Path: utzoo!utgpu!jarvis.csri.toronto.edu!rutgers!tut.cis.ohio-state.edu!quanta.eng.ohio-state.edu!parts.eng.ohio-state.edu!abali From: abali@parts.eng.ohio-state.edu (Bulent Abali) Newsgroups: sci.electronics Subject: Re: Tx with no collector connection? Message-ID: <2540@quanta.eng.ohio-state.edu> Date: 5 Jul 89 02:11:30 GMT References: <3834@phri.UUCP> Sender: news@quanta.eng.ohio-state.edu Reply-To: abali@parts.eng.ohio-state.edu (Bulent Abali) Organization: Ohio State Univ, College of Engineering Lines: 45 In article <3834@phri.UUCP> roy@phri.UUCP (Roy Smith) writes: = First, the DC model: = = = o +6V = | [key to ascii art: < and > are = > resistors, V are ground points, = < 10k C and the transistor is NPN] = > / = | B |/ = +-------| = | |\ = > \ E > < 2.2k | > > < = | > 1k = V < = | = V = = There are, of course, connections to the collecter, but all of them =use a capacitor as a series element, so there is no DC collector connection =at all. How does one go about analysing a DC circuit like this? = = After taking the thevinin equivalent of the 10k/22k voltage divider =you get 1.1V through 1.8 k. Assuming Q is active, you use KVL to write = = 1.8V = 1.8k*Ib + 0.7V + 1k*Beta*Ib =and solve for Ib. This gives you an open-circuited current source of =(Beta+1)*Ib at the collector. Obviously, with Ic=0, Q is cutoff, but =that's not very useful. Am I missing something? Ic=0, therefore Ib=Ie (Ie to be taken out of the emitter). Voltage division at the base gives 6*2.2/(10+2.2)=1.08V. Therefore, 1.08V = 1.8K*Ib + 0.7 + 1K*Ib You have nothing to do with Beta in the DC model, because the transistor is not in the active region. Base acts like a forward biased diode. -=- Bulent Abali Ohio State Univ., Dept.of Electrical Eng. 2015 Neil Av. Columbus, Ohio 43210 abali@baloo.eng.ohio-state.edu