# A curious circuit

• posted

It increases.

John

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Why did you find it interesting?

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(I fixed your drawing, I think... You meant two zeners of 6.1V each, right?)

According to spice, the voltage peaks at about 220 ohms.

There is a balancing act between the zeners and the resistors. When R is

0 ohms, the voltage across the circuit is clearly 10V. As R increases, the voltage increases, that is, until the zener diodes start to conduct. That point is when 5mA * (1k+R) == 6.1V, ie, where R=220 ohms. After that, as R gets larger, more of the current flows directly through the zeners, bypassing R. So, the voltage decreases. When R is infinite, the voltage is 6.1+(2.5mA * 1k) = 8.6V, since the current divides between the two branches.
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Regards,
Bob Monsen
• posted

Posting a circuit I saw in EE Times lo these many years ago.

current source

5 mA

| V1

|

----------

| |

| |

| |

/=====/ >

^ 1K < 1K

/ \\ >

--- <

| R? |

|--/\\/\\/\\--|

| |

/=====/
< ^
1K / \\ 6.2V
< ---

| |

| |

----- -----

/ / / / / /

R? has an initial value of 1K ohms.

Decrease it to 100 ohms.

What happens to voltage V1 as the resistance decreases?

I found it interesting.

PN2222A

• posted

With nodes and wires labeled as shown, write all the immediately known facts: [1] Io = (I1 + I2) = (I3 + I4) (parenthesis for emphasis) [2] I2 + I4 + I5 = 0 [3] I1 + I3 + I5 = 0 (Kirchoff's node rule) [4] V1 = V3 + 6.2 = (V1 - V2) + 6.2 (loop rule; assuming zeners are forward-biased, a dangerous proposition at only 5mA) [6] I5 = (V3 - V2) / R [7] I1 = (V1 - V2) / R1 [8] I4 = V3 / R2 (Ohm's law) Now it's algebra. Add [2] + [3]: (I1 + I2) + (I3 + I4) + 2*I5 = 0 (parenthesis for emphasis) Transitive property with [1]:

2*Io + 2*I5 = 0 Substitute [6]: 2*Io + 2*(V3 - V2) / R = 0 Io = (V2 - V3) / R Take [4] and cancel the 6.2 volt terms: V3 = V1 - V2 --> V1 = -(V2 + V3) --> (V2 + V3) = -V1 (Signs are whacky in this problem, better to keep them straight as we go, and figure it out later. The magnitudes are right.) Substitute modified [4] into our equation: Io = (-V1) / R --> - R * Io = V1. (The current obviously isn't in opposition to V1 nor is R negative, so I got something backwards in the problem. Again, only the quantity matters; sign is a point of view depending on which way your voltmeter probes are. ;-)

...Odd, when I did that on paper, I got V1 = Io * R * R1/R2, which of course comes out the same because R1 = R2 in the problem, but we haven't stated this fact -- see [7] and [8].

If the zener diodes are not turned on, i.e., V1 - V3 < 6.2V and V2 < 6.2V, the problem becomes three resistors in series, and V1 = Io*(R1 + R + R2).

Neither of these equations suggest a maxima with respect to R though, so something must be fishy.

Tim

• posted

You need to analyse it as two circuits based on breakpoints... zeners not conducting, or zeners conducting.

Below ~255 ohms, the zeners aren't conducting.

At zero ohms the voltage is (obviously) 10V.

As the resistor increases you reach a peak of ~11.137V, and the zeners begin conduction.

Any value above 255 ohms the voltage begins falling, being ~8.99V at

10K.

...Jim Thompson

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|  James E.Thompson, P.E.                           |    mens     |
• posted

current source 5 mA | V1 | ---------- | | | | | | /=====/ >

^ 6.2V < 1K / \\ >

--- < | R? | |--/\\/\\/\\--| | | > /=====/ < ^ > 1K / \\ 6.2V < --- | | | | ----- ----- / / / / / /

I found it interesting (as did the original article) because decreasing r? intuitively would seem to reduce the total resistance (or voltage drop) of the network.

Imagine that instead of wires and components, this is a network of roadways with the particular kinds of traffic jams: Opening up a connector road within the roadways would not be expected to increase the driving delay -- more roads, less traffic ... (short term result only).

But here, making a bypass road available increases the delay.

Regards PN2222A

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I read in sci.electronics.design that PN2222A wrote (in ) about 'A curious circuit', on Sun, 6 Nov 2005:

What is the road transport analogue of a zener diode? A traffic light at road works?

--
Regards, John Woodgate, OOO - Own Opinions Only.
If everything has been designed, a god designed evolution by natural selection.
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With ideal components, by inspection the voltage across R will be 1.2v, R = 240O at break and the total 11.2v. With R infinite v = 8.7v

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R =240, the ohms sign became a 0

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Yep, I used a real 6.2V zener model and swept R.

...Jim Thompson

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|  James E.Thompson, P.E.                           |    mens     |
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Equate voltage with delay time, current with traffic volume. The analog would be a road with ample capacity but speed limit so travel time is independent of traffic volume.

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I read in sci.electronics.design that Don Foreman wrote (in ) about 'A curious circuit', on Tue, 8 Nov 2005:

That's a delay line. I equated queue length with voltage (potential) and volume to current.

--
Regards, John Woodgate, OOO - Own Opinions Only.
If everything has been designed, a god designed evolution by natural selection.

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