The circuit in Figure 1 is suited for inductive sources such as a dynamic microphone, transformer or inductance sensor/pick up.
Figure 1 A simple amplifier circuit suited for inductive sources such as a dynamic microphone, transformer or inductance sensor/pick up.
The circuit is versatile; its gain can be varied from 1 to about 130. it also has a wide power range (approximately 5 to 15 V). It uses no electrolytic capacitors, which improves reliability.
A low-impedance dynamic microphone can be connected to the input with or without any matching transformer. The circuit is powered through a two-wire output cable, i.e., the load resistor is separated from the circuit by a (screened) cable.
The maximum gain is about 130. This rather high value is provided by the common emitter input stage (low noise transistor Q1) with dynamic load R3, Q2.
While the maximum gain is achieved when R5=0 Ω, the gain can be reduced by choosing a non-zero value of resistor R5, which introduces a local feedback. Such local feedback also improves the circuit linearity.
For example, when R5=180 Ω, gain is about 60. The emitter follower (Q2) also provides a low output resistance for the circuit.
Like any other device connected to an inductance or long connecting cable, the circuit can be subject to overvoltages on its terminals. Diode D1 protects both input and output to avoid the related damage.
To calculate values of resistors we can begin with values of the power voltage E, collector current of Q1, and value of load resistor R6.
For example let E = 9 V and Ic1 = 0.3 mA, then
R3 = Vbe2 / Ic1 = 0.8 / 0.3 = 2.7 (k) approx.
To provide the needed output headroom we can choose the emitter voltage of Q2 as one-half of the power voltage:
Then we have a simple equation:
Ve2 * R2 / (R1+R2) = Vbe2 + Vbe1,
from which the ratio R1/R2 can be obtained.
Since Ve2 = (Vbe2 + Vbe1)(1 + R1/R2) and doesn’t depend on the value of E, the left-part of the circuit on DC represents a kind of Zener diode, which should be taken into account when you intend to raise the value of E.
Figure 2 shows how the circuit can be modified to get the gain of about 1.
Figure 2 The modified circuit to get a gain of about 1.
—Peter Demchenko studied math at the University of Vilnius and has worked in software development.
Any estimations of THD/power consumption/bandwidth ?
The circuit in Fig 1 is a common emitter with local feedback, so for demanding apps some compromise Gain-THD may be considered. Nevertheless, the active load (Q2) has to make the THD far lower then the one for the simple resistive load. Depending on trans used and signal and feedback levels, the value of THD can be estimated as 0.01% – 0.001%.
Since the circuit in Fig 2 is an emitter follower with the active load, its THD is even lower than THD of simple emitter follower and in the case of large signal depends mostly on the linearity of the transistors used ( 0.001% – 0.0001% ).
As I noted: Ve2 = (Vbe2 + Vbe1)(1 + R1/R2), so the current the circuit draws is equal: (E – Ve2) / R6.
Please explain how D1, which can only conduct if the base of Q1 is somehow driven positive relative to the emitter of Q2, protects against “overvoltages on its terminals.”
I guess the mic could do that a little? though I don’t see how that would be considered “overvoltage”.
The “overvoltage”: here I mean any possible transient which can occur on the _circuit’s_ terminals – due to some electrostatic discharge and/or a long cable or signal source electromagnetic pickup. Shit happens! The diode has to limit possible reverse voltage on the base-emitter junctions…
Really? If THAT’S its purpose, then shouldn’t its anode be connected to common (i.e., transistor collectors and cable shield), not Q1’s base? Then the circuit would be completely protected from negative excursions because they would be shunted directly to common instead of through Q1’s base-collector junction, plus you’d avoid having the diode’s junction capacitance (which could be considerable if it’s a really robust part) eroding the amplifier’s high frequency response.
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