Simple Biasing Circuit

Question

I'd like to formally understand this simple biasing circuit:

Let V⁺ be the supply voltage, V_i be the input voltage at the seemingly unconnected terminal of the capacitor, and let V_o be the output voltage at the junction between the resistors and capacitor. Let S be the impedance unit (i*omega)

using VI relations and Kirchoffs laws: (V⁺ - V_o) / R₁ - V_o/R₂ + (V_i-V_o) * C*S = 0

which after rearranging gives:

V_o = (V⁺/R₁ + V_iCS) / (1/R₁ + 1/R₂ + C*S)

decomposing the numerator it becomes clear that the biasing term is:

= (V⁺/R₁) / (1/R₁ + 1/R₂ + C*S) = V⁺ / (1+R₁/R₂ + R₁CS)

Does the amount of bias really depend on the driving frequency? At DC, S=0 and everything reduces to a voltage divider regardless of the voltage at V_i.

(Sorry for the eye-sore math. Is it possible to do math input on this website like it is on math.stackexchange ?)

Answer 1

quick explanation: The biased voltage can be regarded as a superposition of the contribution from V⁺ (calculated above, called biasing term) and the contribution from V_i (the other term, with V_isC in the numerator):
V_o = V_o,V⁺ + V_o,Vi = V⁺/[R₁(¹/_R1 + ¹/_R2 + sC)] + (V_isC)/(¹/_R1 + ¹/_R2 + sC), where

• V_o,V⁺ = V⁺/[R₁(¹/_R1 + ¹/_R2 + sC)] and
• V_o,Vi = (V_isC)/(¹/_R1 + ¹/_R2 + sC)

When one uses superposition, they redraw the circuit with all other voltage sources shorted and other current sources opened (other than the one being considered). This means that when considering the contribution from V⁺, V_i is grounded, so the frequency[-ies] in the V_o,V⁺ term is that present in V⁺, which should be near zero for a DC source. Using the same arguments, the frequency in the V_o,Vi term is that present in V_i.

Superposition makes sense for many reasons; one of the arguments I've made to justify it to myself is to look at Fourier analysis, which shows that any signal can be decomposed into the superposition of sinusoids, and those sinusoids can be extracted by filtering out the others; the Gibbs phenomenon is often seen in practice as ringing.

To be more precise though, we should take into account the load resistance that would be connected between V_o and ground.

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simplified analysis: The capacitor in this circuit is called a DC blocking capacitor, because it doesn't pass any DC signals. A common and useful technique to analyzing circuits that separate high frequency AC and DC signals like this is to approximate the blocking capacitor as an open circuit to DC signals and short circuit to AC signals. This greatly simplifies analysis of more complicated systems. For mid-band frequencies -- those for which the capacitor presents an impedance comparable, over 5%-10%, to that of R₁||R₂ -- the complicated impedance formula needs to be used. For low frequency signals, where the capacitor impedance is more than ~100·R₁||R₂, the cap can be regarded as an open circuit. Of course, this depends on the sensitivities of your circuitry, but that will be apparent if these considerations are of value.