When testing transistors, how much minimum voltage difference should be seen across the baseemitter junction?
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The same,\n 1. An equivalent circuit of a NPN transistor is two diodes tied anode to anode; one cathode being the emitter, the other the collector, and the junction of the anodes is the base.  \n\n  2. When a NPN transistor is doing-its-thing, there is always a constant 0.6 volt drop between the base and emitter, i.e., the base is always ~ 0.6 volts more positive than the emitter--always! \n\n  3. There is no output at the collector, until the base has reached ~ 0.6 volts and the base is drawing current, i.e., any signal that appears at the base that is not up to ~ 0.6 volts (and not drawing base current), is never seen at the collector.  \n\n  4. The base requires a current, not a voltage to control the collector current.  \n\n  5. The collector is a current source: it does not source a voltage.  \n\n  6. The collector appears to output a voltage when a resistor is connected between it and power.  \n\n  7. The collector is a high impedance when compared to the emitter.  \n\n  8. The transistor can output an amplified signal either from the collector or the emitter (or both).  \n\n  9. When operating with a collector resistor (RL): the output voltage from the collector is an amplified voltage.  \n\n10. When operating with only an emitter resistor (Re): the output voltage from the emitter is not an amplified voltage, because it is always ~ 0.6 volts, below the input (base) voltage--hence the name voltage follower. But because the emitter can source large amounts of current to the "LOAD," it can be said, there was CURRENT amplification. \n\n11. The collector--being high impedance--cannot drive a low impedance load.  \n\n12. The emitter--being a low impedance--can drive a low impedance load.  \n\n13. The voltage gain from the collector is greater than one (Gv > 1).  \n\n14. The voltage gain from the emitter is less than one (Gv < 1).  \n\n15. Both the collector and the emitter: output ~ the same power: E x I = P.
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v=voltage\no=ohm\nm=milliammeter\n\na tester which test those mentioned aboved. don't use it in measuring resistance(ohms) when the circuit is still connected to a source (voltage)
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Figure 8 in the following link shows an open loop test circuit
532709
the minimum current needed to cause the thyristor to latch/switch state
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What's to know?\n\nIt's the typical offset voltage, or DC component of the output.  The signal is AC.
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You can use an integrated circuit, like LTC6906 from Linear Technology which goes up to 1MHz at 2.25 - 5VDC supply,\nsee:  http://www.linear.com/pc/downloadDocument.do?navId=H0,C1,C1010,C1096,P9790,D6998\nAlternative: MAX5094A by Maxim,  see:  http://www.maxim-ic.com/quick_view2.cfm/qv_pk/5017\nI didn't have the time to check the output power, but if it's not enough, a simple emitter stage transistor will do.
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Wow, I can't believe you have been waiting so long for an answer.\n\nThere are a couple of papers you can review:\n\ne.g. http://www.nsti.org/publ/MSM2002/348.pdf#search='frequency%20mosfet%20parameter'\n\nand here also:\nhttp://www.freewebs.com/anishamukherjee/Papers.htm\n\nThere is a tool here, but you need to pay, and I have not used it myself:\n\nhttp://www.highbeam.com/library/doc0.asp?docid=1G1:79591050&refid=ink_tptd_mag_test1&skeyword=&teaser=
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It varies and isn't really predictable like the Moore's Law.\nhttp://www.ask.com/web?q=moore's%20law
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A typical output has a transistor that pulls the output to ground when the output is 0 and a pull up resistor or transistor that drives the output high when the output is 1.\n\nHowever, there are two cases where this might not be desirable. Suppose we have an inverter that we want to use to turn on a 12V light. Sure, you could use a transistor or other switch to do that, but why not use an open collector inverter. This inverter only has the "low side" transistor. So the output is either 0 or floating. If the chip can handle 12V on the input, you can use it to directly turn the light on (it will be on with the output low). The 12V supply would go to the bulb and the other side of the bulb would go to the open collector output.\n\nThe other reason to do this is when you want to drive a lot of outputs into one input. For example, suppose I'm building a CPU. I have an input that causes a register to write its content to the bus. Further, suppose there are 10 reasons you might want to do this. The obvious way is to have a 10 input "OR" gate so that when any of these 10 things occur, the register gets the signal. But an easier way is to make the input active low. Tie a pull up resistor to the input and then drive it with 10 open collector gates. When no one wants to make the register write, the outputs all float and the resistor holds the input high (no action). If any output goes low (or multiple outputs, for that matter), the input goes low and the action occurs. This is often known as wired-OR.\n\nYou might find The Art of Electronics useful. You can read my review of this fine book at http://www.wd5gnr.com/books.htm (I didn't write it, BTW).
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Read this page. I hope it will help you.\nhttp://www.powerdesigners.com/InfoWeb/design_center/articles/MOSFETs/mosfets.shtm
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It depends on the voltage your discussing.  The simplest that comes to mind is two parallel diodes reverse connected.  Silicon diodes will clip at  1.4Vp-p and germanium clip at 0.8Vp-p. \nFor higher voltages  two series zener diodes reverse connected will give you a wide option of clipping voltages. \nIf you specify the clipping volts I can help more.
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Successive approximation A/D's work similar to a binary search algorithm.  The result can be very accurate if the A/D is designed correctly and the rate of the analog signal is relatively slow compared to the clock of the A/D, or a sample and hold block is used.  \nhere is an example of the operation (for this example a 4bit and that the input voltage is 1.2V):\n1.  using a DAC apply 1000 as a reference the  "1" is the test bit (say 1000 =1V)\n2.  Compare analog signal to this reference (result true) therefore the test bit is kept\n3.  Now apply the next test bit and the kept bit 1100 (the one in the 4's place is the new test bit and the one in the 8's place is the kept bit 1100= 1.5V)\n4.  Compare analog signal to this reference (result is false 1.2V <1.5V) therefore test bit is not kept\n5.  Now apply the next test bit and the kept bit 1010 (the one in the 2's place is the new test bit and the "10" in the 8,4's place is the kept bits 1010= 1.25V)\n6. Compare analog signal to this reference (result is false 1.2V <1.25V) therefore test bit is not kept.\n7. Now apply the next test bit and the kept bit 1001 (the one in the 1's place is the new test bit and the "100" in the 8,4,2's place is the kept bits 1001= 1.125V)\n8. Compare analog signal to this reference (result is true 1.2V >1.125V) therefore test bit is kept.\nResult is 1001\n\nHope this helps
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x\n\nthe answer below is just for voltage - see my graph above - it is quite accurate. here it is again, larger: X
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This depends on the logic family employed. For "standard CMOS" which is common today, +5V is 1 and 0V or ground is 0. However, other systems use other voltages. Lower voltages are common today for a few reasons. 3.3V logic, for example, is widely used where 3.3V is a 1 (and 0V is 0). However, it is possible to have negative logic. In addition, some high speed systems like ECL use voltages like -1.75V for a 0 and -0.9V for a 1 (this allows transistors to stay out of saturation which makes for fast switching but poor noise immunity and difficult design).
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The germanium diode have a forward voltage of .3 volts compared to the silicon diode with a forward voltage of .7 voltage.
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please visit\n\nhttp://w3.id.tue.nl/fileadmin/id/objects/E-Atelier/doc/Datasheets/Transistoren/BS170.pdf\n\nfor the detailed report on bs170, N channel EMOSFET
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1.Current in the 385.451 V circuit is bigger.\n\nGod is spelled B-I-G B-A-N-G\n\nPBrax@netscape.com