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enter image description here

Above figure is a BJT transistor emitter follower.

I can understand that in an emitter follower when talking about its input impedance what meant is Vb/Ib and here it is equal to (hfe+1)*R. The idea I think similar to Thevenin equivalent. The voltage applied to circuit input is Vb; and the voltage driven by the source is Ib. So it is easy to understand the meaning of input impedance here for me.

But I couldn't understand what meant by the output impedance of this circuit. What is the meaning of the output impedance here and how can we quantify it in terms of R hfe ect.?

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  • \$\begingroup\$ Assuming normal biasing of the BJT the input impedance at the base node is r,in=hie + Re(1+hfe). This is without considering the bias resistor network in parallel to r,in. \$\endgroup\$ – LvW Dec 24 '15 at 8:42
  • \$\begingroup\$ I think you mean "the current driven by the source is Ib". \$\endgroup\$ – Andy aka Dec 24 '15 at 9:00
  • \$\begingroup\$ Quote: "But I couldn't understand what meant by the output impedance of this circuit." You will remember that a VOLTAGE source is required to have a very low output resistance so that the voltage remains nearly constant - independent on the connected load. Now - when the output resistance of a BJT amplifier is (very) low, the amplified voltage will be relatively independent on any load resistor. The opposite is true for a large output resistance. For this reason, the internal output resistance plays an important role for calculating the influence of any load upon the output signal. \$\endgroup\$ – LvW Dec 24 '15 at 13:03
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But I couldn't understand what meant by the output impedance of this circuit. What is the meaning of the output impedance here and how can we quantify it in terms of R hfe ect.?

You've got to think like a transistor but, to help you, try thinking like a forward biased diode. Let's say your anode voltage is connected to a perfect battery (strong voltage source) then, under very light load conditions on the cathode (say 0.1 mA), the cathode voltage might be 0.5 V lower than the anode: -

enter image description here

If you increased the current a little bit to 0.15 mA, the forward voltage dropped would be about 0.525V. Take the "before" and "after" numbers and work out what the dynamic resistance is: -

Dynamic resistance = change in volts / change in amps = 25 mV/ 0.05 mA = 500 ohms.

If you did this at a higher level of forward conduction (say at 1mA to 1.5mA) you'd get a forward volt change of 10 mV (640 mV to 650 mV). Now, the dynamic resistance has become 10 mV / 0.5 mA = 20 ohms.

This dynamic impedance represents the output impedance of the diode (at the cathode) when the anode is tied to a solid fixed supply voltage.

schematic

simulate this circuit – Schematic created using CircuitLab

If the diode's cathode were connected to a fixed 100 ohm resistor to ground, can you see that the dynamic impedance of the cathode in parallel with the fixed 100 ohms becomes the new output impedance: -

schematic

simulate this circuit

Think about converting the voltage source in series with the dynamic impedance of the diode to a current source like this: -

enter image description here

You see now that the dynamic impedance of the diode is parallel to the 100 ohm and this makes the net resistance smaller. You don't have to consider what value the voltage source is or what the current source amperage is because you reconvert back to the thevenin equivalent and you have the net output impedance.

Note - this added resistor is the emitter resistor when a BJT is considered

So, if the total current drawn from the cathode is ~0.1mA the "source" impedance of the cathode and the combined resistor is 500 ohms || 100 ohms = 83 ohms.

This drops to 20 ohms || 100 ohms (17 ohms) when ~1mA is being taken through the cathode. How does all this relate to a BJT you might be asking. Here's how...

The base emitter junction is a forward biased diode; the base is the anode and the emitter is the cathode but, the clever thing about BJTs is that although there may be "weak voltage" at the base that is easily affected by loading, the collector current doesn't allow this to happen and it is the collector current that replaces the base current as the source of current to the emitter (cathode). Thus, you can still regard the emitter has having the output impedance of a diode when that diode's anode is connected to a strong voltage source.

What I've said is a little oversimplified because there is still a bit of base current taken when providing current to the emitter but, this is usually at about 1% of the level of the pure diode scenario - this boils down to the BJT having a current gain of about 100.

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  • \$\begingroup\$ isnt R1 and D1 in series? doesnt it act as a voltage divider? why is that treated like a parallel circuit when finding output impedance? is that the Thevenin equivalent? \$\endgroup\$ – user16307 Dec 24 '15 at 15:22
  • \$\begingroup\$ They are in series and they do form a divider and yes, it's all about the thevenin equivalent resistance. \$\endgroup\$ – Andy aka Dec 24 '15 at 16:05
  • \$\begingroup\$ you wrote as if they were parallel: "can you see that the dynamic impedance of the cathode in parallel with the fixed 100 ohms becomes the new output impedance" are they in series or parallel? \$\endgroup\$ – user16307 Dec 24 '15 at 16:09
  • \$\begingroup\$ The anode voltage and series diode drop can be converted to a current source in parallel with the dynamic impedance. That dynamic impedance is now in parallel with the 100 ohms. Recalculate the combined impedance then convert back to a voltage source with that new impedance in series. Does that help you understand? \$\endgroup\$ – Andy aka Dec 24 '15 at 16:16
  • \$\begingroup\$ not really im confused. the dynamic resistance of the diode is in series in your drawing but how is that also in parallel i dont get. i cant see the parallelism anywhere \$\endgroup\$ – user16307 Dec 24 '15 at 16:19
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Using ONLY what you have provided, then the output impedance would be Re (the emitter resistor) in parallel with the source impedance/(hfe+1). Since you have no source shown, the output impedance is simply the resistor on the emitter of the transistor. It sounds too simple (and is). However, if you're only using first order approximations, then it's correct.

Edit: The output impedance is the impedance looking into the output (as if it were an input). This is important because the output impedance becomes the source impedance of the next stage, as well as effects the amount of power transfer to a load.

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  • \$\begingroup\$ i dont understand. why is that? what is the meaning of output impedance here? \$\endgroup\$ – user16307 Dec 24 '15 at 6:47
  • \$\begingroup\$ you re saying the results but not explaining why. thats why i asked it. \$\endgroup\$ – user16307 Dec 24 '15 at 6:49
  • \$\begingroup\$ @jjuserjr: The output impedance is the next stage's source impedance. \$\endgroup\$ – Ignacio Vazquez-Abrams Dec 24 '15 at 6:58
  • \$\begingroup\$ The output impedance of the circuit is the impedance to be measured/calculated looking into the output node (and with the input signal set to zero). \$\endgroup\$ – LvW Dec 24 '15 at 8:43
  • \$\begingroup\$ Assuming that you are interested in the output resistance of a working transistor stage , the output resistance impedance r,out is, of course, Re||r,e with r,e=dynamic input resistance at the emitter node in common base configuration. \$\endgroup\$ – LvW Dec 24 '15 at 11:04
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The output impedance of a circuit, comes of considering the same as a two-port network. For that network, different sets of parameters are defined, such as impedance and admittance parameters.

Particularly, the output impedance is defined as the impedance that can be measured by applying voltage and measuring the current on the output terminals when the input signal is zero.
In the case of a transistor as an emitter follower, the output terminals are connected on \$R_E\$.

The output impedance can assimilate the equivalent impedance which is obtained by reducing the circuit to its Thevenin equivalent. Thevenin's theorem is only applicable to linear circuits, so to apply in this case, you must first raise the small signal equivalent circuit. Once obtained this circuit, you can get the equivalent and input impedance.
The output impedance of this stage is the impedance that "see" the next step as the impedance of the signal source.

For this schema, the output impedance is

$$ Z_o = R_E \vert\vert \dfrac{h_{ie}}{h_{fe}+1} $$

where \$h_{ie}\$ depends on the biasing conditions, and is a representative fact of the linearization introduced in the small signal model.

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  • \$\begingroup\$ how did you derive dynamic input resistance which is parallel to Re? \$\endgroup\$ – user16307 Dec 24 '15 at 15:24
  • \$\begingroup\$ \$h_{ie}\$ is parallel from Thevenin's equivalent circuit. You must pasivate all voltage and current sources; then \$V_i\$ is pasivated, i.e. the base is grounded and \$h_{ie}\$ results in parallel with \$R_E\$. \$\endgroup\$ – Martin Petrei Dec 25 '15 at 22:00
  • \$\begingroup\$ I couldnt find any tutorial which explains what you guys saying step by step. "hie result in parallel with RE" but arent they in series? \$\endgroup\$ – user16307 Dec 25 '15 at 22:04
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    \$\begingroup\$ Here is an example of application of Thevenin's Theorem: en.wikipedia.org/wiki/Th%C3%A9venin's_theorem#Example \$h_{ie}\$ in parallel, comes from the derivation of equivalent circuit. \$\endgroup\$ – Martin Petrei Dec 25 '15 at 23:19

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