Resistors getting hot

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

--DF

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

--
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.

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.

--
James T. White

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

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

--
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.

Since you are breadboarding a circuit, why not try this:

: +12V : | : | +12V : \\ | : / R1 | : \\ 47k | : / | : | |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

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.

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

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

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

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

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