currentimpedancesource
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?
A constant power load varies it's impedance on change of input voltage to keep the power constant. A constant impedance load is simply a load that presents an unchanging impedance, like a resistor. An L-Pad is used to change speaker output level whilst maintaining a constant impedance load to the amplifier.
A good example of a constant power load is a switching regulator. Since this has to maintain it's power into it's load, it must draw the same power from it's source even if the source changes voltage.
This is also an example of a negative impedance because in order to maintain the output power, if the voltage in drops, the current must rise (opposite to a standard resistor where the current and voltage rise/fall with each other)
Here is an example circuit, made from a LT1377 boost switching regulator:
Here is the simulation:
The input voltage V(in) (blue trace) starts off at 4V, and gradually rises up to 10V. We can see the power (red trace) stays constant at ~1W over a change of 6V at the input (it's not perfect as it's meant to represent "real life" and not 100% efficient, but it's pretty close)
We can also see the dynamic negative resistance characteristic (green trace) which is due to the input current falling as the voltage rises. Tt falls from ~300mA to ~120mA over the voltage rise from 4V to 10V - don't be confused by the minus sign, that's just the direction of measurement in LTSpice.
The dynamic resistance slope can be roughly calculated by (4V - 10V) / (300mA - 120mA) = -33.3Ω. Looking it another way, 6V / -33.3Ω = -180mA.
There are 2 ways to understand this.
Other than that, practically speaking, HIGH logic state in practice never behaves as ground (0), although at times it may behave as 1.
Best Answer
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.