# Still learning resistors

So i have asked a few questions about resistors, and I have learned quite a lot. Now that I have moved on and bought a multimeter, I have a few more questions.

I have been playing with an ohms calculator as I am currently working on a project requiring external power. I have a 9 volt dc plug that I want to use. I only need 7 volts. When plugging in numbers on the ohms calc, this is what I get:

20ma from arduino 350ohm resistor

= 7volts

So I assume that if I set this up on a breadboard, I will have a 7v power source.

Here is the riddle: I set up the arduino just to light up an led with the 5 volt pin and ground. I do a read with the MM and it reads 4.98, great. So I set up a 330 resistor, but the voltage is the same. I set up 2 10ohm resistors with it for 350 and same voltage. So what the heck is the calculator telling me? Is the voltage less even though the MM says it is still 4.98 volts? I am assuming the current is affected, perhaps providing a lower voltage for the component I am working on? This is still very new to me so I appreciate your guys patience. Thanks. Bob

P.s. Is it possible that even though the MM reads the same volts, the actual voltage is different via by way of reducing amps? If so, what will the reduction of amps do to my components (servos)?

A)Im not too sure what youre asking. B) the 5v pin should always read close to 5v hopefully never more than +/- 0.5v. C) Google “ohms law”

How do you get 350 ohms from 210 (or 2 10 ohms?) resistors?

A good meter will always give the accurate voltage across the two points you are measuring.

[u]Point 1:[/u] Voltage drives current.

If you stick a resistor from a 5v supply to ground, you will have 5 volts across that resistor.

Assuming the supply can source sufficient current, you will then have:

Current = 5/Resistance {This is called a 'Constant Voltage' supply}

If your supply hits its current limit, what will inevitably happen (assuming nothing cuts out or breaks) is that the supply voltage will drop such that the supply voltage becomes

Voltage = [Current Limit] * Resistance {This is called a 'Constant Current' supply}

[u]Point 2:[/u] The voltage may not be exactly 5V from your 5V supply. That rating is nominal, i.e. it will be that voltage +/-some%. If your multimeter is 100% accurate then your supply is 4.98 - i.e. -0.4% from its nominal value.

Tom, so what I think you are saying is, that the resistor does indeed lower the voltage... Even though the MM reads the same voltage? I think.

In lay men's terms... How do I drop a 9v dc to 7 to run a servo? Am I in the ball park by adding a resistor? Am I missing anything here? I don't want to hook this up and damage the servo.

Bstanko6: In lay men's terms... How do I drop a 9v dc to 7 to run a servo? Am I in the ball park by adding a resistor? Am I missing anything here?

You don't use resistors for that. No. You get a voltage regulator IC, a 7805 is easy to do. There are examples aplenty. C30 there may not be entirely necessary. The value of C31 is a bit much, 1-10uF would be better (to me), rated 10V or so.

Yes you can "drop" voltage using resitors in a series voltage divider circuit, but this is really not a suitable way to provide a stable voltage and power output. What you need is a voltage regulator ( 7805) or seperate power supply or combination of both depending on your specific set up. Servos use a lot of power. What power source do you plan to power your servo with? If a resistor or load draws enough current, it can overload the power supply which causes the voltage to drop. Under normal current draw (less than 500ma for usb power or whatever the spec is for the supply) the power supply should be able to keep 5v supply rail fairly close to 5 volts.

Bstanko6: I have been playing with an ohms calculator as I am currently working on a project requiring external power. I have a 9 volt dc plug that I want to use. I only need 7 volts. When plugging in numbers on the ohms calc, this is what I get:

20ma from arduino 350ohm resistor

= 7volts

So I assume that if I set this up on a breadboard, I will have a 7v power source.

I don't understand this. You have a 9vDC power plug, you want 7v. I'm not sure what you were calculating or what a 20mA Arduino has to do with it. ??

Like others have said, if you're thinking of "shedding two volts by using a resistor", it doesn't work that way. Well... according to Ohm's Law it does, but the load (your servo in this case) would have to present a 100% consistent current draw to the power supply, which rarely ever is the case. Thus, the voltage will not be 7v except when the current demand is exactly where you calculated it. Less current, and the voltage will be much higher. More current, and it will be much lower. When the servo is not "on", its current draw will be almost nil and the input voltage will be nearly 9v.

Bstanko6: Here is the riddle: I set up the arduino just to light up an led with the 5 volt pin and ground.

Again, I'm not following. You put an LED between the 5v power supply and ground? BTW, that's a good way to turn your LED into a broken LED. Don't do that without a current-limiting resistor between the LED and either (doesn't matter which) power rail.

Bstanko6: I do a read with the MM and it reads 4.98, great. So I set up a 330 resistor, but the voltage is the same. I set up 2 10ohm resistors with it for 350 and same voltage. So what the heck is the calculator telling me? Is the voltage less even though the MM says it is still 4.98 volts?

You need to be more specific with the phrasing. You "set up a resistor" how/where? In series, between the power and the Arduino? Adding resistance from the power supply is almost never a good thing. Usually, it just prevents the load device from working properly, and possibly heats up and ruins the resistor (depending on how much voltage it drops as heat). Hardly ever useful though.

Bstanko6: P.s. Is it possible that even though the MM reads the same volts, the actual voltage is different via by way of reducing amps?

No. If I'm following your experiment correctly, and you have 5vDC to a resistor, and the resistor to your Arduino, and the Arduino to ground.... what happens is you have two resistors (one a constant value physical resistor, and the Arduino, whose resistance will vary based on its load) forming a voltage divider. The voltage at the point where they meet will vary as the current demands placed on the Arduino vary. With the Arduino doing nothing (not driving any LEDs or anything), its current draw will be very minimal, but would likely still cause a current-limited power supply voltage to drop a little. The voltage at the power supply terminal (north of your resistor) should stay constant though.

SirNickity, let me slow down.

The ohms calculator tells me if I add certain value resistors to a certain current value, then a certain value of voltage is produced. But now I’m being told I have to use a voltage regulator to reduce volts.

So what I am asking is, what is the calculator telling me? If adding resistors does not shed volts as you gents say, then why does adding more ohms resistor to the calc lower the volts? You see, I assume the calc is telling me what resistor “I” would use to lower the volts with a certain current. But now talking with you guys, it seems is not the case. I’m obviously confused about whether volts can be manipulated by resistors.

What you're missing is that current consumed is very rarely a static number. It tends to be HIGHLY variable. So, an Arduino doesn't "consume 20mA of current" in a way that you can bank on it. Rather, currents consumed are often spec'd as "up to", giving you a reasonable guideline for how much current you need to have available to prevent power starvation.

For e.g., if you pretend an Uno is a black box that always consumes precisely 20.0mA, and you add an LED, well, now maybe it consumes 30.7mA. Add a switch with pullup and maybe it consumes 31.5mA. You can't say with any certainty how much it will require at any given moment, as even if you calculate all the possible parts and pieces on the board, current is consumed differently during a clock transition than between ticks of the oscillator. It's a hopelessly moving target.

So, yes, you can use a resistor to dissipate voltage, but using that technique, the current consumed has to be predictable and 100% consistent or the voltage drop across your resistor will vary considerably from what you calculated. This makes the approach useful only when the voltage regulation is not important, but the ability to restrain current flow is desirable -- like a current limiting resistor on an LED for example.

OTOH, you might be interested to know that dropping voltage via resistance is basically what linear voltage regulators do... but the critical difference is they have a built-in feedback mechanism that continuously alters their internal resistance according to the load. As such, when you ask for X mA of current, the internal resistance of the regulator will be so much, and if you ask for twice that, the internal resistance will be correspondingly less. This is achieved by using a partially conducting *tran*sistor rather than a *re*sistor, which allows for the variable control. But the voltage drop is fundamentally the same principle at work.

calculator telling me?

ohms law : Voltage(v) = Current (I) x Resistance (R) . if you know any two variables you can solve for third. V= IxR or I=V/R or R= V/I. The calculator asks for two variables and spits out the third, its your job to know enough accurate info about your circuit to get a reasonable answer. What your doing is this : V= IxR, what you need to be doing is this; R= V/I. What i mean is, generally with leds you know the voltage you are working with (5v supply* ) and you know the max current (typically 20ma) so the variable that you are looking for is the resistor value. not the other way around. (*minus the led's forward voltage which is found in the led datasheet).

"I" would use to lower the volts with a certain current.

LEDs are current driven devices.so we are much more concerned about the current through it than the voltage across. Basically they are "off" until the voltage across the anode and cathode exceeds the forward voltage rating or the voltage. At which point it starts conducting. or "turns on". Once it turns on, it looks like a short and will pass as much current as is available. Without anything to limit the current (ie; resistor) it will promptly be destroyed. In short, the resistor is there to limit current not lower voltage.

You could if desperate do this… but any voltage drop would quickly result in less than desired results.

First image shows a circuit which uses a voltage divider with a 9v input and delivers 7.5v ar 51ma…

2nd image 159ma with adjusted voltage divider to deliver 7v… it should work providing the input voltage stays constant, otherwise buy a regulator or zener.

I appreciate the answers guys. And of course I have learned a lot from them. There is only one issue, why does the calculator tell me I will get 7v if I apply 350 ohm resistor to 20ma? Again I appreciate your answers and don't mean to offend, but you guys sound like Sheldon from Big Bang theory when you talk.

The calc tells me I get a certain voltage when I apply so much resistance to a certain amount of current. Many of you are saying no, use a voltage regulator or divider.

And I'm ok with the answer you gave me, I just don't understand why the volts don't drop on my multimeter when I apply the resistors To the circuit. The calc says it should. I just don't get it. Online doesn't clarify this well either.

So I came to a conclusion...voltage just does not drop from a resistor. Volts are a pre determined variable in the ohms law. Stop me if I am wrong.

Bstanko6: There is only one issue, why does the calculator tell me I will get 7v if I apply 350 ohm resistor to 20ma?

See my previous answer: Voltage drives current.

Your calculator is not saying you will get 7V. It is saying you have to supply 7V across the resistor if you want 20mA to flow.

If you are putting a resistor onto 9V, then putting a meter between ground and the end of the resistor without any other load attached, well, you’ll get 9V. Because the meter itself probably has a resistance of about 10M ohms, there is virtually no current flowing, so there is nearly no voltage dropped across 350 ohms.

Current = 9V/(10M + 350 ohms) = ~ 900uA
Voltage across 350 ohms = 350 ohms x 900uA =~ 315uV
Voltage dropped across meter 10M ohm x 900uA =~ 9V or 8.999685, which your 3-1/2 digit meter will display as 9V.

So the problem is that if current changes, the voltage dropped across the resistor will change. With a servo, current will change wildly, from nearly zero (a few mA of current) when it is not moving, to as much as an ampere of current when it is stalled.

Tom, I see now. The number voltage is what supply i need in reference to 350ohm resistor and 20ma. I see now, I may not have worded that right, but I get it. Thank you all.

Hi, the calculator is telling you how much voltage will be dropped across the resistor, not how much you will be supplied with it.

As has been said you should not use a dropping resistor to supply power in this case.

You are calculating the value of resistor to make [u]5V drop[/u] across it with 20mA. You need to drop 9 - 5 = 4V, redo the calc with 4V, but don't use it in this application.

We hope that this will help you learning basic electronics, ohms law is important and you need a little more guidance in how it works.

Tom... :)

Bstanko6:
So I came to a conclusion…voltage just does not drop from a resistor. Volts are a pre determined variable in the ohms law. Stop me if I am wrong.

Stop, you’re wrong. XD

Bstanko6:
There is only one issue, why does the calculator tell me I will get 7v if I apply 350 ohm resistor to 20ma?

First, your terminology isn’t quite right. Not to be pedantic or nag, but it will help you communicate better if you use conventional terminology. You don’t “apply a resistor to 20mA”, you either put it in parallel or series with something. That “something” is your load. The load can be as simple as another resistor, or as complex as a complete circuit. Let’s start with the former – we’ll use a resistor for the load, and put another resistor in series with it. This forms a voltage divider.

``````        350R     7v    1.2K
9v >---/\/\/\----*----/\/\/\---< Gnd
``````

By using Ohm’s Law, we can determine that 20mA is not to be found anywhere in this circuit. The two resistors together form a complete circuit resistance of 1.55k ohms between 9v and ground. This is about 5.8mA of current draw. With a 350R resistor on top, the load has to have a resistance of about 1.2K for there to be a 2v drop across the 350R resistor, and thus a measurable voltage of 7v at the tap point in the middle.

If I replace the 1.2K resistor with a 100R resistor, then the circuit consumes 20mA. But it also drops 7v across the 350R resistor, and the tap point will then read 2v from ground.

By this simple experiment, you can see that if you replace the 1.2K resistor with a complete circuit, the voltage at the 7v tap point on the graph above will vary depending on the resistance of the load circuit. The current the load consumes is dependent on its resistance, which with anything more complicated than a single resistor, will change over time. Thus, the “7v point” will not be 7v at all times either, it will change in relation to the resistance (current consumption) of the load. You can no longer use a simple voltage divider circuit to “regulate” the 9v down to 7v.

A regulator is purpose built for this. It monitors the voltage at the load and adjusts its internal resistance (represented by the 350R resistor in the graph above) in an opposing fashion to that of the load, so the middle point is always at the voltage you set it to be.