MATLAB amplifier in detail

The model of the SiPM amplifier is a system of 24 equations in 24 variables that has been linearized so that it can be solved by MATLAB.

Circuit diagram

Circuit diagram

The schematic for the amplifier circuit is shown to the right. Click the thumbnail for a larger image. Node voltages and branch currents are marked on the diagram.

Parameters and variables

The MATLAB model has a number of parameters and variables to describe the amplifier circuit, including the 24 unknowns, 4 inputs, and numerous constants.

Input parameters

There are four input parameters:

• Input current: I_in (A)
• Bias voltage: V_b (V)
• Power voltage: V_c (V)
• Frequency: f (Hz)

Unknown variables

There are twenty-four unknown variables. The locations (and directions in the case of currents) are labeled on the circuit diagram. All unknowns are assumed to be of the form

$X(t)=X_{0}e^{i\omega t}+X_{1}\,\!$,

where X₀ gives the amplitude of oscillation, or the AC component, and X₁ gives the DC offset.

• Node voltages: V₁, V₂, V₃, V₄, V₅, V₇, V_out
  • Note: there is no V₆ on this circuit; it was a redundant variable with V_c.
• Resistor currents: I₁, I₂, I₃, I₄, I₅, I₆, I₇, I_t
• Transistor currents: j_b, j_c, j_e, k_b, k_c, k_e
• Capacitor currents: h₁, h₂, h₃

Constants

Resistors

The resistance values are mostly the same as those marked on the actual amplifier itself, however several were changed for better agreement of the model with data. See the article on the actual SiPM Amplifier for details on that circuit.

Component	Resistance
R1	100kΩ
R2	10kΩ
R3	5.6kΩ
R4	1.39kΩ
R5	1kΩ
R6	53Ω
R7	1.55kΩ
Rt	50Ω

Capacitors

The capacitors are not labeled on the amplifier itself or in the documentation supplied with the amplifier, so the following values are guesses as to the capacitances. Note that C₄ does not exist.

Component	Capacitance
C1	100nF
C2	0.1nF
C3	100nF
C5	10nF

Transistors

The transistor parameters used are selections from the Gummel-Poon SPICE model parameters for these two parts.

Parameter	Description	T1 value	T2 value
VT	temperature voltage	0.0259	0.0259
BF	ideal forward maximum β	93	34
NF	forward current emission coefficient	0.99	1.0
IS	transport saturation current	0.24 fA	0.44 fA
ISE	B-E leakage saturation current	2.4 fA	87 fA
NE	B-E leakage emission coefficient	1.46	1.94
RB	zero-bias base resistance	21Ω	5Ω
RE	emitter resistance	0.37Ω	1Ω

Transistor operating point

Each transistor has an operating point, U₁ and U₂. Both are initially assumed to be 0.7V. Under DC conditions (I_in = 0A, f = 0Hz) we iterate on U to refine these two parameters. Each iteration refines the operating point by averaging the current operating point with the associated V_be. V_be is the base-to-emitter voltage of the transistor and is given by

• V_be1 = V₃
• V_be2 = V₇ - V₄.

Derived parameters

The following parameters relate to the transistor constants. They are also constants, but are derived from the more fundamental constants given above. As the fundamental parameters are different for each transistor, there is a different set of derived parameters for each transistor.

• $V_{0}=NF\!\cdot \!Vt\,\!$
• $IBF=IS\!\cdot \!\exp \left({\frac {U}{V_{0}}}\right)\,\!$
• $IBL=ISE\!\cdot \!\exp \left({\frac {U}{NE\!\cdot \!VT}}\right)\,\!$
• $IB=IBF+IBL\,\!$
• $IC=BF\!\cdot \!IBF\,\!$
• $\beta ={\frac {IC}{IB}}\,\!$
• $Q={\frac {IBF}{V_{0}}}\,\!$
• $Z=1+Q\!\cdot \!\left(RB+RE\!\cdot \!BF\right)\,\!$

Equations

There are five categories of equations, which give a set of twenty-four equations in total. Two categories of equations are non-linear and need to be linearized to solve this system as a linear model using matrices.

Resistor voltage drop

The resistor voltage drop equations all take the form

$\Delta V=IR\,\!$

or alternately

$V_{\alpha }-IR=V_{\beta }\,\!$.

They describe the voltage drop associated with current crossing a resistor, according to Ohm's Law. As such, there is one equation per resistor in the circuit.

• R₁: V_b - I₁R₁ = V₁
• R₂: V₂ - I₂R₂ = 0
• R₃: V₄ - I₃R₃ = V₃
• R₄: V₃ - I₄R₄ = 0
• R₅: V₅ - I₅R₅ = V₄
• R₆: V_c - I₆R₆ = V₅
• R₇: V_c - I₇R₇ = V₇
• R_t: V_out - I_tR_t = 0

Node charge flow

Each node must maintain a dynamic equilibrium of charge during steady-state operation. That means that flow of charge (current) into a given node must equal flow of charge (current) out of that same node. Thus the node charge flow equations take the form of

$\sum I=0$

or alternately

$\sum I_{into}=\sum I_{out}$.

There is one such equation per node, and each node already is labeled on the above diagram by the voltage at that point; thus there is one equation per voltage. Additionally, each transistor acts as a node.

• V₁: I₁ = I_in + h₁
• V₂: I_in = I₂ + h₂
• V₃: I₃ + h₂ = I₄ + j_b
• V₄: I₅ + k_b = I₃ + j_c
• V₅: I₆ = I₅ + h₃
• V₇: I₇ = I_t + k_e
• T₁: j_b + j_c = j_e
• T₂: k_e = k_b + k_c

Capacitors

Capacitors relate current and voltage according to the equation

$I=C{\frac {dV}{dt}}$.

As stated above, the unknown voltages and currents are assumed to be of the form

$X(t)=X_{0}e^{i\omega t}+X_{1}\,\!$

so the capacitor equation can be linearized as

$I=i\omega CV\,\!$

where ω = 2πf. This equation works for both AC and DC cases, because in the DC case the derivative on the voltage eliminates any DC bias for the current, but ω = 0 so the equation still holds. There is one such equation for each capacitor.

• C₁ : h₁ = iωC₁V₁
• C₂ : h₂ = iωC₂(V₂ - V₃)
• C₃ : h₃ = iωC₃V₅
• C₅ : I_t = iωC₅(V₇ - V_out)

Transistor current gain

One of the characteristic equations of a transistor is

$I_{c}=\beta I_{b}\,\!$.

There is one such equation associated with each transistor.

• T₁: j_c = β₁j_b
• T₂: k_c = β₂k_b

Transistor exponential response

Another characteristic equation of transistors is

$Z\!\cdot \!I_{b}=IS\!\cdot \!\exp \left({\frac {V_{be}}{V_{0}}}\right)\,\!$.

This equation is linearized by performing a Taylor expansion up to the first degree, which gives

$Z\!\cdot \!I_{b}=Q\!\cdot \!(V_{0}+V_{be}-U)\,\!$.

Under AC conditions this equation is modified by defining V₀ = U. There is one such equation for each transistor

• T₁: Z₁j_b = Q₁(V₀₁ + V₃ - U₁)
• T₂: Z₂k_b = Q₂(V₀₂ + V₇ - V₄ - U₂)

Solution

The solution (that is, V_out) is found by first iterating as described above to find the transistor operating points to the desired precision, then solving under AC conditions to find the correct V_out. "Solving" (both during iteration and for the final answer) involves running the 24-equation matrix through MATLAB and selecting out the solution generated for the V_out variable. For responses, see the article on the SiPM Amplifier.