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Q1 The transistor is designed so as to have two emitter circuits in parallel Q1 is modeled by the offset diode model, as shown in Figure 1049 We shall now consider each of the four cases 1 v1 = v2 = 0 V With the emitters of Q1 connected to ground and the base of Q1 at 5 V, the BE junction will clearly be forward biased and Q1 is on This result means that the base current of Q2 (equal to the collector current of Q1 ) is negative, and therefore Q2 must be off If Q2 is off, its emitter current must be zero, and therefore no base current can ow into Q3 , which is in turn also off With Q3 off, no current ows through R3 , and therefore vout = 5 vR3 = 5 V pdf compression library c# C# Compress PDF SDK: Compress , shrink PDF file size in C# .net ...
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Detect if a PDF File is a Portfolio in C# , VB.NET · Compress PDF Images in C# , VB. .... of the images in the PDF document by using the following code example. how to compress pdf file size in c# Windows 8 How to Compress PDF in C# sample in C# for Visual ...
8 Jun 2018 ... Developers can compress PDF file size in C# . Reduce size with image and content in PDF, and be able to delete annotations and metadata in ... 2 v1 = 5 V; v2 = 0 V Now, with reference to Figure 1049, we see that diode D1 is still forward-biased, but D2 is now reverse-biased because of the 5-V potential at v2 Since one of the two emitter branches is capable of conducting, base current will ow and Q1 will be on The remainder of the analysis is the same as in case (1), and Q2 and Q3 will both be off, leading to vout = 5 V 3 v1 = 0 V; v2 = 5 V By symmetry with case (2), we conclude that, again, one emitter branch is conducting, and therefore Q1 will be on, Q2 and Q3 will both be off, and vout = 5 V 4 v1 = 5 V; v2 = 5 V When both v1 and v2 are at 5 V, diodes D1 and D2 are both strongly reverse-biased, and therefore no emitter current can ow Thus, Q1 must be off Note, however, that while D1 and D2 are reverse-biased, D3 is forward-biased, and therefore a current will ow into the base of Q2 ; thus, Q2 is on and since the emitter of Q2 is connected to the base of Q3 , Q3 will also see a positive base current and will be on To determine the output voltage, we assume that Q3 is operating in saturation Then, applying KVL to the collector circuit we have: VCC = IC3 R3 + VCE3 or IC3 = and vout = VCC IC R3 = 5 22 10 3 22 10 3 = 5 484 = 016 V These results are summarized in the table below The output values are consistent with TTL logic; the output voltage for case (4) is suf ciently close to zero to be considered zero for logic purposes VCC VCE3 VCC VCE sat 5 02 = = = 22 mA RC RC 2,200. c# pdfsharp compression InflaterDynHeader, PdfSharp .SharpZipLib.Zip. Compression C# ...
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28 Feb 2018 ... C# How to compress PDF images. There is no doubt that the PDF file which contains plenty of images will enlarge the file and consume ... Note Although Microsoft chose to use a class for their String data type, it does not always act like a class involved, the method demonstrated in this example to determine whether transistors are on or off leads to very simple analysis Since in logic devices one is interested primarily TTL NAND gate may be found in the accompanying CD-ROM You may wish to validate the saturation assumption for transistors Q2 and Q3 by measuring VCE2 and VCE3 in the simulation The analysis method employed in Example 1011 can be used to analyze any TTL gate With a little practice, the calculations of this example will become familiar The Check Your Understanding exercises and homework problems will reinforce the concepts developed in this section . pdf compress in c# how to optimize pdf using c# programming. - CodeProject
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