Calculating Voltage Drop

I don't want to give you the answer as much as I want to show you the pathway to the answer. The question you gave sounds a little like a homework question, so just to be sure I want you to learn how to solve the problem, but I will at least point you into the right direction.

Actually, I'm just trying to pick up a new hobby a little late in life but so far, it's been fun. As far as my question is concerned, I just used a single resistor I soldered in-line to drop the speed of a 12vdc fan so would I check the resistance through the fan also?

"Actually, I'm just trying to pick up a new hobby a little late in life but so far, it's been fun."

Never too late!

First, let's figure out the total circuit resistance.

I = E/R, let's solve for R, the total circuit resistance. We know I and E. Rearranging the equation to solve for R we get:

R = E/I = 13.8/0.46 = 30 Ω. What?

That is the first clue something is wrong with the problem. So, is the total circuit current from the power supply really 0.46 Amps? Is the resistor really 30 Ω?

The VOM should show 0.46*30= 13.8(=supply voltage) not 9.5 volts, something wrong with it. if your current and voltage measurements are correct.

Here is a picture of the test showing the voltage drop after the resistor.

My power supply provides 13.8 volts before the resistor and 9.52 volts after the resistor at the fan lead.

Here is a picture of the resistor which is orange/black/black/gold and according to my chart I found online it is a 30 ohm 5%. Sorry I didn't include this in the last post.

Hi Longbow,

I think this picture gives all the data required.

What you are measuring here is the voltage drop across the fan: 9.52 volts. Therefore, if your power supply is supplying 13.8V under load (in other words, if you moved the red test lead to the right side of the resistor, the meter would read 13.8v), then the voltage drop across the resistor is 4.3v. We know the resistor is 30 ohms, so the current in the circuit (which is the same throughout) is (4.3v/30ohms) .143 A. This is about right for such a fan.

The power dissipated in the resistor would be the current (.143 A) times the voltage dropped across it (4.3V): therefore .616 watts. Your resistor looks like it is rated for 1/2 watt, so it will get quite hot if you run the fan for long this way. If you are just experimenting temporarily, it will be OK.

The power dissipated in the fan is (.143 x 9.5) 1.36 watts. The total circuit power is (.143 x 13.8) 1.97 watts. You can see that the resistor is using up a lot (relatively) of power and sending it off as waste heat, so in a production application, you would usually not use a resistor to control a voltage like this, if you were concerned about battery life, etc.

Re: the amperage reading on your meter: I'd guess you may have had the leads plugged into the higher amperage jacks? The digits 14 are about right, so I'd guess the lead positioning was incorrect -- either that or the meter is whacko.

So, I'm guessing you've got a 30Ω resistor in series with a fan? First thing, measure the actual current through the circuit with your VOM set on dc amps. Then measure the actual voltage across the resistor, that is, put the black lead on one resistor lead and the red lead on the other resistor lead. Then IR = V should be true. The rest of the voltage is dropped by the fan itself. Come back and tell us how it's going.

"So, I'm guessing you've got a 30Ω resistor in series with a fan?"

Yes, that's right.

"First thing, measure the actual current through the circuit with your VOM set on dc amps."

Ok. I set my meter on 2m DC and ran the circuit thru the meter and the reading was 0.014 which I think is 14 milliamps.

"Then measure the actual voltage across the resistor, that is, put the black lead on one resistor lead and the red lead on the other resistor lead."

This measures 4.3vdc which is what I had when subtracted the supply voltage from the 9.5 end result.

Maybe I need to ask my question in another way. Let's say I have a fan, etc that operates at 9 volts and my only power source is a 13.8vdc power supply. How would one calculate the resistor needed to drop the voltage down to 9 volts? That is the math problem I can't figure out.

GA ^{_{and no "homework crack"}}

One should not put a current meter in series in such circuits as the impedance of the meter itself in the most sensitive scales is quite large. Measuring the voltage drop across a known resistor (the 30 Ω one in this case is adequate).

You mention that your voltage supply has 13.8V -open measurement (that means not any load attached to it, just the voltmeter which is a negligible load).

If your supply was an ideal one, that wouldn't have any internal resistance. The fact that you measure only 9.5V across the resistor shows that the 4.3V voltage drop (13.8V - 9.5V = 4.3V) is due to the internal resistance of the supply. For that configuration, your power supply has an internal resistance of 4.3V/0.317A=13.58 ohm.

Because the voltage drop across the 30 ohm resistor is 9.5V, the current through the circuit is about 0.317A.

That means that, in order to get the "target" current of 0.46A through the resistor, you need a higher voltage (preferably variable and able to yield 0.46A) power supply.

Excuse me, since I am only a second year student, but I think I can help. I think your meter must be shot, because if it is set to "2m" and displaying 0.014 , you are only reading 14 micro amps. 2m is a setting that will measure up to 2 milliamps (=.002 amps) , so a readout of ".014" should translate as .000014 amps. I am confused about this.....

But anyway aside from that confusion, after reading others work, it seems to me that you are getting a 9.5 volt output to the fan with the 30 ohm resistor, so there is 4.3 volts across the resistor. However you want to drop 4.8 volts across a resistor. Well if the current is .142 amps (4.3/30) then just try 4.8/.1426 which equals 33.66 , so you may just need a slightly larger resistor, just like TVP45 suggested a 36 ohm should work out.

Man it was tough figuring out what your problem was, at first I thought you had a 9.5 voltage drop across the resistor, so did a couple others from what I saw. However after reading everything I see that it is just a matter of you needing a little more resistance. Thanks for the challenge

I hope I am right/helpful, good luck!

"so a readout of ".014" should translate as .000014 amps. I am confused about this....."

Good catch, but how likely could you get a fan to do work with .00015 Watts of power?

This is a good example where stepping back and reexamining the data tells you that mA was probably not the actual setting on the meter.

The resistance of the fan will not be constant, but, TVP's 36Ω won't be far wrong.

Only two questions/problems remain.

Where did you get the original reading of 0.46 Amps from?

What power is that resistor? It looks like ½ or ¼ Watt to me.
Your most important formula is Ohms law (V=IR). But, second is W=IV (Power=CurrentxVoltage)
0.143x4.3=0.615 Watts (so I bet that resistor is getting quite hot).

I guessed 1 Watt and from TP's numbers 0.6 W loss. But hot yep.

Power! Oh boy are you right!

Longbow, you need a resistor that is power rated at least 2x your power dissipation!

Now that this is settled, a wee trip down memory lane.

How many remember calculating filament heater resistors like this? If you ever worked on the old 5 tube, ac/dc radios...

What in Gods name are you talking about Grandpa?!

(hahaha... just joking, please don't take offense)

That's too old school for me!

It's like kissing a beautiful woman on New Year, the telling is never as good as the doing.

Neat design !!! Share the schematics !

Yahlasit

The design is based off of the Williamson tube amp from the mid fifties. I started with a version here:

PP KT88

I made a number of proprietary changes, which I do not want to share until I determine the commercial fate of this beast (I have had a number of requests to sell them).

The VTVKT88 amp I cited has a number of little things that I felt were not the best way to go about the problem, particularly with the power supply, but it still is a great amp as shown. I would argue that the toroidal transformers have some advantages, but suffer some disadvantages in the bass that the article does not mention. For that reason I selected more traditional EI transformers and a more modern power supply solution, while still keeping to the spirit of the valve era.

I used one IC (a 555 timer), which is not part of the signal path, to provide a start up delay of 60 seconds for the B+ circuit to allow the tube filaments to reach operational temperature before DC power is applied. The net effect increases tube life.

I remember a 4 tube job I had with series strung filaments. It used a 50C5 (50 volts), a 35W4 (35 volts) and a couple of 12AX7? AU7? anyhow two 12 volt tubes for a total of 109 VAC filament voltage... a lo.o.o.o.o.ong time ago.

Bill

the formula is V=I*R where V is the voltage drop across the resistor,I is the current flowing through the resistor and R is the resistance of the resistor.

except 9.5 volt all other values are correct.so there might b problem with ur VOM or some other components are present in ur ckt.

You can not treat a DC brushless fan like yours as a resistive load, you can add a resistor to limit voltage to it thanks to a voltage drop, but at the cost of wasting power on your limiting resistor, You should use a voltage regulator part number LM7809 or MC7809, Lead 1 is V.input Lead 2 is ground and Lead 3 is V.output.

I know you're trying to run the fan at its rated voltage; but just as a coment for future reference; These fans have a built-in driver circuit on a little board below the center label. And voltage changes are not recommended as a method of modifying its speed.

Good luck !

Yahlasit

Yahlasit,

.........The 78xx series are linear regulators. Do you think one of these will waste less energy than a resistor?

He could just use diodes as forward drops and not suffer low current on start up.

Hi Randall,

These regulators are certainly linear, something like a zenner + transistor, wich is more efficient than a zenner + resistor, wich in an open load condition, still consumes about the same power as loaded (increasing load current decreases zenner current, 'till V.z. is reached).

A 78xx not only wastes less energy, but also keeps better voltage output in spite of input variations above 10.5 V approx. and up to V.Max of the regulator (varies with manufacturer, but is around 30 VDC).

Yahlasit

The point is they are not switching power supplies:-

You've still (like the resistor) got a relatively constant 4.8 Volts across and 136 mA through the series pass element, and, hence the same power dissipated in that element as in the resistor.

I agree, they are not switching power supplies (even though they're compact and powerful, I avoid them whenever possible, because they generate too much noise), and about your argument, it makes sense; you've already made me hesitate, I'll have to do the experiment, or save me time and show me some numbers (not too complex please).

Yahlasit

To all that replied to this post I'll say thanks. My first problem, after much ado, was my meter which was 100ma off in it's reading which was where the confusion was coming from for me. None of Ohm's Law was working for me!!!! After borrowing a meter from work, all the number fell in place and I could figure the actual voltage drop. I know it's a short ride but it almost drove me nuts as to why it wouldn't figure out. As for the fan/resistor....I was just doing a simple test trying to refresh myself on ohm's law and being able to figure it out for myself. So, all in all it was a learning experience for me AND I get to buy a new meter!!! Thanks again to everybody and kind regards.