jenglish@tardis.Tymnet.COM (Jim English) (11/08/89)
I know this is probably elementary knowledge, but I'll ask
anyway. What would I need to do to a simple circuit to lighten
the current on it. Basically, the power (DC) comes in, passes
through a resistor, a transistor, and drives a light display.
The resistor and transistor get quite hot. I am using a 5 watt
resistor, and it hasn't helped. Do I put a cap across the
resistor? ...Thanks...
--
Jim English MD-IPC | JENGLISH@F74.TYMNET.COM or jenglish@tardis.tymnet.com
(214)637-7406 Dallas | UUCP: ...!{ames,pyramid}!oliveb!tymix!tardis!jenglishyahoo@unix.cis.pitt.edu (Kenneth L Moore) (11/09/89)
In article <763@tardis.Tymnet.COM> jenglish@tardis.Tymnet.COM (Jim English) writes:
==>Basically, the power (DC) comes in, passes
==>through a resistor, a transistor, and drives a light display.
==>The resistor and transistor get quite hot. I am using a 5 watt
==>resistor, and it hasn't helped.
==>Jim English
If the resistance is appropriate for the circuit, try a higher
wattage resistor with the same value of resistance. This will
not decrease the current but the resistor will be better able
to disspate the excess energy as heat.
Remember that Power = I*I*R, so if you increase the wattage without
changing R, you increase the capability to carry current.
The same argument applies to the transistor.
Another alternative is to redesign the circuit.
--
yahoo@unix.cis.pitt.edu (Kenneth L Moore)
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@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@myers@hpfcdj.HP.COM (Bob Myers) (11/10/89)
>the current on it. Basically, the power (DC) comes in, passes >through a resistor, a transistor, and drives a light display. >The resistor and transistor get quite hot. I am using a 5 watt The resistor is there to limit the current; increasing the resistance will reduce the current. When calculating the value of resistor to use and its power rating, use Ohm's Law (V=IR) to determine the current passed by the resistor for a given voltage across it (in re-arranged form, Ohm's Law becomes I = V/R), and then P = I^2 x R for the power dissipated by the resistor. Use *at least* the next larger standard power rating (i.e., if you find that the resistor is dissipating 3W, use no smaller than a 5W resistor). One problem is that the voltage across the resistor may not be a constant for your circuit - it won't be, for example, if the light in question is an incandescent bulb rather than an LED. You need to provide a better description of the circuit for more specific recommendations. Bob Myers KC0EW HP Graphics Tech. Div.| Opinions expressed here are not Ft. Collins, Colorado | those of my employer or any other myers%hpfcla@hplabs.hp.com | sentient life-form on this planet.
davidc@vlsisj.VLSI.COM (David Chapman) (11/10/89)
In article <763@tardis.Tymnet.COM> jenglish@tardis.Tymnet.COM (Jim English) writes: >I know this is probably elementary knowledge, but I'll ask >anyway. What would I need to do to a simple circuit to lighten >the current on it. Basically, the power (DC) comes in, passes >through a resistor, a transistor, and drives a light display. >The resistor and transistor get quite hot. I am using a 5 watt >resistor, and it hasn't helped. Do I put a cap across the >resistor? The normal way to reduce the current is to use a larger resistor (in ohms, not in watts). Unfortunately, this also reduces the voltage to the display. Hmmmm, maybe not. If the transistor is doing any voltage regulation then it might not be so bad. Try a somewhat higher resistor (increase in 5-10% steps) and see what happens. Stick with the 5W size, though. The wattage is the amount of energy the part can dissipate without failing. It doesn't _necessarily_ mean it would be cooler if it had a higher wattage (but you don't really want to know about thermal conductance :-). The transistor definitely should NOT be hot, though. You might want to compute the current through it to make sure you're not exceeding its limits. The easiest way is probably to measure the voltage across the resistor. If you're still having problems, post the input voltage, the resistance, and the output voltage required (to the display) and we'll take a look at it. -- David Chapman {known world}!decwrl!vlsisj!fndry!davidc vlsisj!fndry!davidc@decwrl.dec.com
dag@hp-lsd.COS.HP.COM (David Geiser) (11/11/89)
Put another resistor in series with the existing one.
toddpw@tybalt.caltech.edu (Todd P. Whitesel) (11/11/89)
Looks like you have two easy (?) choices: increase the resistor value so there will be less current and therefore less heat. get another resistor/transistor pair and run them in parallel so each only dissapates half the power (you'll have to fix the resistor values) and therefore half the heat. Todd Whitesel toddpw @ tybalt.caltech.edu