At 11:30 AM 2/4/2004 -0600, you wrote: > Now, I will let my ignorance shine. I have used lots of >bipolar transistors like the 2N3055 in power supplies. They show an >emitter voltage of the base voltage minus 1 silicon junction drop such >that the output is about .6 volts lower than the base bias. Yes. > What kind of voltage drop does a properly-biased source >follower show? It's the Vgs required to get the source current (which is the same as the drain current, of course). Have a look at this datasheet: http://www.ee.nmt.edu/~wedeward/EE443L/FA99/IRF510.pdf Figure 7 (transfer characteristics). Suppose the source load is 3 ohms and you have 3 volts across it (1A). Typically, then the Vgs must be about 5V, so that will be the "drop". You'll have 8 volts on the gate relative to ground. At 1 volt on the load, the Vgs will be between 3.5 and 4.5V, depending on temperature. So, the "drop" increases at higher load currents (it does on a BJT as well, just not as noticeably). > For that matter, do vacuum tubes have a certain voltage drop >between the grid bias and the cathode? I haven't actually ever tried >such a circuit. Vacuum tubes are very similar to MOSFETs (they are depletion mode, but other than that..) > One other little thing since I opened this can, while a >follower circuit can make a good power supply regulator, you must pay >close attention to the power dissipation of the device you are using. >Theoretically, one could put 100 volts on the collector of a >transistor, bias the gate at 5 volts relative to ground and get 4.4 >volts out the emitter. The trouble is that that transistor dissipates >the energy needed to run whatever is on the emitter as pure heat. It >is like a big resistor dropping 95 volts across it so it won't take >much of a load under those conditions to get that transistor very, >very hot. Yes. > When you use a FET or bipolar transistor as a switch and turn >it fully on, the voltage drop across it is extremely small like maybe >a volt or two relative to the power supply voltage. That means the >power dissipation is orders of magnitude smaller than if it is >operating in linear mode. Can be 10's of mV if it's a big power MOSFET running at reasonable current, or a BJT running at low current with a lot of base current. > One thing I remember from DC circuits class that floored me >for a while was the fact that except for the case where the transistor >drops exactly half the power supply voltage, there are always two >impedances for any given power dissipation. You must solve I^2*R for >R. In other words, the quadratic equation. Sure. For example, if you have a battery and you want 10mW out of it, you can put very light load on it so the voltage is almost the OC voltage, or short the hell out of it so the terminal voltage is almost zero and you just get the 10mW that way. The maximum power transfer out of the battery is when the external load = the internal resistance. U > I remember making a fool out of myself in the class because it >just didn't take in my brain for a day or so. But it did sink in, obviously, and stuck. Best regards, Spehro Pefhany --"it's the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com -- http://www.piclist.com#nomail Going offline? Don't AutoReply us! email listserv@mitvma.mit.edu with SET PICList DIGEST in the body