1 11
    43. 4 7 2 12
    44. 4 7 6 13
    45. 4 10 1 0
    46. 5 5 1 15
    47. 5 7 5 0
    48. 6 5 2 17
    49. 6 7 1 18
    50. 6 8 2 19
    51. 6 9 2 20
    52. 6 10 2 0
    53. 7 6 2 22
    54. 7 9 1 0
    55. 8 7 4 24
    56. 8 7 8 25
    57. 8 8 1 0
    58. 9 7 7 0
    Last edited: Aug 16, 2014
  2. RichardO

    Well-Known Member

    May 4, 2013
    I don't think that pulsing the LED is going to save you much power. The brightness of an LED is pretty much proportional to the average current flowing through it. So, if the average pulsing current is the same as the DC current then the brightness will be the same.

    However, pulsing the LED does seem to appear brighter to the human eye. I don't have numbers on this but I don't expect there to be more than a few percent increase in apparent brightness when pulsing.

    A few more quick comments...

    First. The LTspice model you are using seems to be for the bipolar transistor version of the 555 timer such as the NE555.

    Second. The NE555 draws current for a lot of stuff inside. According to the data sheet, this ends up looking like about 2 K ohms (typical) resulting in about 4.5 mA at 9 volts.

    Third. A CMOS version of the 555 such as the ICM7555 or TLC555 draws about 50 uA at 9 volts. The idle current is mostly from the 100 k ohm (instead of 5 K) bias resistors.

    Fourth. So, what you really want to use is the CMOS version of the 555.

    Lastly. Don't forget to include the timing resistor current in your current draw calculations. If you a large value timing resistor then you can pretty much ignore this current. For instance, a 1 Meg ohm resistor would only add a few microamps to the current draw. (I would not go much over 1 meg ohm and I would never use an electrolytic capacitor for timing because of the high leakage current).
  3. Alec_t

    AAC Fanatic!

    Sep 17, 2013
    Unless your DMM is a true-RMS type the readings probably can't be relied on at the ~700 Hz pulse rate.

    Can you post the .asc file of the simulation?
  4. hrs

    Thread Starter Member

    Jun 13, 2014
    Are you refering to the 180k and 4.7k resistors and the idle current calculation? How would I account for them in the calculation?

    I've attached to .asc file.
  5. RichardO

    Well-Known Member

    May 4, 2013
    The power supply current specification for the 555 is for the idle condition -- that is with the chip is powered on but not oscillating. The idle current leaves out two factors. The first is the current used by the IC when charging and discharging internal capacitances at high frequency. You can ignore this at your low frequency. The other is the current used to charge the timing capacitor. This current is set by the timing resistors that you select. The total current drawn by the timer circuit is the sum of the idle current and the average charge current.

    To use minimal current, I would increase the 180K to 1.8M and the 4.7K to 47K and reduce the value of the timing resistor to . The average current through a timing resistor is:
    (Power_supply_voltage /2) / Charge_timing_resistor.

    The "Charge_timing_resistor" is timing resistor that is connected to the power supply. In your case it is the 4.7K which connects to the power supply through the diode. So, the 4.7K draws about 2 ma average current and a 47 K would draw about 1/10 that.

    You can reduce the current further by driving the diode from the 555 output and the positive power supply. You need to swap the 4.7K and the 180K resistors so that the 180K is now the charging resistor (instead of the 4.7K).

    This saves current by reducing the charging current and driving the LED more efficiently from the 555 output. A 555 timer -- whether bipolar or CMOS -- drives to ground better than to the supply. Because of this, you can increase the value of the LED's current limiting resistor and still get the same current through the LED.

    One other change. Since the LED is being driven to ground, the 555 Discharge pin can be used to drive the LED. This reduces interaction between the LED drive and the timing.

    I attached my version of the flasher.
    hrs likes this.
  6. wayneh


    Sep 9, 2010
    I retract that useless statement. If your only current-limiting is resistive, then there is no pulsing scheme that can save power versus continuous current, while maintaining the same perceived brightness. This is especially true because the intensity versus current plot for an LED is not linear and tends to bend over as the current increases. You get more brightness bang for your power buck at lower current.

    Pulsing is still used when precise color is an issue. LED color varies with current.

    Awww nuts, I see I've ruined my previous post. No big loss I suppose.
    Last edited: Aug 18, 2014
    hrs likes this.
  7. crutschow


    Mar 14, 2008
    For the same average current, pulsing can actually increase the power in the resistor since the power is the square of the current. Thus, for example, with a 50% duty-cycle with twice the pulsed current (1/2 the resistance) for the same average current, then the resistor power would be double (2^{2} * 1/2).

    To reduce the power with PWM you need an inductor in series with the LED and a free-wheeling diode to ground.
    hrs likes this.
  8. hrs

    Thread Starter Member

    Jun 13, 2014
    I think I get most of the points made so far, except for
    Is driving to ground vs driving to supply to be understood as pin 3 (output) acting as a sink vs a source?
  9. MikeML

    AAC Fanatic!

    Oct 2, 2009
    The only way you can save power while pulsing a led is to use a SMPS configured as a constant-current driver. The inductor in the SMPS raises the efficiency of the driver to ~85% vs dissipating power in a resistor, beit pwmed or linear...

    All of the 555 circuits are just constant-voltage PWM circuits (not proper constant-current LED drivers).
  10. RichardO

    Well-Known Member

    May 4, 2013
    Yes, the output voltage while sinking current is usually less than a volt for the bipolar 555. That same bipolar 555 will only get to within about 1 1/2 volts of the positive power supply even while sourcing small currents. At large currents, like 100 ma, this voltage can be as high a 2.25 volts.

    The CMOS part has similar limitations but at lower currents.