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So I've been learning about current source circuits - this one in particular. I'm told that an ideal current source should give out the current no matter what happens to the load. But it should also have a ideally infinite internal impedance.

Rb, the 2 diodes and Re ensure that the current Ic remain constant - but I'm not sure how to prove that this circuit satisfies the high internal impedance requirement.

Any guidance?

current source circuit

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  • \$\begingroup\$ Remember the meaning of the slope of the transistor`s output characteristic curves Ic=f(Vce) for a set of constant currents Ib. \$\endgroup\$ – LvW Oct 27 '14 at 7:56
  • \$\begingroup\$ Also when hunting for clues on this one note it would traditionally be referred to as a current sink. If the current remains constant for varying voltage then the impedance is infinite. You can plot a V/I curve and see the working part of the curve is vertical at the set current (the actual slope is high and not vertical if you simulate or calculate with real-world components). \$\endgroup\$ – KalleMP Oct 27 '14 at 10:27
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    \$\begingroup\$ "an ideal current source should give out the current no matter what happens to the load. But it should also have a ideally infinite internal impedance." Using the word "but" makes it sound like you think these are conflicting goals. Actually, infinite output impedance is why an ideal current source has constant output current. \$\endgroup\$ – The Photon Oct 27 '14 at 16:08
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schematic

simulate this circuit – Schematic created using CircuitLab

This circuit is a bit similar with the well known Widlar current source. I only give the small signal diagram, and my result (the calculation procedure is not difficult, but it will be painful to write it here). If you think the circuit is right, then you can do some calculations with it, and compare with my result.

$$ R_{o} = r_{o}[1+(R_{e}||(r_{\pi} + R_{b}||(r_{d1}+r_{d2})))(g_{m}+1/r_{o})] $$

If \$(1/r_{o}) \ll g_{m}\$, then

$$ R_{o} \approx r_{o}(1+g_{m}(R_{e}||(r_{\pi} + R_{b}||(r_{d1}+r_{d2})))) $$

From the equation of \$R_{o}\$, you can analysis how the components can affect your output impedance.

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I'm not all that good at single transistor circuits. But to analyze this I think you need to get beyond the simple transistor model. The first thing to look at might be the Early voltage (or Early effect.) As different loads are put on the collector, there will be different Vce voltages. And the that will change Ic for a given Vbe. (The Ebers-Moll equation.) Mind you I'm not sure how you fold this back into the circuit and calculate an output impedance. (Hopefully someone more knowledgeable will speak up.)

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In my first comment I have mentioned the slope of the Ic=f(Vce) set of curves. With respect to G. Herold`s answer I like to make clear that a measure of this this slope is the so-called Early voltage.

However, it is to be noted that the output resistance of the shown circuit is further increased due to the negative feedback effect caused by the emitter resistance Re. The additional factor to be considered is (1-LG) with loop gain LG=-gm*Re(gm:transconductance).

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The small-signal output impedance, \$z_{out}\$, can be determined by adding a small-signal voltage source at the output node (the collector of the BJT) and calculating the resulting small-signal change in current. Then \$z_{out} =v_c/i_c\$.

schematic

simulate this circuit – Schematic created using CircuitLab

The first step I would take to calculate \$i_c\$ is to replace the BJT with the hybrid-pi model.

schematic

simulate this circuit

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