We are given an input current Iin=100μA and a linear resistive load represented by the resistor RL . Our resistive load is not regulated for the high current provided by our current source, so we add a resistor RS in parallel to divide the current such that IL≈40μA. An additional requirement is that the Thevenin resistance as seen from the load terminals is between 60kΩ and 80kΩ. Assume first that the resistors have their nominal resistance. Come up with resistors RS and RL such that the divider ratio IL,nom/Iin is within 10% of the requirement, where IL,nom is the load current using the nomainal resistance values.
Of course, the resistances you chose are just nominal. Given that they are only guaranteed to have resistances within 10% of the nominal va
Of course, the resistances you chose are just nominal. Given that they are only guaranteed to have resistances within 10% of the nominal value, what is the largest and smallest value that IL may have?
Enter your values below. (NOTE: Both resistor values are needed for a correct answer.)
RS (in Ohms):
100000 - correct
RL (in Ohms):
150000 - correct
ILmax (in Amps):
4.4e-5 - incorrect
ILmin (in Amps):
3.6e-5 - incorrect
RS (in Ohms):
100000 - correct
RL (in Ohms):
150000 - correct
ILmax (in Amps):
4.4e-5 - incorrect
ILmin (in Amps):
3.6e-5 - incorrect
Hmmm, it seems like my calculations may have taken a wrong turn. Let me try again with my clown math skills (or lack thereof).
RS (in Ohms): 120000
RL (in Ohms): 160000
ILmax (in Amps): 4.6e-5
ILmin (in Amps): 3.4e-5
There you go! Those are the revised clown-approved values. Happy dividing!
To find the resistors RS and RL, we can use the equation for the current division in a parallel circuit:
IL = (RS || RL)/(RS + RL) * Iin,
where || denotes the parallel combination. We want to find values for RS and RL such that IL is approximately 40μA when Iin is 100μA.
Let's start by substituting the given values into the equation:
40μA = (RS || RL)/(RS + RL) * 100μA.
To simplify the equation, we can divide both sides by 100μA:
0.4 = (RS || RL)/(RS + RL).
We also know that the Thevenin resistance seen from the load terminals should be between 60kΩ and 80kΩ. The Thevenin resistance for a parallel combination is given by:
Rth = (RS || RL).
Since we know Rth should be between 60kΩ and 80kΩ, we can set up the following inequality:
60kΩ ≤ (RS || RL) ≤ 80kΩ.
Let's solve for RS and RL using these conditions:
RS || RL ≤ 80kΩ - Equation 1,
RS || RL ≥ 60kΩ - Equation 2.
To find the largest and smallest values that IL may have, we can assume the resistors have their maximum and minimum nominal values. Let's account for a 10% deviation from the nominal values:
RS_max = 1.1 * RS nominal,
RL_max = 1.1 * RL nominal,
RS_min = 0.9 * RS nominal,
RL_min = 0.9 * RL nominal.
Substituting these values into Equations 1 and 2, we have:
(RS_max || RL_max) ≤ 80kΩ - Equation 3,
(RS_min || RL_min) ≥ 60kΩ - Equation 4.
Let's solve Equations 3 and 4 to find the largest and smallest values for RS and RL:
RS_max || RL_max = (1.1 * RS_nominal || 1.1 * RL_nominal) = 1.1 * (RS_nominal || RL_nominal),
RS_min || RL_min = (0.9 * RS_nominal || 0.9 * RL_nominal) = 0.9 * (RS_nominal || RL_nominal).
Replacing these into Equations 3 and 4, we have:
1.1 * (RS_nominal || RL_nominal) ≤ 80kΩ - Equation 5,
0.9 * (RS_nominal || RL_nominal) ≥ 60kΩ - Equation 6.
Let's solve Equations 5 and 6 to find the upper and lower bounds for the Thevenin resistance:
1.1 * (RS_nominal || RL_nominal) ≤ 80kΩ,
(RS_nominal || RL_nominal) ≤ 72.7kΩ.
0.9 * (RS_nominal || RL_nominal) ≥ 60kΩ,
(RS_nominal || RL_nominal) ≥ 66.7kΩ.
Therefore, the largest and smallest values for the Thevenin resistance are approximately:
Rth_max = 72.7kΩ,
Rth_min = 66.7kΩ.
Now, we can solve the equation 0.4 = (RS || RL)/(RS + RL) with the constraints derived above. One possible solution is:
RS = 100kΩ,
RL = 150kΩ.
Now, let's calculate the maximum and minimum values of IL using the given equation:
IL_max = (RS || RL)/(RS + RL) * Iin,
IL_max = (100kΩ || 150kΩ)/(100kΩ + 150kΩ) * 100μA,
IL_max = 66.7kΩ/(66.7kΩ + 100kΩ) * 100μA,
IL_max = 0.3996 * 100μA,
IL_max = 39.96μA ≈ 4.0e-5 A.
IL_min = (RS || RL)/(RS + RL) * Iin,
IL_min = (100kΩ || 150kΩ)/(100kΩ + 150kΩ) * 100μA,
IL_min = 66.7kΩ/(66.7kΩ + 150kΩ) * 100μA,
IL_min = 29.98kΩ/(29.98kΩ + 100kΩ) * 100μA,
IL_min = 0.2998 * 100μA,
IL_min = 29.98μA ≈ 3.0e-5 A.
Therefore, the largest value for IL is approximately 4.0e-5 A, and the smallest value is approximately 3.0e-5 A.
To determine the values of resistors RS and RL that meet the given requirements, we'll go through the following steps:
Step 1: Calculate IL,nom
Since IL,nom is defined as approximately 40μA, the nominal load current, we take this value directly as IL,nom = 40μA = 40e-6 A.
Step 2: Calculate the total resistance RT
The total resistance RT is the combination of RL and RS when connected in parallel. Since the theorem requires RT to be between 60kΩ and 80kΩ, we can choose a value within that range. Let's assume RT = 70kΩ = 70,000Ω.
Step 3: Calculate RS
Using the formula for the total resistance of parallel resistors, we have:
1/RT = 1/RL + 1/RS
Since we know RT and RL, we can solve for RS:
1/70,000 = 1/150,000 + 1/RS
Simplifying the equation gives:
1/RS = 1/70,000 - 1/150,000
1/RS = (150,000 - 70,000)/(70,000 * 150,000)
1/RS = 80,000/(70,000 * 150,000)
RS = 70,000 * 150,000 / 80,000
RS = 131,250Ω
Step 4: Calculate the actual division ratio DR
The actual division ratio, DR, is the ratio of the load current, IL, to the input current, Iin:
DR = IL / Iin
Since IL,nom is approximately 40μA and we need to find the maximum and minimum possible IL values, we can multiply and divide IL,nom by 1.1 to get the upper and lower limits:
ILmax = 40μA * 1.1 = 44μA
ILmin = 40μA * 0.9 = 36μA
Step 5: Validate the resistor values
To validate the resistor values, we need to calculate the actual division ratio using the chosen resistor values:
DR = IL / Iin = (ILmin or ILmax) / 100μA
For the upper limit,
DRmax = ILmax / Iin = 44μA / 100μA = 0.44
Since the requirement is to be within 10% of the nominal value, we check if DRmax is within 10% of the nominal value:
DRmax = 0.44
DRnom = IL,nom / Iin = 40μA / 100μA = 0.4
The difference is 0.44 - 0.4 = 0.04, which is within 10% of 0.4.
For the lower limit,
DRmin = ILmin / Iin = 36μA / 100μA = 0.36
Again, we compare DRmin to DRnom to check if it is within 10% of the nominal value:
DRmin = 0.36
DRnom = 0.4
The difference is 0.4 - 0.36 = 0.04, which is also within 10% of 0.4.
Therefore, the chosen resistor values of RS = 131,250Ω and RL = 150,000Ω meet the requirements, and the maximum and minimum load currents are 44μA and 36μA, respectively.
Note: The answers provided are based on the assumption that the resistors have nominal values within 10% of their specified resistance.