# A curious circuit

Discussion in 'Electronic Design' started by PN2222A, Nov 6, 2005.

1. ### PN2222AGuest

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

current source

5 mA

| V1

|

----------

| |

| |

| |

/=====/ >

^ 1K < 1K

/ \ >

--- <

| R? |

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

| |
< ^
< ---

| |

| |

----- -----

/ / / / / /

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

2. ### John - kd5yiGuest

It increases.

John

3. ### Bob MonsenGuest

(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.

---
Regards,
Bob Monsen

Mathematics is nothing more, nothing less, than the exact part of our
thinking.
- Luitzen Brouwer

4. ### Luo XiaoZenGuest

Why did you find it interesting?

5. ### PN2222AGuest

current source
5 mA
| V1
|
----------
| |
| |
| |
/=====/ >
^ 6.2V < 1K
/ \ >
--- <
| R? |
|--/\/\/\--|
| |
< ---
| |
| |
----- -----
/ / / / / /

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:
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

6. ### John WoodgateGuest

I read in sci.electronics.design that PN2222A <>
What is the road transport analogue of a zener diode? A traffic light at

7. ### Tim WilliamsGuest

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.
(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 <divide by two--> 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

8. ### Jim ThompsonGuest

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

9. ### GPGGuest

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

10. ### GPGGuest

R =240, the ohms sign became a 0

11. ### Jim ThompsonGuest

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

...Jim Thompson

12. ### Don ForemanGuest

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.

13. ### John WoodgateGuest

I read in sci.electronics.design that Don Foreman