# Loaded Voltage Divider - How to solve it?

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#### Inquisitive

##### Super Moderator
Ok, this is not homework per se. I'm reading a textbook for personal learning. Working through the problems. I have come across this problem I'm having trouble trying to solve. What value should R2 be in the above figure to set up a diode current of 0.25mA?

Can somebody walk me through this?

Thanks!

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#### Ian Rogers

##### User Extraordinaire
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To start you need to find the voltage required to drive this diode....

i is constant so 5000 * .25mA = 1.25v.

Add the voltage drop of the diode @0.6v = 1.85v

Now you can find out how much current will be running through the 30k resistor

12v - 1.85v = 10.15v

10.15 / 30k = 0.338mA

You need to deduct 0.25mA from this... = 88.33uA.

now work out the last resistor 1.85v / 88.33uA = @ 20k...

#### Inquisitive

##### Super Moderator
Ok. I know that the voltage drop across the diode is 0.7V. So, that explains one discrepancy I had.

So, 5k = 5000 Ohms, .25mA = 0.00025A
5000 Ω * .00025A = 1.25 V
Vd 0.7V + 1.25V = 1.95 V
12 v Total - 1.95 V R3 Leg = 10.05 V
10.05 V / 30000 = 3.35*10^-3 = 3.35^-7

This is where I'm lost. Been out of school too long. Lost in the exponent manipulation.
When I subtract .25mA from this, the values don't make sense to me.

Insight or suggestions anyone?

#### ericgibbs

##### Well-Known Member
hi Inq,

This is incorrect: 3.35^-7

Its already in the correct units as 3.35*10^-4

E

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#### Inquisitive

##### Super Moderator
Ok, that being the correct units.

I should have:
10.05 V / 30000 = 3.35*10^-3

Now I need to deduct my intended new .25mA?
3.35*10^-3mA+.25mA = .25335mA

1.95 V / .25335 = 7.69 or 7.7V

1.95 / 7.69 = .2535 (Lost again?)

By the way. The textbook claims that the final R2 value should be 23 k Ohms.

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#### Inquisitive

##### Super Moderator
I have a calculator that displays engineering mode ie... "0.335μ" instead of scientific "* 10^-4"

Makes my life easier....
Got a model you would recommend? For those of us who are out of school? Last edited:

#### WTP Pepper

##### Active Member
Ok, this is not homework per se. I'm reading a textbook for personal learning. Working through the problems. I have come across this problem I'm having trouble trying to solve. What value should R2 be in the above figure to set up a diode current of 0.25mA?

Can somebody walk me through this?

Thanks!
Without giving you the answer direct, make a Thevenin equivalent circuit using R1 & R2 then work on this equivalent resistance, new voltage and R3 to calculate the current through the diode.

#### MrAl

##### Well-Known Member
Hello there,

To the OP:

The simplest method to use here to calculate the required currents and/or voltages is to simply replace the diode with a voltage source which is approximately equal to the diode voltage drop when it is passing a current of the required current. Assuming this is a regular rectifier diode, the assumed value is usually 0.7 volts. So you would remove the diode, and replace it with a battery (positive terminal goes to the 'anode' connection, negative terminal to the 'cathode' connection) of 0.7 volts. You would then analyze the circuit just like any other two source circuit.

If you happen to have the curve for the diode (we dont always have this information anyway) you can use that to 'upgrade' the analysis using an ad hoc method of approximation that involves changing the assumed voltage drop (the battery) and recalculating, then repeating until you get the desired accuracy. But the temperature also affects the diode voltage drop so we often just use a first approximation and let it go at that. But if you already know the diode current you're one step ahead because you can look at the diode curve and find the voltage (which will be approximate anyway) and then use that in the calculations. Since you know the diode current is 0.25 milliamps, you're one step ahead of the game already.

You could also build up an equation that includes the diode curve, but i assume you dont want to have to go through all that. So start with the assumed voltage drop of 0.7v and see what you can do with it. If you need help with a two source circuit that's not a problem either.

If you'd like to check your result, you can use this formula for the diode current:

i=(Vs*R2-Vd*R2-Vd*R1)/(R2*R3+R1*R3+R1*R2)

where

Vs is the source voltage (12v here),
Vd is the diode voltage (we said we would assume is 0.7v),
i is the current through the diode (you said would be 0.000250 amps here),
R1 through R3 are the resistors.

You would plug in all the values and compute the result to check your own result.
You could 'cheat' and solve that equation for R2, but you should try to develop a solution yourself first Last edited:

#### Inquisitive

##### Super Moderator
Solved. Thanks for the help everyone.

This is how I solved it.

0.7 V = diode voltage drop
12 Vin
.25mA is known or .00025A
V = I * R so, .00025 * 5000 Ω = 1.25 V
1.25 V + 0.7 V = 1.95 V on the R3 leg
12 V - 1.95 V = 10.05 V on the R1 leg
I=V/R so, 10.05 V / 30000 Ω = 3.35*10^-4
3.35*10^-4-.00025A = 8.5*10^-5
1.95 V / 8.5*10^-5 = 2.294117*10^4

Now putting this into readable form. 22941.17 or rounding it off to the nearest 1000. The answer would be 23 k Ω

Thanks to all those who helped. Much appreciated!

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#### MikeMl

##### Well-Known Member
I get 16.5K to 22.7K

look what happens when I asked LTSpice to solve it. The x=axis is the value of R2. Plots show the voltage across the diode, and the current through the diode as a function of R2 and temperature. Green=-50C, red=0C, lt. blue=50C, dk. blue=100C.

Note that the assumption that the voltage across the diode is 0.7V leaves quite a bit to be desired... A diode makes a very good temperature sensor...

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#### Inquisitive

##### Super Moderator
For the purposes of the textbook this was a solve it on paper with calculator exercise. So far the text only mentions a voltage range of 0.6V - 0.7V (voltage drop). So, that is what I had to work with.

I find the LTSpice method interesting. Handy to know the temperature variations and their effect on the circuit. Had no idea that you'd get that range of answers.

Thanks, Mike

#### MrAl

##### Well-Known Member
Hi,

Im happy to hear you solved it, and solved it using your own approach. That's often the best way. Also good is to understand the more general way to solve it, which is using the analysis for a two source circuit and coming up with an equation for R2.

Another approach similar to yours is to use the fact that the (assumed) diode drop is 0.7 and since the known current is 0.000250 amps we know the resistance of the diode is 0.7/0.000250 Ohms, which makes it equal to 2800 ohms. So we can then replace the diode with a resistor of 2800 Ohms which gives us a circuit made entirely of pure resistances.

0.7v is usually the assumed voltage drop of a diode that is not assigned an actual part number for in these kinds of problems. If we have an actual part number then we might be able to find the Is, N, and Rs at least for the diode and use an exponential or logarithmic equation for the diode.

Circuits like this one are always fun to solve. What else is interesting is when you have a few diodes in the circuit with their own resistors and are in parallel and in series with another resistor, and you have to use the exponential model of the diode.

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#### MikeMl

##### Well-Known Member
Here is how I would proceed algebraically:

The voltage drop across R3 = IR = 0.00025*5000 = 1.25V
Therefore, the voltage at the node at the junction of R1 and R2 is 0.7 (the diode drop) +1.25 = 1.95V
The current into that node is I(R1) = (12-1.95)/30K = 0.000335A =335uA

The sum of the currents into that node must be zero, so 335u = 250uA + I(R2), or I(R2) = 335uA-250uA = 85uA
Therefore, R2 = E/I = 1.95/85u = 0.0229megΩ = 22.9KΩ.

#### Ian Rogers

##### User Extraordinaire
Forum Supporter
Here is how I would proceed algebraically:

The voltage drop across R3 = IR = 0.00025*5000 = 1.25V
Therefore, the voltage at the node at the junction of R1 and R2 is 0.7 (the diode drop) +1.25 = 1.95V
The current into that node is I(R1) = (12-1.95)/30K = 0.000335A =335uA

The sum of the currents into that node must be zero, so 335u = 250uA + I(R2), or I(R2) = 335uA-250uA = 85uA
Therefore, R2 = E/I = 1.95/85u = 0.0229megΩ = 22.9KΩ.
Did that in post #2.

#### MrAl

##### Well-Known Member
Hi,

Some very good ideas for alternate ways of doing it.

The two source method is a general method that applies to any number of sources (or in this case diodes). Each source is shorted out one at a time and the analysis is done for one of the currents or whatever is interesting. The responses are added up and equated, then the solution is formed. If we need to solve for a given variable (such as a resistor value) we can then do that with the resulting equation(s).

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