The Triode & Amplification
A grid that controls a big current.
Lesson at a glanceShow the key points
- i_p = k(V_p + μV_g)^(3/2).
- r_p, g_m and μ, with μ = r_p g_m (41.9).
- Gain A = μ/(1 + r_p/R_L) (41.10).
Put a wire mesh between a diode’s cathode and anode and a small voltage on it controls a big current. That made the triode the first amplifier.
The triode
A triode is a diode with a wire grid between cathode and anode: three terminals. Thermions pass through gaps in the grid to the anode. With the grid at zero and the anode positive it behaves like a diode. A negative grid voltage Vg repels the electrons and reduces the plate current; a positive one helps them along. Because the grid is close to the cathode, a change in Vg has much more effect than the same change in Vp:
Characteristics and constants
Plate current is measured as Vp and Vg are varied with rheostats. Plots of ip against Vp at constant Vg are the anode (plate) characteristics; ip against Vg at constant Vp, the mutual characteristics. A triode is normally operated on the straight parts of these curves.
When both voltages change, Δip = ΔVp/rp + gmΔVg. Keeping ip fixed (Δip = 0):
A triode has rp = 10 kΩ and gm = 2.5 mA/V. Find μ, and the change in plate voltage that offsets a 1.0 V drop in grid voltage.
The triode as an amplifier
A small varying signal vs (say from a microphone) is added to the grid voltage: ΔVg = vs. A load resistor RL is in the plate circuit. With the operating point on the linear part of the mutual characteristic, the plate current follows the signal, and the voltage across RL changes by v₀ = RLΔip, a much larger copy of the signal.
The gain
With a battery V in the plate circuit, Vp = V − RLip, so ΔVp = −v₀. Then
The triode above (μ = 25, rp = 10 kΩ) drives a 30 kΩ load. Find the voltage gain and the output for a 0.40 V signal.
- 1Triode: a grid between cathode and anode; i_p = k(V_p + μV_g)^(3/2).
- 2Constants r_p, g_m, μ with μ = r_p g_m (41.9).
- 3Amplifier gain A = μ/(1 + r_p/R_L) (41.10).
Check yourself
Keep exploring
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