Simple Thevenin circuit help

  • Engineering
  • Thread starter nathsen
  • Start date
  • #1
7
0
Thread moved from the technical forums, so no Homework Help Template is shown
Hi, I am struggling to simplify this into a thevenin circuit. Firtsly I am unsure how the zener diode works and why the R1 is not required.(so does it create a voltage drop across the whole loop and , how does it work exactly?) Also unsure if the resistors are in series or not. I think you are menat to determine if the same current is flowing through each resistor to know whether they are in series or not, so if you assume the battery is at the 6.2Vwould you say that R3 and R4 are in series which you add, then the three remaining are in parallel?
The actual measured values are:
R2=5.6KO
R3=2.7KO
R4=68KO
R5=3.3KO
Rth=4.67Ko
Thanks
 

Attachments

Answers and Replies

  • #2
gneill
Mentor
20,913
2,862
Hi nathsen. A zener diode provides a nearly constant potential drop across itself so long as it conducts a current greater than its threshold value (it varies from device to device, but generally just a few mA are required to put it into its normal operating conditions). When its operating in this region it behaves much like a fixed voltage source with a pretty low internal resistance (usually below 20 Ohms and typically in the single digits).

You should see if you can find a V-I curve for a zener to take a look at so that you can get a feel for what it does.

For your purposes treat it like a 6.2 V DC voltage source. It's resistance will be much smaller than all the other resistances in the circuit so its a good bet that it can be ignored.

R1 and the actual voltage source are ignored because they can't be "seen" behind the zener's fixed potential and low resistance.
 
  • #3
7
0
Hi nathsen. A zener diode provides a nearly constant potential drop across itself so long as it conducts a current greater than its threshold value (it varies from device to device, but generally just a few mA are required to put it into its normal operating conditions). When its operating in this region it behaves much like a fixed voltage source with a pretty low internal resistance (usually below 20 Ohms and typically in the single digits).

You should see if you can find a V-I curve for a zener to take a look at so that you can get a feel for what it does.

For your purposes treat it like a 6.2 V DC voltage source. It's resistance will be much smaller than all the other resistances in the circuit so its a good bet that it can be ignored.

R1 and the actual voltage source are ignored because they can't be "seen" behind the zener's fixed potential and low resistance.
Thanks for your help. So assuming it's a 6.2 V DC source, under my understanding I just simplify the remaining resistors now.
Where am I going wrong? I'm thinking R3 and R4 are in series, then add the resultant in parallel to R2 using product over sum, then that resultant product rule over sum to R5?
 
  • #4
gneill
Mentor
20,913
2,862
I think you're having trouble recognizing the parallel and series components. The circuit diagram is laid out in a manner that can make it tricky.

Take several colored pens and trace the wires of each node with a different color (or use some other means to uniquely identify the wiring associated with separate nodes). Components that are in parallel will be connected to the same pair of colors. Remember, when you suppress the voltage source (zener) it's replaced by a wire, so both of its connections become part of the same node.
 
  • #5
7
0
I think you're having trouble recognizing the parallel and series components. The circuit diagram is laid out in a manner that can make it tricky.

Take several colored pens and trace the wires of each node with a different color (or use some other means to uniquely identify the wiring associated with separate nodes). Components that are in parallel will be connected to the same pair of colors. Remember, when you suppress the voltage source (zener) it's replaced by a wire, so both of its connections become part of the same node.
Thanks very much for your reply. I've tried to follow your advice (see picture). Is that correct in theory? I am still not getting the value desired but is it through other means rather than how I'm simplifying the circuit!?
 

Attachments

  • #6
gneill
Mentor
20,913
2,862
Remember that when the voltage source is suppressed it is replaced by a short circuit (wire). So your pink and green nodes should become all one color.
upload_2016-4-26_16-5-54.png
 
  • #7
7
0
Remember that when the voltage source is suppressed it is replaced by a short circuit (wire). So your pink and green nodes should become all one color.
View attachment 99811
Ok so i've gone over the green with pink. Now I need to identify components connected to "same pair of colours". So actually Resistor 3 and 4 are both connected to pink and blue which means they are in parallel. And 2 and 5 are also in parallel? Sorry i'm getting a mental block over this!
 
  • #10
gneill
Mentor
20,913
2,862
Sometimes redrawing the circuit in a slightly different way can yield insights; you pick up relationships that you missed before.

If you move R2 and the zener to different locations in the same wire (so no topological changes) as shown below, then you should be able to see that the circuit on the left can be morphed in to the circuit on the right:

upload_2016-4-27_21-37-54.png


If you then suppress the zener as you would a voltage source, what do you get for the Thevenin resistance of the network?
 
  • Like
Likes cnh1995

Related Threads on Simple Thevenin circuit help

  • Last Post
Replies
1
Views
1K
Replies
1
Views
1K
  • Last Post
Replies
2
Views
2K
Replies
3
Views
1K
Replies
2
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
6
Views
740
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
1K
Replies
4
Views
3K
Top