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How are these resistors in series and parallel

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I was going through the biasing resistor calculation for RS-485 and can across this AppNote (https://www.renesas.com/us/en/document/apn/an1986-external-fail-safe-biasing-rs-485-networks).

On page 3, you can see below:

1712209338960.png


I just want to understand how the 2 RB resistors are considered to be in series and their final value is considered to be in parallel with RT2?

A resistor is said to be in parallel when both is ends are connected to another resistor. But how are the two RB resistors considered to be in series? And their equvivalent resistance finally in parallel with RT2?

Can someone explain?
 
Hi,

let´s make it more simple and just use two resistors.
* R_t = Resistor at top
* R_b = Resistor at bottom
* V_c = voltage at the center node

Let´s use 2 resitors 1k each.
Connected in series, one end at +5V (R_t), the other end at 0V (R_b).
Now the total series resistance is R_t + R_b = 2 * 1k = 2k.

The current through the series of resistors is 5V/2k = 2.5mA. It also os the current thorugh each resistor.
The voltage across R_t: V_t = 2.5mA * 1k = 2.5V
The voltage across R_b: V_b = V_c = 2.5mA * 1k = 2.5V

What we have done now: We created a 2.5V voltage source. (unloaded)
Now let´s connect a load to this new voltage source.
And let the load pull down the node by exctly 1.0V (in a way that V_c becomes 1.5V)

Now let´s calculate the load current:
* What happens with the current at R_b. Since V_B is 1.5V now, the current through R_b: I_b = V_b / R_b = 1.5V / 1k = 1.5mA.
So R_b current dropped from 2.5mA down to 1.5mA, he difference = 1.0mA needs to flow through the load.
* What happens to R_t? The voltage across R_t increases from 2.5V to now 3.5V ... resulting in a current increase from 2.5mA to 3.5mA.
Thise difference in current of 1mA ALSO needs to flow through the load.
Now we have a load current of 1mA + 1mA = 2mA.
...
And as we decided earlier the V_c voltage dropped by 1.0V.
So our 2.5V voltage source has a "source" impedance of V_drop / I_Load = 1.0V / 2mA = 500 Ohms.
(The circuit is electrically identical to a 2.5V voltage source with a series resistance of 500 Ohms)

--> This is the impedance (resistance) of the center node.

******
Since the - what we called - "load" current is determined by R_t and R_b, and you have to add both currents .... the resulting current act as if the resistors are connected in parallel:
I_L = delta_I_t + delta_I_b = you need to alculate the source impedance by

R_source = 1 / ( 1/R_t + 1/R_b) ... which is the formula of two paralleled resistors.


******

My recommendation:
Use a free circuit simulation tool and play around with different Rs and see what happens to the voltages and the currents. And do some calculations.


Klaus
 
Hi,

let´s make it more simple and just use two resistors.
* R_t = Resistor at top
* R_b = Resistor at bottom
* V_c = voltage at the center node

Let´s use 2 resitors 1k each.
Connected in series, one end at +5V (R_t), the other end at 0V (R_b).
Now the total series resistance is R_t + R_b = 2 * 1k = 2k.

The current through the series of resistors is 5V/2k = 2.5mA. It also os the current thorugh each resistor.
The voltage across R_t: V_t = 2.5mA * 1k = 2.5V
The voltage across R_b: V_b = V_c = 2.5mA * 1k = 2.5V

What we have done now: We created a 2.5V voltage source. (unloaded)
Now let´s connect a load to this new voltage source.
And let the load pull down the node by exctly 1.0V (in a way that V_c becomes 1.5V)

Now let´s calculate the load current:
* What happens with the current at R_b. Since V_B is 1.5V now, the current through R_b: I_b = V_b / R_b = 1.5V / 1k = 1.5mA.
So R_b current dropped from 2.5mA down to 1.5mA, he difference = 1.0mA needs to flow through the load.
* What happens to R_t? The voltage across R_t increases from 2.5V to now 3.5V ... resulting in a current increase from 2.5mA to 3.5mA.
Thise difference in current of 1mA ALSO needs to flow through the load.
Now we have a load current of 1mA + 1mA = 2mA.
...
And as we decided earlier the V_c voltage dropped by 1.0V.
So our 2.5V voltage source has a "source" impedance of V_drop / I_Load = 1.0V / 2mA = 500 Ohms.
(The circuit is electrically identical to a 2.5V voltage source with a series resistance of 500 Ohms)

--> This is the impedance (resistance) of the center node.

******
Since the - what we called - "load" current is determined by R_t and R_b, and you have to add both currents .... the resulting current act as if the resistors are connected in parallel:
I_L = delta_I_t + delta_I_b = you need to alculate the source impedance by

R_source = 1 / ( 1/R_t + 1/R_b) ... which is the formula of two paralleled resistors.


******

My recommendation:
Use a free circuit simulation tool and play around with different Rs and see what happens to the voltages and the currents. And do some calculations.


Klaus

This is a really helpful answer for me. Thank you very much. The example you gave is for a node right.

In my question, we have two nodes. A and B (from the Figure 2 in the above image). In that case, the equivalent parallel resistance of the two RBs, will be placed across RT2, I understand. Thank you.
 
Hi,

it´s basically the same.

Either: Force a change of current, measure the change of voltage then use Ohm´s law.
Or: Force a change of voltage, measure the change of current then use Ohm´s law.

R = delta_V / delta_I

Klaus
 

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