This was inspired by an analogous mechanical conundrum using springs and string:
Two simple circuits, each using 10k resistors and 10V zeners, each fed 1mA constant current - which voltage is higher, output1 or output2?
The right hand circuit is the same as the left one apart from one resistor (R5) being shorted out.
For the inspiration: The Spring Paradox - YouTube
Most people's intuition is that shorting the resistor will lower the impedance of the whole circuit.
Two simple circuits, each using 10k resistors and 10V zeners, each fed 1mA constant current - which voltage is higher, output1 or output2?
The right hand circuit is the same as the left one apart from one resistor (R5) being shorted out.
For the inspiration: The Spring Paradox - YouTube
Most people's intuition is that shorting the resistor will lower the impedance of the whole circuit.
The second circuit has two series Zeners from CCS to ground, so the CCS output is at 2 x 10V = 20V.
The first circuit fixes the voltage at the junction of the top and center 10k at 10V wrt ground.
The other Zener fixes the voltage difference between the CCS node and the junction of the bottom and center
10k resistors at 10V. Since the resistors are equal, they divide the 10V equally across each 10k, at 5V each.
Then the voltage on the bottom resistor is the known 10V (at the junction of the top and center 10k)
less the known 5V across the middle 10k, or 5V.
Then the CCS voltage is that 5V on the lower resistor, plus the left Zener 10V, or 15V.
The first circuit fixes the voltage at the junction of the top and center 10k at 10V wrt ground.
The other Zener fixes the voltage difference between the CCS node and the junction of the bottom and center
10k resistors at 10V. Since the resistors are equal, they divide the 10V equally across each 10k, at 5V each.
Then the voltage on the bottom resistor is the known 10V (at the junction of the top and center 10k)
less the known 5V across the middle 10k, or 5V.
Then the CCS voltage is that 5V on the lower resistor, plus the left Zener 10V, or 15V.
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I remember what might be a much earlier take on this kind of idea.
It might have been in an ancient copy of ETI perhaps dated to the early 70's and was a network intended to model traffic flow and the idea was to show that reducing circuit impedance to model wider roads capable of handling more traffic (it was an R and Zener network) actually caused less current overall to flow (and so increased congestion).
It was a neat 'trick', particularly to a youngster who devoured any and all old electronics mags he could get his hands on 😉
It might have been in an ancient copy of ETI perhaps dated to the early 70's and was a network intended to model traffic flow and the idea was to show that reducing circuit impedance to model wider roads capable of handling more traffic (it was an R and Zener network) actually caused less current overall to flow (and so increased congestion).
It was a neat 'trick', particularly to a youngster who devoured any and all old electronics mags he could get his hands on 😉
There is a large web site full of old Wireless World mags.
WIRELESS WORLD: UK technical magazine 1913-2008
I bought a DVD with most of the old Maplin magazines and project books off ebay.
WIRELESS WORLD: UK technical magazine 1913-2008
I bought a DVD with most of the old Maplin magazines and project books off ebay.
According your reasoning, the current on the middle resistor is 5V / 10k = 0.5 mA. But this can't be true, because the top resistor carries the sum of currents of the middle resistor + bottom zener. Likewise the bottom resistor carries the sum of currents of the middle resistor + top zener. That means the currents are not equal on the three resistors (the middle resistor carries less), neither are the voltages.
Output2 is simple.
1mA and 10k gives 1mA, all the current goes to the resistors nothing for the zeners, output is 1 x (10k+10k) = 20V.
Output1 is more difficult.
1/3mA goes to the zener and 2/3mA to R1 (or R2).
Output1 is then 10V + 2/3 x 10k = 16,666666 V or 16 2/3 V
and VR5 = 10/3 V.
Mona
1mA and 10k gives 1mA, all the current goes to the resistors nothing for the zeners, output is 1 x (10k+10k) = 20V.
Output1 is more difficult.
1/3mA goes to the zener and 2/3mA to R1 (or R2).
Output1 is then 10V + 2/3 x 10k = 16,666666 V or 16 2/3 V
and VR5 = 10/3 V.
Mona