Bipolar Voltage Divider

[Tucson Annie]

Hi Guys,

This isn't really a homework question, but it's close enough to fit this category I'm studying DC Series-Parallel analysis and have a voltage divider circuit with loads attached. Now, if that's all it was, I can figure it out using the 10 percent bleeder current rule....(I was taught to add the load currents and multiply them by 10 percent to find the bleeder current and then add the bleeder current to the load current to get the total current).

But due to the placement of the reference in this diagram, I'm thrown for a loop so to speak. I am not sure how to go about solving the circuit because of it.....here is the schematic:

**broken link removed**

Now, I'm not supposed to use mesh or nodal or Thevinen methods yet so anyway, any help will be greatly appreciated! As I say, if it wasn't for the position of the ground reference I'd know how to do this....

Annie

 

[Hero999]

Here's how I'd do it:

Calculate the value of R1 for a voltage drop of 3V with a current of 23.53mA using Ohm's law.

Work out the resistance of Load A and B.

Work out the required potential divider ratio.

I'd probably remove R2 and use the value of Load A as the top resistor (comprising of R3 and load B).

Rearranging the potential divider formula, I'd calculate the required value for the lower resistor

I'd rearrange the formula for calculating two resistors in parallel to get the value for R3.

This will only work if the current flowing through load A and B is constant.

 

[Tucson Annie]

Thank you, Hero!

So I don't have to add the current that is flowing through the negative load to the total current through R1? Hmmmm. The reason this problem is a problem for me (so to speak) is this....my AC/DC Electronics Professor posted something very similar to the above on the board last week. I drew the circuit in my notes but did not take good enough written notes to see what he had given us up front and what I had calculated out....so, now I am at a loss if he throws it at me on quiz tomorrow. I haven't been able to find any similar problems in my books or on the web...just the one above....I guess it must not be super common....

Thanks again,

Annie

 

[Hero999]

I have never heard of the 10% bleeder rule, does it mean the current though the potential divider should be approximately ten times the load?

To simplify what I wrote previously, I'd work out the load resistances then calculate the values of R1, R2 and R3 to get the required voltages.

R2 isn't needed because the current though load A is higher than load B. If R2 is removed, the current through R1 will be equal to the current through load A.

 

[MrAl]

Hi there Annie,


Here's another way to approach this circuit problem...


Since the only problem you seem to be having is that the reference voltage
is in a 'weird' place (ie connected in a way you are not used to seeing)
you can fix that in a heartbeat by simply moving the ground to the bottom
of R3 which is also the bottom of V1 (the supply voltage). You then
have a circuit with a supply voltage that is probably closer to what you are
more used to seeing, so you can solve the network the way you usually do.
After you solve the network you can then move the ground back and adjust
the voltages by using simple subtraction. This is possible with this circuit
because moving the ground does not change the voltage drops, only the
reference node.

Here's an outline of the process:

1). Move the ground to the lower end of V1 (same as lower end of R1).
2). Label the voltage at the top of R2 as VA and the voltage at the top
of R3 as VB.
3). Solve for VA and VB the way you normally do.
4). Move the ground back to the original location, then calculate the voltage
at the bottom of R3 as v=-VB and the voltage at the top of R2 as v=VA-VB.


BTW, if the two 'loads' are resistive you can calculate the resistance first before you start, then
put that resistance in parallel with the other resistance then go from there.

Note that in this post VB is being underlined automatically and that is not intentional. It may be
the browser doing that so ignore that if you see the underlining.

 

[Hero999]

I agree it's a good idea to move the reference to make it easier to calculate.

Regarding the underline of VB. The new forum software has a new feature which explains initialisms when the mouse cursor is moved over them. Unfortunately it's not very smart so it always translates VB to vBulletin when it could mean something else (Visual Basic) or be meaningless.

 

[Tucson Annie]

Hi Guys,

Thanks for the input! You know what drives me crazy? That different text books describe the same circuits in different ways. The fact that you haven't heard of bleeder currents is not surprising because it isn't in any of my other textbooks, just the one I'm using in class now..also, some textbooks use conventional current flow and some use electron flow, that drives me crazy!....so, here is what I am going to do....I am going to ask my instructor how to solve this and then I'll post what his answer is...

Thanks again...

Annie

 

[MrAl]

Hello again,


Annie:
That's a good idea because with many courses the answers depend partly on what the instructor is looking for anyway.

Hero:
Ha ha, i see what you mean, geeze that sure is strange
Some quick tests: vb vB Vb VB

 

[Hayato]

Well,

LOAD A and R2 are submitted to the same potential (12V), because they are in parallel.

LOAD B and R3 follow the same case above, but they are submitted to - 20 V.

Current of LOAD A + R2 = Current of LOAD B + R3 = Current of R1 (they are in parallel).

The total battery voltage is 35 V. So, if tou hava a 12V drop at LOAD A, a 20V drop at LOAD B, your drop at R1 will be 35 - (12 + 20) = 3V

Following the KCL and KVL:

12/R2 + 23.53mA = 20/R3 + 17.65 mA

12/R2 - 20/R3 = -5.88 mA (If we substract the current passing through R3 from the current passing through R2 we must have a 5.88 mA excess, to balance the currents).

Now, if I understood the bleeder rule:
(23.53 + 17.65)*0.1 = 4.118 mA

Let's take that value as the reference, as far I understand, you need at least a 10% of current bleeding, so:

IR2 = 4.118 mA
IR3 = 4.118 mA + 5.88 mA = 9.998 mA


Total current at LOAD A and R2 = 4.118 + 23.53 = 27.648 mA
Total current at LOAD B and R3 = 9.998 + 17.65 = 27.648 mA
KCL met!

R2 value = 12V/4.118mA =~ 2,914 ohms
R3 value = 20V/9.998mA =~ 2,000.4 ohms
R1 value = 3V/27.648mA =~ 108.5 ohms

 

[Hero999]

Here's how I'd work it out, it's easier than I thought:

R2 is not used.
R1 = (35-12-20)/23.53mA = 127.5R
I(R3) = 23.53mA - 17.65mA = 5.88mA
R3 = 20/5.88mA = 3.4k

It might not be what your lecturer is looking for though as I don't understand the 10% bleeder rule.

 

[Hayato]

The thing is, it is very arbitrary. Agreed with Hero.

Since the bleeder rule was used to help bias electron tubes, it is very arbitrary.

The important thing is always follow the KCL and KVL, so that your circuit does not present any blackholes (energy disappearing) or whiteholes (energy summoning) =P.

 

[Russ Hensel]

What course are you taking ( and where is it being offered )? Just curious.