6.301 Solid-State CircuitsRecitation 24: Charge Control w/DemoProf. Joel L. DawsonWe’re almost at the finish line! Charge control is the last major topic that we will cover this term.Why charge control? Hasn’t the hybrid-π model served us well this whole semester? It certainly has.But to see some of the holes that it has left, ask yourself the following questions:• How does one analytically treat a device that has entered saturation?• Are there any dynamics associated with suddenly changing regions of operation?• What really happens if we suddenly cut-off the base current of a BJT?The list goes on, and the point is that the answers to these questions can be useful. Transistors, afterall, aren’t used only for small signal amplification. Sometimes they are useful as switches, either forrealizing digital logic or in digital-to-analog converters.We should emphasize that charge control, like open circuit time constants, is not good for giving usprecise numerical answers. It does give us design insight, telling us, for instance, what topologiesshould yield high switching speeds.To review, recall that in lecture we looked at finding Ic(t) in the following situation:Since everything depends on qF, we first figure out qF(t):iB=qFτBF+dqFdt= IBu(t)↑ICIBu(t )iC=qFτFiB=qFτBF+dqFdtiE= −qF1τF+1τBF⎛⎝⎜⎞⎠⎟−dqFdt6.301 Solid-State CircuitsRecitation 24: Charge Control w/DemoProf. Joel L. DawsonPage 2Using basic differential equations, we solved this and found:qF=τBFIB1 − e−t /τBF( )IC=qFτF=τBFτFIB1 − e−t /τBF( )=βIB1 − e−t /τBF( )Now, for the class exercise we’re going to solve this a different way.CLASS EXERCISE: Consider the hybrid-π modelDerive the collector current as a function of time. You may find the following reminders helpful:(1)gmrπ=βF(2)rπCπ=βFgm⎛⎝⎜⎞⎠⎟gmτF( )=βFτF=τBF(Workspace)gmVπIC(t )↓CπVπrπ↑IBu(t )6.301 Solid-State CircuitsRecitation 24: Charge Control w/DemoProf. Joel L. DawsonPage 3The equivalence is a little surprising, isn’t it? Think about why this worked out…noticing that gmrπand rπCπ are both independent of collector current.Very well. We promised design insight as a result of our derivations, and we’re coming to a firstexample. According to our charge control model, turning on a transistor is simply a matter of placingthe right amount of charge in the base.When we tried the following circuit in lecture:There was a limit to how rapidly we could change the amount of charge in the case specifically:dqFdt≤VIRBBut what if we could arrange to put charge into the base very quickly? Consider:→V0RLRBIB−+vI(t )vI(t ) = VI+ VBE( )u(t )V0RLRBiCB−+vI(t )vI(t ) = VI+ VBE( )u(t )CB6.301 Solid-State CircuitsRecitation 24: Charge Control w/DemoProf. Joel L. DawsonPage 4Hmm. Conceptually, the voltage across the capacitor changes instantaneously from zero to VI:iCB= CBdVCBdt= CbVIδ(t)Therefore, the base current can be written:iB(t ) =VIRBu(t ) + CBVIδ(t )Using charge control, we haveiB=qFτBF+dqFdt=VIRBu(t) + CBVIδ(t)The solution isqF(t) =τBFVIRB1 − e− t /τBF( )u(t) + CBVIe− t /τBFNotice that if we choose CB=τBFRB,qF(t) =τBFVIRBu(t)6.301 Solid-State CircuitsRecitation 24: Charge Control w/DemoProf. Joel L. DawsonPage 5iCreaches its final value instantaneously! For this reason, CB is sometimes called a speed-upcapacitor. Step responses for different values of CB would look like the following:Real circuits do exactly this! CB may require some trimming, but this is a trick that works in thereal world.ICtCBtoo highCBoptimumCBtoo low6.301 Solid-State CircuitsRecitation 24: Charge Control w/DemoProf. Joel L. DawsonPage 6Charge control also gives us insight into “emitter switching.” That is,Challenge: Find IC and IB. We’ll talk more about this case later.IEu(t