EECS 105 Fall 2003, Lecture 8 Lecture 8: Capacitors and PN Junctions Prof. Niknejad Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 8 Prof. A. Niknejad Lecture Outline Department of EECS Review of Electrostatics IC MIM Capacitors Non-Linear Capacitors PN Junctions Thermal Equilibrium University of California, Berkeley EECS 105 Fall 2003, Lecture 8 Prof. A. Niknejad Electrostatics Review (1) Electric field go from positive charge to negative charge (by convention) +++++++++++++++++++++ −−−−−−−−−−−−−−− Electric field lines diverge on charge E In words, if the electric field changes magnitude, there has to be charge involved! Result: In a charge free region, the electric field must be constant! Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 8 Prof. A. Niknejad Electrostatics Review (2) Gauss’ Law equivalently says that if there is a net electric field leaving a region, there has to be positive charge in that region: +++++++++++++++++++++ −−−−−−−−−−−−−−− Electric Fields are Leaving This Box! E dS Recall: V E dV V dV Q / Department of EECS Q Q E dV E dS V S University of California, Berkeley EECS 105 Fall 2003, Lecture 8 Prof. A. Niknejad Electrostatics in 1D Everything simplifies in 1-D dE E dx dE dx ( x' ) E ( x) E ( x0 ) dx' x x 0 Consider a uniform charge distribution E (x) Zero field boundary condition (x ) 0 x1 Department of EECS x x ( x' ) E ( x) dx' 0 x 0 0 x1 x1 University of California, Berkeley EECS 105 Fall 2003, Lecture 8 Prof. A. Niknejad Electrostatic Potential The electric field (force) is related to the potential (energy): d E dx Negative sign says that field lines go from high potential points to lower potential points (negative slope) Note: An electron should “float” to a high potential point: d d Fe qE e dx 1 Fe e e dx 2 Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 8 Prof. A. Niknejad More Potential Integrating this basic relation, we have that the (x) potential is the integral of the field: ( x) ( x0 ) E dl dl C In 1D, this is a simple integral: x ( x) ( x0 ) E ( x' )dx' ( x0 ) E x0 Going the other way, we have Poisson’s equation in 1D: d 2 ( x) ( x) 2 dx Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 8 Prof. A. Niknejad Boundary Conditions Potential must be a continuous function. If not, the fields (forces) would be infinite Electric fields need not be continuous. We have already seen that the electric fields diverge on charges. In fact, across an interface we have: x E dS E S 1 E1 (1 ) 1 2 E2 S Qinside Qinside x 0 0 1E1S 2 E2 S 0 E2 ( 2 ) S E1 2 E2 1 Field discontiuity implies charge density at surface! Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 8 Prof. A. Niknejad IC MIM Capacitor Bottom Plate Top Plate Bottom Plate Contacts Thin Oxide Q CV By forming a thin oxide and metal (or polysilicon) plates, a capacitor is formed Contacts are made to top and bottom plate Parasitic capacitance exists between bottom plate and substrate Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 8 Prof. A. Niknejad Review of Capacitors E dS + − Vs E dl E0tox Vs Q E dS E A 0 +++++++++++++++++++++ −−−−−−−−−−−−−−− Q Vs E0 tox Vs Q A tox Q E dS Q CVs A C tox For an ideal metal, all charge must be at surface Gauss’ law: Surface integral of electric field over closed surface equals charge inside volume Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 8 Prof. A. Niknejad Capacitor Q-V Relation +++++++++++++++++++++ Q y −−−−−−−−−−−−−−− Vs Q( y ) y Q CVs Total charge is linearly related to voltage Charge density is a delta function at surface (for perfect metals) Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 8 Prof. A. Niknejad A Non-Linear Capacitor +++++++++++++++++++++ y Q −−−−−−−−−−−−−−− Vs Q( y ) y Q f (Vs ) We’ll soon meet capacitors that have a non-linear Q-V relationship If plates are not ideal metal, the charge density can penetrate into surface Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 8 Prof. A. Niknejad What’s the Capacitance? For a non-linear capacitor, we have Q f (Vs ) CVs We can’t identify a capacitance Imagine we apply a small signal on top of a bias voltage: Q f (Vs vs ) f (Vs ) df (V ) vs dV V Vs Constant charge The incremental charge is therefore: df (V ) Q Q0 q f (Vs ) vs dV V Vs Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 8 Prof. A. Niknejad Small Signal Capacitance Break the equation for total charge into two terms: Incremental Charge df (V ) Q Q0 q f (Vs ) vs dV V Vs Constant Charge df (V ) q vs C vs dV V Vs df (V ) C dV V Vs Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 8 Prof. A. Niknejad Example of Non-Linear Capacitor Next lecture we’ll see that for a PN junction, the charge is a function of the reverse bias: Q j (V ) qN a x p 1 Charge At N Side of Junction V Voltage Across NP Junction b Constants Small signal capacitance: C j (V ) Department of EECS dQ j dV qN a x p 2b 1 1 V b C j0 1 V b University of California, Berkeley EECS 105 Fall 2003, Lecture 8 Prof. A. Niknejad Carrier Concentration and Potential In thermal equilibrium, there are no external fields and we thus expect the electron and hole current densities to be zero: dno J n 0 qn0 n E0 qDn dx n dno q d no E0 no 0 dx kT dx Dn kT dno dn0 d0 Vth n0 q n0 Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 8 Prof. A. Niknejad Carrier Concentration and Potential (2) We have an equation relating the potential to the carrier concentration kT dno dn d0 Vth 0 n0 q n0 If we integrate the above equation we have n0 ( x) 0 ( x) 0 ( x0 ) Vth ln n0 ( x0 ) We define the potential reference to be intrinsic Si: 0 ( x0 ) 0 Department of EECS n0 ( x0 ) ni University of California, Berkeley EECS 105 Fall 2003, Lecture 8 Prof. A. Niknejad Carrier Concentration Versus Potential The carrier concentration is thus a function of potential n0 ( x) ni e0 ( x ) / Vth Check that for zero potential, we have intrinsic carrier concentration (reference). If we do a similar calculation for holes, we arrive at a similar equation p0 ( x) ni e 0 ( x ) / Vth Note that the law of mass action is upheld n0 ( x) p0 ( x) ni2 e 0 ( x ) / Vth e0 ( x ) / Vth ni2 Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 8 Prof. A. Niknejad The Doping Changes Potential Due to the log nature of the potential, the potential changes linearly for exponential increase in doping: n0 ( x) n0 ( x) n0 ( x) 0 ( x) Vth ln 26mV ln 26mV ln 10 log 10 ni ( x0 ) ni ( x0 ) 10 n0 ( x) 0 ( x) 60mV log 10 10 p0 ( x ) 0 ( x) 60mV log 10 10 Quick calculation aid: For a p-type concentration of 1016 cm-3, the potential is -360 mV N-type materials have a positive potential with respect to intrinsic Si Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 8 Prof. A. Niknejad PN Junctions: Overview The most important device is a junction between a p-type region and an n-type region When the junction is first formed, due to the concentration gradient, mobile charges transfer near junction Electrons leave n-type region and holes leave p-type region These mobile carriers become minority carriers in new region (can’t penetrate far due to recombination) Due to charge transfer, a voltage difference occurs between regions This creates a field at the junction that causes drift currents to oppose the diffusion current In thermal equilibrium, drift current and diffusion must balance Department of EECS p-type NA −+−+−+−+−+− − −−−−−− V ++++ −+ +−+−+−+−+− + ND n-type University of California, Berkeley EECS 105 Fall 2003, Lecture 8 Prof. A. Niknejad PN Junction Currents Consider the PN junction in thermal equilibrium Again, the currents have to be zero, so we have dno J n 0 qn0 n E0 qDn dx dno qn0 n E0 qDn dx E0 dno dx kT 1 dn0 n0 n q n0 dx Dn dpo Dp kT 1 dp0 dx E0 n0 p q p0 dx Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 8 Prof. A. Niknejad PN Junction Fields p-type n-type NA ND p0 N a p0 ( x) J diff E0 x p0 ni2 n0 Na xn 0 ni2 p0 Nd n0 N d J diff E0 – –++ Transition Region Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 8 Prof. A. Niknejad Total Charge in Transition Region To solve for the electric fields, we need to write down the charge density in the transition region: 0 ( x) q( p0 n0 N d N a ) In the p-side of the junction, there are very few electrons and only acceptors: 0 ( x) q( p0 N a ) x p0 x 0 Since the hole concentration is decreasing on the pside, the net charge is negative: N a p0 Department of EECS 0 ( x) 0 University of California, Berkeley EECS 105 Fall 2003, Lecture 8 Prof. A. Niknejad Charge on N-Side Analogous to the p-side, the charge on the n-side is given by: 0 ( x) q(n0 N d ) 0 x xn 0 The net charge here is positive since: 0 ( x) 0 N d n0 n0 N d ni2 n0 Na J diff E0 – –++ Department of EECS Transition Region University of California, Berkeley