Sheet 1 of 6

Cascode BJT Circuit

A popular circuit for use at high frequencies is the , as shown below:-

B e2 c2 C Vout Q2

c1 Zout b2 b1 RL Q1 Vin A

e1 Zin

Gain of Q1 due to load of Q2

1 CE = - gm1RL1 RIN (CB stage ) = gm2

1 ∴ A V 1 = - gm1 which assuming are matched = 1 gm2

AI1 = - β

Miller capacitance

CSHUNT = CBC ()1− Av = CBC (1− (−1)) CSHUNT = 2CBC

The miller capacitance across the input of the CE stage is double the Base collector capacitance of Q1. Sheet 2 of 6

Input Impedance of Q1 at point A is

β ICQ k.T -23 −1 RIN = where gm = ; VT = where k = Boltzmans constant = 1.3807x10 JK gm VT q q = Electron charge = 1.6022x10-19C T = Temperature in Kelvin gm = (mS)

Voltage Gain Av

Voltage gain of CE stage is ~ 1 (due to low output RL) so the voltage gain of the amplifier will be from the common-base stage:

VOUT β2.ib2.RL β2.RL RL ICQ VA VA 1 A V = = = ≈ = . = (as = rce ) VIN ib2 (β2 + 1)rbe2 (β2 + 1)rbe2 rbe VT ICQ VT gm

Current Gain Ai

Current gain of CB stage is ~ 1 so the current gain of the amplifier will be from the common- emitter stage.

Current gain (Ai) = Ai of the CE stage (Ai of CB stage = 1) = β

Output Impedance

ROUT = RL

Of course as the voltage gain of the whole amplifier is dependant on the load resistor (and ultimately rce2), then adding an active load (eg ) will allow high voltage gain. Sheet 3 of 6

Cascode Simulation

For comparison a common-emitter stage was simulated on ADS to measure voltage gain.

Using the HF3127B array we can calculate the voltage gain of the circuit at low frequencies. If we assume a supply voltage of 5V and a device current of 5mA, we can calculate the value of the load resistor (We also assume that we want half the supply voltage across the device so that VC = 2.5V).

VCC - VC 5 - 2.5 RL = = -3 = 500Ω ICQ 5x10

Device Data VA = 50V ; VT = 25mV ; β = 100

-3 VT 25x10 AV= - gm.R L = - .RL = - -3 .500 = 2500 in dB' s = 10 * log(2500) = 34dB ICQ 5x10

Vc - Vbe 2.5 - 0.7 Base collector bias resistor = = -3 = 36KΩ ICQ /β 5x10 /100

At low frequencies we would expect a voltage gain of 34dB rolling off to a gain of 0dB at the fT of the device (~ 8GHz).

Below is shown the simulation and result from ADS.

R V_DC R3 SRC1 DC R=500 Ohm Vdc=5.0 V DC DC1 I_Probe R I_Probe1 R4 R=27000 Ohm

C AC

B AC DC_Block E AC1 V_AC DC_Block1 Start=1.0 MHz SRC4 Stop=9000 MHz Vac=1 V HFA3127B_npn Step= Freq=freq X2

Note a DC block is used as the AC source has a DC of 0V. During simulation the value of the base-collector resistor was reduced to ensure IC = 5mA. And the resulting data of voltage gain vs frequency for the single-stage common-emitter amplifier.

Sheet 4 of 6

DC.vc2 DC.I_Probe1.i 2.289 V 5.423mA

m2 m1 freq=1.000MHz freq=1.696GHz dB(AC.vc2)=35.944 dB(AC.vc2)=15.335

40m2

30 ) 20 m1 vc2 . C A

( 10 B d 0

-10 0123456789 freq, GHz

The predicted low-frequency voltage gain agrees well with our initial calculation.

However, we have not included a source resistance – this will effectively form a potential divider with the base bulk resistor rb, and lower the voltage entering the device and hence lowering the gain.

Now a 50-ohm resistor has been added to the ideal input voltage source.

AC DC AC DC AC1 DC1 Start=1.0 MHz Stop=4000 MHz Step=

V_DC R BJT_Model SRC1 R3 BJTM2 Vdc=5 V R=500 Ohm NPN=yes Br=1e1 Cjc=3.98e-13 Re=1.848 PNP=no Ikr=5.4e-2 Vjc=9.7e-1 Rc=1.14e1 R I_Probe Bf=1.036e2 Isc=1.6e-14 Mjc=2.4e-1 Kf=0 R5 I_Probe1 Ikf=5.4e-2 Nc=1.8 Xcjc=9e-1 Af=1 R=50 Ohm Ise=1.686e-19 Var=4.5 Fc=5e-1 Kb= Ne=1.4 Nr= Cje=2.4e-13 Ab= V_AC Vaf=7.2e1 Tr=4e-9 Vje=8.69e-1 Fb= SRC4 Nf= Eg=1.11 Mje=5.1e-1 Ffe= Vdc=0.817 V Tf=10.51e-12 Is=1.84e-16 Cjs=1.15e-13 Lateral=no Vac=1 V Xtf=2.3 Imax= Vjs=7.5e-1 AllParams= Freq=freq Vtf=3.5 Xti=3 Mjs=0 BJT_NPN Itf=3.5e-2 Tnom= Rb=5.007e1 BJT2 Ptf=0 Nk= Irb= Model=BJTM2 Xtb=0 Iss= Rbm=1.974 Area= Approxqb=yes Ns= RbModel=MDS Region= Temp= Mode=nonlinear Sheet 5 of 6

The resulting plot now shows how gain has rolled off faster due to the .

DC.vc2 DC.I_Probe1.i 2.486 V 5.028mA

m2 m1 freq=1.000MHz freq=1.692GHz dB(AC.vc2)=34.820 dB(AC.vc2)=8.738 40m2

30 ) vc2 . 20 C A (

B m1 d 10

0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 freq, GHz

To improve the high frequency response and stability we now add a common-base stage to form a Cascode amplifier. The common-base stage is biased in saturation and therefore there will be little voltage drop across the collector-emitter junction.

AC DC AC AC1 DC Start=1.0 MHz DC1 Stop=4000 MHz Step= V_DC R SRC1 R3 Vdc=5 V R=500 Ohm

I_Probe BJT_Model I_Probe1 BJTM2 NPN=yes Br=1e1 Cjc=3.98e-13 Re=1.848 BJT_NPN PNP=no Ikr=5.4e-2 Vjc=9.7e-1 Rc=1.14e1 BJT3 V_DC Bf=1.036e2 Isc=1.6e-14 Mjc=2.4e-1 Kf=0 Model=BJTM2 SRC5 Ikf=5.4e-2 Nc=1.8 Xcjc=9e-1 Af=1 Area= Vdc=3 V Ise=1.686e-19 Var=4.5 Fc=5e-1 Kb= Region= Ne=1.4 Nr= Cje=2.4e-13 Ab= R Temp= Vaf=7.2e1 Tr=4e-9 Vje=8.69e-1 Fb= R5 Mode=nonlinear Nf= Eg=1.11 Mje=5.1e-1 Ffe= R=50 Ohm Tf=10.51e-12 Is=1.84e-16 Cjs=1.15e-13 Lateral=no Xtf=2.3 Imax= Vjs=7.5e-1 AllParams= V_AC Vtf=3.5 Xti=3 Mjs=0 SRC4 Itf=3.5e-2 Tnom= Rb=5.007e1 Vdc=0.817 V Ptf=0 Nk= Irb= Vac=1 V Xtb=0 Iss= Rbm=1.974 Freq=freq Approxqb=yes Ns= RbModel=MDS BJT_NPN BJT2 Model=BJTM2 Area= Region= Temp= Mode=nonlinear

Sheet 6 of 6

The red plot shows the voltage gain of the CE stage showing that the gain of the first stage is now a lot lower than the single CE stage amplifier. The blue plot shows the new voltage gain response of the Cascode amplifier and at marker 2 (1.69GHz) the gain has improved from 8.7dB to 12.6dB, as the Milller effect has been reduced.

DC.vc2 DC.I_Probe1.i 2.528 V 4.944mA

m1 m2 freq=1.000MHz freq=1.692GHz dB(AC.vc2)=34.935 dB(AC.vc2)=12.631

40m1

30 ) ) 2 c vc1 v . .

C C 20 A A

( ( m2 B B d d 10

0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 freq, GHz