We know that the noise power generated in a resistor is proportional to temperature. If we have a resistor with a zero thermal coefficient so that the resistance is constant with the temperature, does the temperature of the resistor increase due to the thermal noise? (No, not even sub-infinitesimally! / rap) In other words, does the resistor noise create noise voltage or current in the resistor, which in turn heats the resistor to a higher temperature, which in turn increases the noise, which in turn heats the resistor to a higher temperature, etc.?
(If you have a 1-M resistor with a BW of 1 MHz, V = 125 µV, and I = 125 pA, its self-heating would be about 15 fW. In a 250-mW resistor, with nominal heatsinking, this would cause a temperature rise of 60 femto degrees. I don’t think we have to worry, even if it did do self-heating! If you have a different R value, the dissipation in femtowatts will be about the same, even if the BW does get bigger. /rap)
My guess is that the noise power calculated by P = 4 × k × T × BW assumes an infinite heatsink maintained at temperature T. Without an infinite heatsink, then, there will be some increase in temperature, albeit small, depending on the thermal resistance. With a practical value of thermal resistance, the temperature would no doubt only increment by a minute amount. However, theoretically, if the thermal resistance between the resistor and the heatsink was very large, the resistor temperature could increase until the resistor burns out.
(Fat chance! If the temperature rose from 25°C to 125°C or 225°C, the resistor would radiate its watts away—no heatsinking required. /rap)
I figured I would get your insight before I calculated the temperature increase by some limit method or, perish the thought, simulation. This might generate an interesting problem in which one would calculate the thermal resistance at which the resistor power would dissipate a specific amount of power
On the other hand, it would seem that if you sealed the resistor in a vacuum with a thermal resistance approaching infinity, the resistor could get pretty hot, perhaps 10 picodegrees of temperature rise! The heat power would have to go somewhere or the temperature would reach infinity, so we should be able to harvest some of the heat and use it to do work. But, no, this is not going to cause the world to come to an end. The positive feedback will not send it off to a crazy temperature, which leads me to think there is something wrong with my theory. (I gotta admit, I can’t tell you exactly what is wrong with your theory, but something is! /rap) Have you ever thought about this?
I have tried as hard as possible to avoid thinking about it! I’m going to go home and pour a drink—or two. I’ll let you know if I learn anything. Rap
What is your opinion about biasing the phototransistor base input to change the response of a phototransistor to light? I am trying to maximize the transistor’s sensitivity to light (digital on-off), but I also think it may be possible to use the base input to increase the speed in which the phototransistor turns off. I do not think I can do both easily.
If you pull down on the base of the phototransistor, that helps it turn off—for example, from the base (pin 6) of a 4N28. But that will make it much less sensitive to the first little bit of photocurrent. It won’t be sensitive to small photocurrents—maybe not at all.
You may be able to feed the phototransistor’s output to a linear amplifier and then to a nonlinear amplifier. But I think you are right that there is no simple way to do both of the things that you wish for. Maybe you can buy a higher gain of phototransistor, but they are often slower—slower to turn on, and slower to turn off.
Write down exactly what you want each little bit (increment) of current to do. But it sounds as if you want a linear amplifier with high gain, plus the advantages of high digital gain (fast on/off). It is usually hard to do both! Best wishes. Rap
I’m working on a home project and need a programmable (waveshape) ac source (50 to 300 V, up to 10 A). Since it is for home, I need a cost-effective way to obtain one. Any suggestions?
Audio power amplifiers that can put out 1500 W aren’t cheap. Put two of them pushpull (bridge output) into a big transformer, which also isn’t cheap, because you can find audio amplifiers that would put out 100 V at 30 A (as into 4 O) but not 300 V into 10 A. V × I is not interchangeable, unless you can find a suitable transformer.
You want 300 V RMS? That’s a BIG power transformer. Get a dozen 300-W audio amplifiers. Put each output into a mid-sized transformer to give 30 V × 10 A on the secondary. Stack 10 of those transformer secondaries. Stand back! Rap
comments invited! firstname.lastname@example.org —or:
Mail Stop D2597a, national Semiconductor P.o. Box 58090, Santa clara, ca 95052-8090
BOB PEASE obtained a BSEE from MIT in 1961 and is Staff Scientist at National Semiconductor Corp., Santa Clara, Calif.