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Resistors getting hot

Discussion in 'Electronic Basics' started by MarkMc, Apr 19, 2006.

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  1. MarkMc

    MarkMc Guest

    Hi I'm a bit of a novice, and I just tested the circuit below on a
    breadboard, and I find that the RL and R2 resistors are getting
    incredibly hot. They are 0.25W rated. I measured the current through
    R2 to be 20mA, and voltage across R2 was 10v.

    Using P = IV, I make that 0.2W, and using P = I^2R, I get 0.19W, both
    in spec as far as I can see. I wonder if I'm doing something else

    RL will eventually be a relay, but on the breadboard, I've made this
    470R and an LED.

    | | | +12v
    .-. .-.10k|
    | | | | |
    RL| | | | |
    '-' ___ '-' |<
    +-|___|-+ -| Q2
    6.8k | 10k |\
    ___ |/ |
    On/Off Sig o-|___|-+ -| Q1 |
    .-. |> .-.
    | | | | |R2
    10k| | | | |
    '-' | '-'
    | | |
    | | |
    === === LED V ->
    GND GND -

    Q1 = BC107B (NPN)
    Q2 = BC177B (PNP)
    RL = 470R
    R2 = 470R

    RL and R2=470 both 0.25W both get very hot
    (created by AACircuit v1.28.5 beta 02/06/05

  2. Deefoo

    Deefoo Guest

    In spec doesn't mean that it can't be hot.

  3. MarkMc

    MarkMc Guest

    Right, ok. Does this mean that there's nothing obvious that I'm doing

    Perhaps I should leave the cct running for a few hours to see if the
    resistors cope?

  4. John Fields

    John Fields Guest

    If Q1 is fully saturated with, say, 0.3V between the collector and
    emitter, then with 11.7V across 470 ohms you'll have:

    E² 136.89
    P = --- = -------- = 0.291 watts
    R 470R

    which is over the rating of the resistor, so you should use a 1/2
    watt resistor.

    For R2, assuming the same Vce(sat) for Q2 and 2V for the LED means
    that the resistor would see 9.7V and it would dissipate about 200mW,
    which is in spec, but it'll get hot. Use a 1/2 watt resistor if the
    high temperature bothers you.
  5. Mark,

    You've done the math right (to within the tolerance of your voltage and
    current measurements). Standard practice is to ensure that resistors
    are not dissipating more than 1/2 of their rating. Operating resistors
    very hot tends to result in permanent changes to their resistance.

    While you can simply use 1/2W resistors, you could also change the
    circuit so that RL and R2 are formed by having two resistors in series
    or parallel so that the dissipation is shared (being careful not to
    space them too closely). For example two 250 Ohm, 1/4 watt resistors in
    series or two 1000 Ohm, 1/4 watt resistors in parallel are both
    equivalent to a single 500 Ohm, 1/2 watt resistor.

    Another way would be to increase the resistance of RL and R2 so that
    your LEDs are only drawing 10 ma. That would reduce the LED brightness,
    but likely still be bright enough to be seen. Using 1K resistors, the
    power dissipation would then be 0.1W in each.

    Good luck.
  6. MarkMc

    MarkMc Guest

    Thanks for the responses.

    RL was originally designed to be a relay, which I just substituted for
    the 470R and an LED on the breadboard. I don't have the relay specs
    with me, so as long as I haven't goofed on the sums there, I'm not too
    worried about RL, but I'll re-check anyway.

    R2 - interestingly, I did try two 1k's in parallel, but they still
    seemed to get quite hot, which I found odd. I'll try adjusting Ice to
    10mA, and adjust the base and R2 resistors accordingly, to see how that

    Thanks for the P = E^2/R, I didn't know that formula. For some reason
    I was thinking that a lower resistance would give lower power
    dissipation, but it's obvious that this will then cause more current to
    flow through R2.

  7. John  Larkin

    John Larkin Guest

    Get rid of RL - it serves no function now - and it won't get hot.

    Resistors usually get hot, too hot to touch, at rated power. It's
    common to derate then maybe 2:1 to improve reliability and avoid
    discoloring PC boards longterm.

  8. John Fields

    John Fields Guest

    Well, he _did_ say he was going to replace it with a relay, but the
    circuit's _way_ overkill anyway. He can do the whole thing like

    | | |
    [470R] |K |
    |A [1N4001] [COIL]
    [LED] | |
    | | |
    IN>--+--[R2]--B 2N4401
    | E
    [R1] |
    | |

    and maybe even get rid of R1, depending on what his input signal
    source looks like.
  9. Since you are breadboarding a circuit, why not try this:

    : +12V
    : |
    : | +12V
    : \ |
    : / R1 |
    : \ 47k |
    : / |
    : | |<e Q2
    : +-------| BC177B
    : | |\c
    : | |
    : |/c Q1 |
    : OnOff---| BC107B |
    : |>e |
    : | \
    : | / R3
    : | \ 470
    : \ /
    : / R2 |
    : \ 15k |
    : / |
    : | ---
    : | \ / D1
    : gnd --- LED
    : |
    : |
    : |
    : gnd

    Fewer parts. No dissipation to speak of, beyond your current limit
    resistor (R3, in my diagram), which will still consume some 200mW or
    so and should probably be 1/2W or better. R1 and R2 here will be
    about 1mW or less, in terms of dissipation. Similar, for Q1 and Q2,
    which should be only a few mW each.

    I am assuming here that your On/Off signal is a 0V or 5V source. The
    base will only pull a microamp or so, so that shouldn't be a problem.

  10. Jason S

    Jason S Guest

    Mark, just thought I'd mention here that if LED brightness would be a
    concern with the 1K resistor like James suggested here, you could try
    high-brightness LED's... they draw slightly more current than the
    conventional 'diffused' LED, but you can increase the resistance to
    eliminate the resistor from overheating and still get the LED bright. If
    you have one, try it - You could even go higher than 1K if you wanted to, as
    long as you're happy with the brightness and heat problem.

  11. MarkMc

    MarkMc Guest

    Hi All

    In response to a couple of the replies, I'm not fussed about the
    brightness of the LED, and as was stated, the circuit is overkill for
    what it's trying to do. I think I got over-excited about finally
    understanding the difference between PNP and NPN transistors!

    The input signal is a +12 or 0v on/off signal from a series of NOR and
    NAND gates. This portion of the circuit was being boarded up and
    tested in isolation of the logic gates.

    Ok, I found the relay datasheet. The relay is model FRS8, @12vdc, coil
    resistance is 320 ohms, rated current 37.5mA, and "power consumption"
    is 0.45 approx. Is there anything special I need to consider for
    driving this via my BC107B transistor? I've calculated to get 37.5mA
    Ice, I need about 1.8mA Ibe, giving a base resistor of 6.8k. How does
    that sound?

    Many thanks
  12. ehsjr

    ehsjr Guest

    Rule of thumb - drive the base with about Ic/10
    to saturate the transistor hard. So you could change
    to 3.3K base series R, and the base R to ground to 33K
    from 10K. Also, put a diode backwards (banded end to
    +) across the relay.

  13. MarkMc

    MarkMc Guest

    Hi Ed

    Somebody once told me to drive Ibe with Ice/Hfemin/10, which is what I
    used to set the base resistor. Is that what you meant, or do you use a
    different rule of thumb?

    I used an Hfe of 200 as the minimum, so 37.5mA/20 gives 1.8mA

    Regarding the pull-down resistor, is there any rule of thumb to
    determine what it should be? I just used something bigger than the
    base resistor, which isn't particularly scientific!

    Thanks for the hint about the reverse-biased diode, I do know about
    that one, but didn't show it on the diagram.

    With the LED and 47R resistor in parallel with my 320R relay and
    reverse-biased diode, how do I calculate the current through the relay?
    Is there a voltage drop across an LED for example?

  14. ehsjr

    ehsjr Guest

    Hi Mark,

    Actually, the rule of thumb I'm talking about just uses
    10 for the gain for saturation, so e.g. 3.75 mA Ib gives
    37.5 Ic. No need to look up the xsistor Hfe, which is not
    a fixed value in the first place.

    A rule of thumb for the pulldown is 10x the series base R.

    The current through the relay will be (Vcc-Vcesat)/Rcoil,
    regardless of what you put in parallel with it. In a parallel
    circuit, each leg draws Vcc/Rleg current irrespective of
    what other parallel legs draw. The diode connected backwards
    across the relay won't conduct while the relay is energized.
    The other parallel leg - the LED & resistor - will conduct
    while the relay is energized, but will not affect the current
    through the relay as long as Vcc stays the same.

    For all intents and purposes with this circuit, you can ignore
    Vcesat to get ballpark, and it's just 12/320. Or if you want
    to try to be more exact, you can attempt to extrapolate Vcesat
    from the curve on the datasheet. It looks to me to be around
    100 mV, so the formula would be 11.9/320.

    There's nothing critical about the circuit, so you get
    to play "fast and loose" using rules of thumb and
    approximations (ie ignoring Vcesat)

    I'm not sure what you meant by the led and 47R resistor in
    parallel with the relay. 47R is way too low, if you intend
    to use it in series to limit the current through the LED.
    Maybe you meant 470 R? That would limit the current to ~22 mA
    assuming a typical red LED. A typical red LED drops about 1.8
    volts at about 20 mA. There is a variation in what it drops,
    depending on how much current is drawn. That drop is usually
    referred to as Vf (forward voltage).

  15. MarkMc

    MarkMc Guest

    Hi Ed

    Thanks for the interesting 'rules of thumb' and tips re vcesat.

    Yes, I think I did mean 470R, but I've now decided on 1k, which seems
    to stay cool enough without hurting the LED brightness too much.

    Thanks again for all the help.

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