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sorry if this seems very easy but I am looking for a little help as I have only just begun to learn EE. My homework problem has me create a power supply using a common reference point only a 24 V source. Only resistors are involved. The goal is to have outputs of +12 V, -12 V, and +5 V. In addition, the power source must supply 80 mW. I've got the outputs of -12 V and +12 V figured out, but have been struggling endlessly to get the +5 V output correct (as well as figuring out the 80 mW). The below schematic is what I have so far:

schematic

simulate this circuit – Schematic created using CircuitLab

My inclination is to say "divide the 12 V using voltage division," but I haven't figured out how to make it work with the ground. Some advice would be much appreciated. Thank you.

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    \$\begingroup\$ Is 80mW meant to be the total loading of all three supply rails? \$\endgroup\$ – replete Jan 28 '17 at 4:35
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    \$\begingroup\$ Power supply usually means a device that keep one or more stable voltages available. In your case the voltages are +12V, -12V and +5V. Stable means "no voltage drop until the load is more than allowed". This needs voltage regulator circuits. It can be achieved by resistors only approximately (= by allowing an enormous power loss and having a well stable input) \$\endgroup\$ – user287001 Jan 28 '17 at 4:56
  • \$\begingroup\$ @user287001 the OP has stated that this is a homework and only resistors are involved \$\endgroup\$ – Claudio Avi Chami Jan 28 '17 at 5:27
  • \$\begingroup\$ When you make a divider, usually you use the 10/1 rule. That means, if the current through the divider is ten times the current through the load, the divider output is stable enough, namely, the current that the load 'steals' from it doesn't change the output voltage a lot. What you did for +/-12 is OK (I have not checked the values of the resistors, but the idea is OK). You can make the 5V with another divider from +12V \$\endgroup\$ – Claudio Avi Chami Jan 28 '17 at 5:29
  • \$\begingroup\$ Sorry that it's not clear. I'll copy and paste the problem as it appears on the assignment if it helps: Using a single 24 V voltage source and some resistors, design a power supply to meet these requirements: (i) by connecting between and a common reference point and different resistors, a user can get +12 V, -12 V and +5 V outputs (ii) the power supplied by the source is 80 mW Hint: consider a voltage divider configuration. \$\endgroup\$ – Indigo22 Jan 28 '17 at 22:08
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Your specifications for the load weren't clear enough, so let's talk about your voltage divider first. If you want a +12V, a -12V and a +5V output, you need three resistors instead of two, with the sum of the resistance values of R1 & R2 being equivalent to the resistance value of R3. Your +12Vdrop is the sum of individual drops across R1 & R2 in series, your -12Vdrop is the drop across R3 with the ground before it, while your 5Vdrop will be the drop across R1 alone.

Your schematic should look like this:

enter image description here

(I used kΩ resistors to minimize power consumption through the divider although you can use resistors of smaller value as long as they are proportional this way).

The total resistance value of the entire circuit is 7.2kΩ, while supply voltage is 24V. You can see that our +12V drop is the drop across both R1 and R2 in series, meaning both the resistors' values should have the sum of half the total resistance value. 2.1kΩ+1.5kΩ = 3.6kΩ, which is half of 7.2kΩ or the said total resistance value. This means that half of the supply voltage is the voltage drop across the two resistors, hence +12V. The drop across R3, having a value of 3.6kΩ, should also have a 12V drop across it, as its resistance value is half of the total resistance value of the entire circuit. The ground is always defined as 0V, so since R3 comes after it, there has to be a negative voltage drop across R3, hence -12V. The +5V division is provided by R1, with a value of 1.5kΩ which is 1/4.8 of the total resistance value, therefore the voltage drop across it will be 1/4.8 of the supply 24V, hence the +5V drop across it. Looking at the schematic, the +5V division is embedded within the +12V division because this, together with the 2.1kΩ resistor, provides half the total resistance value of the circuit to provide the +12V.

Again I need more clarification about the load. I need to know which voltage divider the load is connected to in order to start with that 80mW being talked about. Hope I helped.

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  • \$\begingroup\$ No, the answer is wrong. You have to DELIVER 80mW to the load, not to consume them on the divider. Also, where would be the GND signal at the output? What you have done is +12, -12 and -7V. The divider for the 5V must be on the upper half, not on the lower one. \$\endgroup\$ – Claudio Avi Chami Jan 28 '17 at 9:22
  • \$\begingroup\$ I'm sorry, must've mistaken OP. Fixing the post... \$\endgroup\$ – mjtsquared Jan 28 '17 at 10:08
  • \$\begingroup\$ But exactly what load is being talked about? Do all three voltage dividers each have their own load consuming 80mW? And isn't the load going to have a specific resistance value as power consumption is based on resistance of the load? \$\endgroup\$ – mjtsquared Jan 28 '17 at 10:35
  • \$\begingroup\$ Sorry but it is still not good. The 5V need to have a common ground with the +12 and -12V. And the resistor values you chose are way too big. It is not clear to me from what voltage the 80mW are delivered. \$\endgroup\$ – Claudio Avi Chami Jan 28 '17 at 14:04
  • \$\begingroup\$ Sorry that it's not clear. I'll copy and paste the problem as it appears on the assignment if it helps: Using a single 24 V voltage source and some resistors, design a power supply to meet these requirements: (i) by connecting between and a common reference point and different resistors, a user can get +12 V, -12 V and +5 V outputs (ii) the power supplied by the source is 80 mW Hint: consider a voltage divider configuration. \$\endgroup\$ – Indigo22 Jan 28 '17 at 22:09
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Assuming that no power is actually supplied to the load, your first task is to calculate the dropping resistors. Use $$P =\frac{V^2}{R} $$ where V is 24.

Now divide that resistance in half, and use 2 such resistors as per your schematic.

Finally, divide your R2 into 2 resistors set up as a voltage divider, with the sum of the resistors equal to the R2 value, with 12 volts divided down to 5 volts. (Hint - the ratio of the two will be 5:7, right?)

EDIT - Since you have made it clear that the 80 mW figure applies "to the load", it is also clear from your schematic that the 80 mW figure must apply to the 5 volt supply only, since as long as R1 and R2 are identical the supply voltages will be equal to +/- 12 (assuming equal loads). So, since $$P = iV $$ $$i=\frac{P}{V} =\frac{.08}{5} = .016 = 16\text{ mA}$$Then the 5 volt load can be modelled as 312.5 ohn resistor, and you can figure a voltage divider which will produce that. Note that this resistor network will need to be completely separate from your R1/R2 network, and the reference points cannot be common. In other words, there will not be a common ground point for +5 and +/-12. The reason is that if you do so, the load resistances (which are not specified, and so must be assumed to be arbitrary) will interact with the 5 volt network and change the 5 volt level.

All in all, this is a very confusing, poorly defined problem, and unless you have failed to pass on information which was contained in the problem as it was given to you, I suggest that you are dealing with a bad course.

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  • \$\begingroup\$ I don't understand your power calculation. The questions says that 80mV must be supplied to the load. Although it is not clear how the power is divided between the several output voltages \$\endgroup\$ – Claudio Avi Chami Jan 28 '17 at 6:04
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From OP question:

the power source must supply 80 mW

Which is different from:

(ii) the power supplied by the source is 80 mW

It is a poorly designed question. A power supply supplies power to a load. Three voltages implies three loads. If you draw any current from either +12V, +5V or -12V, all voltages and power will be different. So it is not a power supply!

\$ R_T = \frac {V^2} {P_T} = \frac {(24 V)^2} {80 mW} = 7.2k\Omega\$

schematic

simulate this circuit – Schematic created using CircuitLab

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