6.301 Solid-State Circuits Recitation 25: More on Charge Control and Space-Charge Lasers Prof. Joel L. Dawson First, let’s do some cleanup from the last recitation. We said that we would treat the problem of emitter switching as an example of how to use the charge control equations. Note that from our experience with Lab 2, we might expect this common base stage to be relatively fast. What does charge control have to say about that? Use charge-control equations for emitter current. Notice that I follow my own sign convention here…I choose to define current flow out of the emitter as positive. iE= qF1!F+1!BF"#$%&'+dqFdt= IEu(t) If we solve this, we find that qF=!IE1 " e" t /!( ) Where ! is defined by 1!=1!F+1!BF Writing this out: !=!F!BF!F+!BF="!F2!F+"!F=""+ 1!F=#F!F IEu(t )!iC6.301 Solid-State Circuits Recitation 25: More on Charge Control and Space-Charge Lasers Prof. Joel L. Dawson Page 2 So the time constant involved is smaller than !F! Recall that the time constant associated with base current drive was !BF. Now we have all we need to calculate iC: iC=qF!F="FIE1 # e# t /"F!F( ) And for iB, we apply KCL (again remembering the sign convention that I have chosen): Initially, then, the base current = iE! CLASS EXERCISE Consider the following circuit: At time t=0, we open switch S1. Derive and sketch IC as a function of time. (Workspace) iB= iE! iC= IE!"FIE1 ! e! t /"F#F( )iB+ iC= iE!S1!ICiC=qF!FiB=qF!BF+dqFdtiE= "qf1!F+1!BF#$%&'("dqFdtIB6.301 Solid-State Circuits Recitation 25: More on Charge Control and Space-Charge Lasers Prof. Joel L. Dawson Page 3 Take a moment to remind yourself of what is physically happening. We have assumed that recombination in the base is negligible. Thus, the collector current continues just as long as it takes to clear qFout of the base via reverse injection. Space Charge Layers This, unfortunately, is where things get messy. The upside is that working through the math here will give you a very detailed understanding of what goes on in a bipolar transistor. Recall that the BC and the BE pn junctions have capacitances associated with their depletion regions. We are able to define a nonlinear, voltage-dependent capacitance associated with these depletion regions, and relate charge and voltage as dQ = C(V )dV To figure out how much charge it takes to get to a voltage V0, we must integrate dQ0Q0!= C(V )dV0V0!Q0= C(V )dV0V0! As a check, we note that with a capacitance that is independent of voltage, we have Q0= CdV = C dV = CV00V0!0V0! Just as we expect. Now for this BJT, we have qVEqVC6.301 Solid-State Circuits Recitation 25: More on Charge Control and Space-Charge Lasers Prof. Joel L. Dawson Page 4 We have to store charge qVC on Cµ, and qVE on Cje (part of C!). This charge has to be supplied to the base, and that is why these capacitances are relevant to us here. For Cµ, we have Cµ=Cµ01 ! VBC/"C( )mC= Cµ0"CmC"C! VBC( )! mC If we’re curious about the charge on Cµ, we must perform the integration dQVC= Cµ(V )dV0VBC!0QVC! QVC= !Cµ0"CmC1 ! mC#$%&'("C! VBC( )1! mc)*++,-..0VBC= KC"CmC/!"C! VBC( )mC/)*+,-. Where KC=Cµ0!mC1 " mC and mC!= 1 " mC That’s a lot of work, and for what? Well, for one, we can now write down the charge on these nonlinear capacitors as a function of VBE and VBC. A typical mC for a base-collector junction is 1/2, and for a good base-emitter junction is 1/3. We have QVC= KC!C1 / 2"!C" VBC( )1 / 2#$%&QVE= KE!E2 / 3"!E" VBE( )2 / 3#$%&6.301 Solid-State Circuits Recitation 25: More on Charge Control and Space-Charge Lasers Prof. Joel L. Dawson Page 5 And the additional charge control terms iB=dQVCdt+dQVEdtiC= !dQVCdtiE= !dQVEdt Ultimately, these monstrous expressions for QVC and QVE will be useful because, once we compute the total charge necessary to get to a particular VBC or VBE, we will sweep the nonlinearity completely under the rug. This calls for an example. Consider the following circuit with a base current drive. We ask, how long does it take to get out of cutoff? In order to get out of cutoff, VBE must equal 600mV. In order to get VBE to 600mV, we must charge QVE and QVC appropriately. The total amount of charge needed is !QVE= QVE(0.6V ) " QVB(0V ) = 3.8 pC!QVC= QVC("9.6V ) " QVC("10.2V ) = 0.6 pC RL!iBiB= 200µARL= 1k!VCL= 10.2V"F= 100"R= 5#F= 1ns#R= 5ns#S= 29ns6.301 Solid-State Circuits Recitation 25: More on Charge Control and Space-Charge Lasers Prof. Joel L. Dawson Page 6 Total charge that must be supplied by iB to get out of cutoff: 4.4pF. And here is where we sweep aside the nonlinearity. We say that !t =!QiB=!QVE+ !QVCiB=4.4 pC0.2mA= 22ns Good for a ballpark