# Resistors getting hot

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

1. ### MarkMcGuest

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
wrong.

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 -
|
|
===
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 www.tech-chat.de)

Regards,
Mark

2. ### DeefooGuest

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

--DF

3. ### MarkMcGuest

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

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

Regards,
Mark

4. ### John FieldsGuest

---
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. ### James T. WhiteGuest

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. ### MarkMcGuest

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
works.

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.

Regards,
Mark

7. ### John LarkinGuest

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.

John

8. ### John FieldsGuest

---
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
this:

+12>------------+------+---------+
| | |
[470R] |K |
|A [1N4001] [COIL]
[LED] | |
| | |
+------+---------+
|
C
IN>--+--[R2]--B 2N4401
| E
[R1] |
| |
GND>-+----------+

and maybe even get rid of R1, depending on what his input signal
source looks like.

9. ### Jonathan KirwanGuest

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.

Jon

10. ### Jason SGuest

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.

Jason.

11. ### MarkMcGuest

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
Mark

12. ### ehsjrGuest

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.

Ed

13. ### MarkMcGuest

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?

Regards,
Mark

14. ### ehsjrGuest

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).

Ed

15. ### MarkMcGuest

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.

Regards,
Mark