Help with a simple voltage divider please

Discussion in 'The Projects Forum' started by Shorted-Fuse, Jan 4, 2007.

1. Shorted-Fuse Thread Starter New Member

Jan 2, 2007
2
0
Hello all. Happy New Year. This is a good site with a lot of information.

It seems like it's always the simple projects that are confusing. Here is a simple voltage divider circuit to power a device from a vehicle. Details
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12-14v DC power source
5-8v DC power requirements
1 watt power draw

Requirements
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1. Components must fit inside the 12v plug
2. Minimize current draw on the battery when the device is powered on or off

Calculations
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1 watt @ 5v = 200ma
1 watt @ 6v = 166ma (used as a reference)
1 watt @ 8v = 125ma

14v power source used for the calculations below

Option1: Eliminate R1 and connect the load in series with R2.

Problem: When the load is powered off, there is infinite resistance across the open circuit which will have 14v potential to the load. There is a risk of damaging the device when it is first powered on.

Option2: To minimize the current draw on the battery, choose R1 so that almost all of the current is going through the load ... like 10K. Then, choose R2 to handle the appropriate voltage and current.

a) if V(load) = 5v, then R(load) = 5 / .200 or 25 ohms
if R1 is 10K, then R(parallel) = 1 / ( (1/R1) + (1/Rload) ) or 1 / ( (1/10K) + (1/25) ) or 24.93 ohms
I(parallel) = I(R1) + I(load) or .5ma + 200ma or 200.5ma
if V(parallel) = 5v, then V(R2) = 9v (14v used for calculations)
R2 = V(R2) / I(load) or 9 / .2005 or 44.88 ohms
P(R2) = V(R2) * I(load) or 9 * .2005 or 1.8 watt

b) if V(load) = 6v, then R(load) = 6 / .166 or 36 ohms
if R1 is 10K, then R(parallel) = 1 / ( (1/R1) + (1/Rload) ) or 1 / ( (1/10K) + (1/36) ) or 35.87 ohms
I(parallel) = I(R1) + I(load) or .6ma + 166ma or 166.6ma
if V(parallel) = 6v, then V(R2) = 8v
R2 = 8 / .1666 or 48.02 ohms
P(R2) = 8 * .1666 or 1.33 watt

c) if V(load) = 8v, then R(load) = 8 / .125 or 64 ohms
if R1 is 10K, then R(parallel) = 1 / ( (1/R1) + (1/Rload) ) or 1 / ( (1/10K) + (1/64) ) or 63.59 ohms
I(parallel) = I(R1) + I(load) or .8ma + 125ma or 125.8ma
if V(parallel) = 8v, then V(R2) = 6v
R2 = 6 / .1258 or 47.69 ohms
P(R2) = 6 * .1258 or .75 watt

Problem: The value of R1 is so high that there is still 14v potential across R1. Again, there is a risk of damaging the device when it is first powered on.

Option3: Choose R1 and R2 to divide the voltage with the device powered off.

a) if V(load) = 6v, then R(load) = V(load) / I(load) or 6v / .166 or 36 ohms
if R1 is also 36 ohms, then I(R1) = 166ma
P(R1) = 6 * .166 or 1 watt

R(parallel) = 1 / ( (1/R1) + (1/Rload) ) or 1 / ( (1/36) + (1/36) ) or 18 ohms
I(parallel) = I(R1) + I(Rload) or 166ma + 166ma or 333ma

if V(parallel) = 6v, then V(R2) = 8v
R2 = V(R2) / I(parallel) or 8 / .333 or 24 ohms
P(R2) = 8 * .333 or 2.6 watt

When the load is powered off, V(R1) = V(total) * R1 / (R1 + R2) or 14 * 36 / 60 or 8.4v

Problem: The current draw is twice the load. 50% efficiency
There is still 166ma of draw when the device is powered off.

I am writing a computer program to go through all of the possible combinations of R1 and R2 to minimize the current draw, and still provide the proper voltages whether the device is powered on or off.

I am also looking into a 6v regulator if I can find one small enough. Any other ideas? Many thanks.

2. Papabravo Expert

Feb 24, 2006
12,284
2,723
It is error to think that your circuit represents good design practice. The proper way to take a higher voltage to a lower voltage is to dispense with the resistors and use a voltage regulator. Just forget this voltage divider nonsense for this application. It is only useful for low power applications like biasing and reference voltage generation. It is not meant to deliver power to loads. Stop wasting your valuable time.

3. Shorted-Fuse Thread Starter New Member

Jan 2, 2007
2
0
You are correct. Since the output voltage does not need to vary, I could use an LM7806 to provide +6v regulated output. Small capacitors could be added across the input and output to reduce any variations and maintain a constant voltage level. And, the components are small enough to fit inside the 12v plug. Thanks.

4. hgmjr Retired Moderator

Jan 28, 2005
9,029
219
One thing to bear in mind as you go forward with the design using a linear regulator (a wise decision btw) that the difference between the input voltage and the output voltage multiplied by the current drawn by the load will result in heat dissipation in the 7806. The 1 Watt being delivered to the load will result in (12-5)*0.2 = 1.4 Watts dissipated by the 7806.

You will need to make sure that the device is provided with an adequate heatsink.

hgmjr

5. Viper-7 New Member

Jan 7, 2007
1
0
Wouldn't the same amount of heat be dissipated by the resistors? I mean the LM78xx regulators are basically just a set of switched resistors internally right?

Also from previous experience, you can safely get away with about 10w heat being dissipated from a TO-220 casing (typical for LM78xx regulators) without a heatsink.

Also, if you are worried about space, the same regulator in a TO-252 casing will easily fit anywhere and should still dissipate 2-3w of heat passively no worries

6. kubeek Expert

Sep 20, 2005
5,584
1,093
The datasheet says that the TO220 thermal resistance from junction to ambient is 65K/W.

The temperature protection kicks off at cca 150°C, so with 25°C ambient tempereature the maximal power to be dissipated is dT/Rth = (150-25)/65 = 1.92W with no heatsink.

So you cannot dissipate 10W using a TO220 device without heatsink and fan, and I have the same doubts about the TO252 case.