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# Voltage Divider Verification

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#### EN0

##### Member
Hey Everyone,

Just want to make sure I'm doing this right.

Suppose I have a voltage divider like in the first attachement. You'll see that R1 = 680Ω and R2 = 20k giving me an output voltage of 8.7V [Vout = R2/(R1+R2)]. Now, to calculate the output current of the voltage divider, you do the following: Iout = Vin/(R1+R2). So my current output is approximately 435μA. Let's say I want to have 2.5μA for my output, I simply add a resistor like in the second attachment. What would the output current be now? Also, wouldn't there be a voltage drop across R3 so I would have to make the voltage a bit larger to compensate for that drop? Ignoring the voltage drop, this is what I did: 435μA - 2.5μA = 437.5μA. So, 8.7V / 437.5μA = 19885.71429Ω ≈ 20k.

Please let me know if I'm doing this right! This is to get the correct base current and voltage to an NPN transistor.

#### Attachments

• VoltageDividerAnalysis1.JPG
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• VoltageDividerAnalysis2.JPG
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Hi Austin.

The way I would approach this is to first just look at just the voltage divider R1 and R2.

As you have already worked out, that by itself will give an open circuit output voltage of 8.70406 Volts (when the output is not loaded).

The next thing you need to do is work out the source impedance at the junction of R1 and R2, which is just R1 and R2 in parallel. That works out to 657.6 ohms.

The power supply source at the junction of R1 and R2 now looks like a perfect 8.70406 volt battery in series with 657.6 ohms.

You can now use this Thevenin equivalent to calculate how the voltage divider is going to behave when further loaded by "something else" connected to the junction of R1 and R2 as a load.

Last edited:
First of all the 435μA is not a output current.
It is a current that is flow form Vs1 trough R1 and R2.
Vs1+--->R1--->R2--->-Vs1
So output voltage is equal Vout=I2*R2 = Vs1/(R1+R2) * R2 = R2/(R1+R2) x Vs1=8.7040619V
And the output current without a load is 0A and I1=I2

And now if we connect the load that draw 25uA of current we get this situation:
The solution can be find quite easy with a little help from algebra and KCL, KVL.
Vcc - I1*R1 - I2*R2 = 0
I1 = I2 + Ip

Vcc - (I2+Ip)*R1 - I2*R2 = 0
Vcc - I2*R1 - Ip*R1 - I2*R2 = 0
Vcc - I2*(R1+R2) -Ip*R1 = 0
Ip*R1 + I2*(R1+R2) = Vcc
I2 = ( Vcc - Ip*R1 ) / (R1+R2)

Vout=I2*R2= [ R2*( Vcc - Ip*R1 ) ] / (R1+R2)

And if you still want Vout=8.7V then you must correct R1.
New R1=(Vcc-Vout)/(I2+Ip)=(9V-8.7V)/(453uA+25uA)=0.3V/478µA=627Ω=620Ω

Last edited:
Hey Everyone,

Just want to make sure I'm doing this right.

Suppose I have a voltage divider like in the first attachement. You'll see that R1 = 680Ω and R2 = 20k giving me an output voltage of 8.7V [Vout = R2/(R1+R2)]. Now, to calculate the output current of the voltage divider, you do the following: Iout = Vin/(R1+R2). So my current output is approximately 435μA. Let's say I want to have 2.5μA for my output, I simply add a resistor like in the second attachment. What would the output current be now? Also, wouldn't there be a voltage drop across R3 so I would have to make the voltage a bit larger to compensate for that drop? Ignoring the voltage drop, this is what I did: 435μA - 2.5μA = 437.5μA. So, 8.7V / 437.5μA = 19885.71429Ω ≈ 20k.

Please let me know if I'm doing this right! This is to get the correct base current and voltage to an NPN transistor.

Hi,

If i understand your question right you want to reduce the current through
the voltage divider while still maintaining 8.7v output.

RL=V/I
which is:
RL=8.7/0.0000025=3.48 megohms.

To get the lower resistor.
To get the upper resistor take the same 2.5ua and Vcc-8.7v to get
that resistor:
RU=(Vcc-8.7)/0.0000025=(9-8.7)/0.0000025=0.3/0.0000025=120k.

You should realize however that any small load connected to the
output of the divider will change the output voltage. If the load
connects to ground it will reduce this voltage, if it connects to
Vcc it will increase this voltage.

You should also realize that if that 9v supply is a battery that will
eventually drain down and the output will drop because of that too.

Last edited:
This transistor is biased completely wrong. It is almost saturated. The collector is +9V and the base is +8.7V.

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