Matthias' Blog

A Gilbert cell multiplier is a circuit that takes two inputs as differential currents and produces a differential current that is proportional to the product of the two inputs Let X and Y be arbitrary values between 1 and -1. Let I x and I y be arbitrary non-zero currents. The combined emitter current of the left differential pair, \(I_{le}=I_y\left(1+Y\right)\), and the combined emitter current of the right differential pair, \(I_{re}=I_y\left(1-Y\right)\). Let V 1 be the collector voltage of the leftmost transistor and V 2 be the collector voltage of the rightmost transistor. Let I s be the saturation current of the base-emitter junction. Let V t be the thermal voltage, about 0.025865 volts. Part 1 Consider the leftmost transistor. Note that its base and collector are connected together. Recall that the base current is \(I_s\left(\frac{e^{V_1/V_t}-1}{\beta}\right)\). Therefore the total current flowing through through the transistor is \(I_s\left(\frac{\beta+1}{\beta}\l...

Consider the following oscillator circuit: For a circuit to oscillate, it must meet the following criteria: Total loop gain is 1 The total phase shift is a multiple of 2π radians One can see that the op-amp is wired as an inverting amplifier, so it has a phase shift of π radians. Therefore, the phase shift network consisting of capacitors C 1 , C 2 , C 3 and resistors R 1 , R 2 , R 3 must provide the remaining π degrees of phase shift in order for the circuit to meet criterion 2. Let \(f\) be the frequency at which the circuit oscillates. The impedance of the capacitors at frequency \(f\) is \(Z_c=\frac{1}{2\pi jfC}\), where j is the imaginary unit and \(C\) is the capacitance of all the capacitors. Let \(R\) be the resistance of all the resistors except R f . It is also known that given two linear devices with impedances Z 1 and Z 2 (either real or complex impedances): their total impedance when wired in series is \(Z_1+Z_2\) their total impedance when w...

This page will describe how to calculate the time it takes for a capacitor to charge from one voltage to another through a resistor. Let C be the capacitance of a capacitor being charged from a voltage source with voltage V through a resistor of resistance R. Suppose the capacitor starts with 0 volts across it and is allowed to charge. The voltage \(V_c(t)\) can be modelled as a function of time (t): $$ V_c(t) = V - Ve^{-\frac{t}{RC}} $$ This equation can be solved for t to find the time it takes for a capacitor to charge from 0 volts to \(u\) volts $$ u = V - Ve^{-\frac{t}{RC}} $$ $$ \frac{u}{V} = 1 - e^{-\frac{t}{RC}} $$ $$ 1 - \frac{u}{V} = e^{-\frac{t}{RC}} $$ $$ \ln\left(\frac{V-u}{V}\right) = -\frac{t}{RC} $$ $$ RC\ln\left(\frac{V}{V-u}\right) = t $$ Suppose one is only interested in the amount of time T it takes for a capacitor to charge from \(l\) volts to \(u\) volts. One would calculate the time for it to go from 0 to \(u\) and subtract the time for it to go from 0 to \(l...

Suppose there are a pair of transistors with their emitters connected together to a current source of magnitude I. Let \(I_a\) and \(I_b\) be the collector currents, \(I_{ab}\) and \(I_{bb}\) be the base currents, and \(V_e\) be the emitter voltage. Let \(V_a\) and \(V_b\) be the base voltages of the transistors. The following equation represents the base current: $$ I_{bb} = I_s\left(\frac{e^{V_{BE}/V_t}-1}{\beta}+\frac{e^{V_{BC}/V_t}-1}{\beta_r}\right) $$ where \(I_s\) is the saturation current, \(\beta\) is the current gain of the transistor, \(\beta_r\) is the reverse gain, and \(V_t\) is the thermal voltage (about 0.025865 volts) If the collectors are pulled high and \(V_{BC}\) is negative, then we can ignore the second term because the B-C junction is reverse biased. $$ I_{bb} = I_s\left(\frac{e^{\left(V_b-V_e\right)/V_t}-1}{\beta}\right) $$ The following equation represents the collector current $$ I_b = I_s\left(e^{V_{BE}/V_t}-1\right) - I_s\left(e^{V_{BC}/V_t}-1\right)\l...