![]() The alternate would be to use logic level FET's and get rid of the issue all together. The HFE of the 3906 should be high enough that it would be possible to reduce the base bias current low enough to make sure the transistor is turned on *AND* the LED's in the first half are off. This should be mitigated though the use of a larger base resistor on the second (and to balance) first driver transistors. (this is so because the voltage the collector of the original 2N3906 will now "always" be near VDD)Ĭlearly much more current is available when the original xstr is driven ON (with a low to its base) - however "if" user's Leds are efficient they may illuminate as a small ILLEGAL current path has been created!Īhh! Your right! There is a clear path through the emitter->base diode to the LEDs in the first half of the display through the second half driver. I fear that 2 xstr rather than the 3 xstr approach I (clearly I believe) outlined will likely cause the "slight" but illegal illumination of the original digit. SeniorEE wrote:Signing off w/this post - you should have enough. I'm more of a programming/digital guy and transistors are not my forte. A "zero" on my pin turns on a digit with a 2N3906, but what transistor should I use to have the same effect but driven with a "one" instead? That way I would be able to drive the two digits with a single I/O pin. ![]() Problem is, I don't know what to use to replace the 2N3906. I don't worry about "light bleed", since turning off the segments before switching will prevent any "bleed" between the two digits. A zero on my pin would drive the first digit, a one would drive the 2nd digit. What I was thinking is to drive the two digits with a single I/O pin instead of two (driving the 2N3906 in figure 6). My problem is that I don't have two pins to drive the two digits of my LB-302VF long story short, I'm short of one I/O pin to drive two 7-segment displays. I bought an ATtiny84 but forgot about an input/output when I bought it. Note that the circuit has a 5 V supply.I need a bit of help with a project I'm doing. Determine the input common-mode range.You can verify this by calculating the gain yourself if you like. Note that all the output circuitry is made up of CC amplifiers that have a voltage gain close to unity. Thus, determine the DC open-loop voltage gain (DC differential gain).This stage is defined as the gain from the base of Q16 to the base of Q23. This stage is defined to end as the input to the base of Q16, including the impedance seen looking into the base of Q16 as a load to the first stage. Put the results in a table and show all calculations. You may assume that B 100 for all transistors, (although in reality, ß is not constant but depends on the current). Note that the transistor output resistance ro is approximately given by Va/Ic. Find Ic, ro and gm, for every transistor.While you are progressing through these calculations try to understand the purpose of each transistor. ![]() All of this work needs to be integrated in your report. Your work should be complete without referencing the textbook or notes for figures or equations. When appropriate, show the small signal equivalent circuit you are using. Parameter Early Voltage (VA) B (Assume constant, although it normally depends on current) Is (recall that Ic = IgeVbe/VT) NPN 80 V 100 6.73 fA 0.6 V 5 V PNP 20 V 100 1.41 fA 0.6 V 5 V VBE Vcc The BJT used are NPN Transistors (2N3904) and PNP Transistors (2N3906) for the simulations, these tran- sistors are used to approximate the functional 741. The table below lists some assumptions for NPN and PNP parameters that will help you with your prelab calculations. You may assume that B = 100 for all transistors, (although in reality, B is not constant but depends on the current). ![]() Note that the transistor output resistance r, is approximately given by VA/Ic. Find Ic, r, and gm, for every transistor.
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