# Shunt Resistor help

Discussion in 'Homework Help' started by tristonian, Mar 24, 2012.

1. ### tristonian Thread Starter New Member

Mar 23, 2012
2
0
Hello aac people,

I have a very basic question that has stumped me for the last two days despite countless searching. I'm sure the answer is very basic and a clear explanation would help out.

I an attempt to teach myself electronics I studied up on resistors from a couple resources (art of electronics and this site). I then have been trying to apply my knowledge with a very basic circuit: the U-pad. I followed along this website that explains the theory behind pads and has instructions on how to make your own: http://www.uneeda-audio.com/pads/

Here's the problem:
I don't understand how the load (mic preamplifier in this case) sees the shunt resistor as the source impedance (150 ohms in this example) and not the total resistance of the 2 series resistors (1k each in this example) and the the shunt resistor.

I do understand the voltage divider portion of the circuit and that the voltage drop occurs as a product of the ratio between the series resistor and the series resistor.

I don't understand what is happening by bridging the + and - signal in this balanced circuit other than it is allowing current to pass from + to - and vice versa.

I don't understand how to calculate the thevenin equivalent resistance of the U-pad

Any help, clarification, or links would be greatly appreciated.

Thanks,
T

2. ### tristonian Thread Starter New Member

Mar 23, 2012
2
0
Would love to get just a link that can help me unfold this mystery that I hear is so simple! Reread resistor section in the DC Chapter on this site. Any guidance is appreciated

T

3. ### WBahn Moderator

Mar 31, 2012
23,384
7,096
Without hunting through that guy's website, I don't know exactly what circuit you are referring to.

If you have two 1kohm resistors in series and the pair are connected between points A and B, then a source connected to A and B will see 2kohm as the effective resistance. For instance, imagine connecting a fixed voltage source of 10V across A and B and you will get 5mA of current flowing. Now, put an additional 200ohm resistor between points A and B (known as 'shunting' the original load since 'shunting' merely implies the redirection of something around something else) and the source will see the parallel combination of the two. The source will still provide the 5mA to the two 1kohm resistors, but it will also provide 50mA to the 200ohm resistor. From the source's perspective, it was hooked up to a load that resulted in a total of 55mA flowing, from which it would conclude that it is connected to a 182ohm resistor.

In general, if two resistors that are significantly far apart in value (one is more than, say, ten times the value of the other), then the series combination will look like something just a little bigger than the bigger of the two while the parallel combination will look like something just a little smaller than the smallest of the two.

4. ### mlog Member

Feb 11, 2012
276
36
The Thevenin resistance of the U-pad is found the same way as any other Thevenin resistance. From the right side, look into the left side with the input voltage shorted. You have R2 in parallel with R1.

$R_t_h=\frac{R_1 \cdot R_2}{R_1+R_2}$

Last edited: Apr 19, 2012
5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
When I studied attenuator pads my understanding was that the inserted pad input & output resistances were meant to match the original source and load resistances. In other words the source & load "matching" conditions were maintained albeit with an included attenuation factor for the inserted pad.

I'm not sure the linked site actually follows that convention in the design process.

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
The somewhat ancient tome "Reference Data For Radio Engineers" has a useful chapter (ch 9 in the version I look at) on matched / unbalanced attenuator design.

You can also find clever design links like this one

http://chemandy.com/calculators/matching-t-attenuator-calculator.htm