# 1.5 volt bulb won't light in my circuit newbie

#### Audioguru

Sep 24, 2016
3,650
Remember seeing an old fashioned flashlight? It used a 2.4V incandescent light bulb and two HUGE D cells in series. The high current used by the light bulb reduced the 3V when new battery voltage to 2.4V.
The cells in a 9V battery are like tiny AAAA cells.

#### stspringer

May 10, 2019
116
What you need is a resistor in series with the bulb.

If the bulb draws 300mA at 1.5V we can use Ohms law to find the resistance of the bulb when it is lit. This is different than the resistance when it is not lit because, as the filament heats up, its resistance increases. So, using Ohms law:

V = I * R
R = V / I

R = 1.5 / 0.3 = 5 Ohms.

Now, we want to make a voltage divider using the bulb itself as the lower resistor so that the voltage across the bulb is 1.5V.

Using the equation of a voltage divider:

Vout = Vin * R2 / (R1 + R2)

replacing R2 with the bulb resistance, and the two voltages we get:

1.5 = 9 * 5 (R1 + 5)

1,5 * (R1 + 5) = 9 * 5

1.5 * R1 + 1.5 * 5 = 9 * 5

1.5 * R1 = 9 * 5 - 1.5 * 5

1.5 * R1 = 35

R1 = 35 / 1.5 = 25

Use a 27 Ohm resistor in series with your bulb and your bulb will light. With a fresh battery it might light for 20 minutes.

Edit: If you have successfully used LED, you should have already known how to calculate this.

The way it is usually expressed for LEDs is

R = (Vin - Vf) / If

in this case:

R = (9 - 1.5) / 0.3 = 25

Edited to fix original calculation and show alternate calculation.

Bob
Thank you Bob,
The bulb lit but the 27 ohm resistor started smoking. Did I miss something?

#### BobK

Jan 5, 2010
7,682
Well, I didn’t do the power calculation.

P = I * V

The current is 0.3 and the voltage across the resistor is 9- 1.5 = 7.5

So the power is 2.25 Watts. You need a resistor rated at 3W or more.

Bob

#### Audioguru

Sep 24, 2016
3,650
If the 9V battery kept its voltage at 9V (it won't) when overloaded with the high current, then the total resistance is 27 ohms + 5 ohms= 32 ohms and the current is 9V/32 ohms= 281mA.
The heating is called Power Dissipation and the formula is (P= I squared x R) so the 27 ohms resistor heated with 2.13W.
You need a huge 5W resistor. What is the power rating of your smoking resistor?
You are killing the 9V battery with such a high current so it also got hot.

#### AnalogKid

Jun 10, 2015
2,804
A linear voltage regulator is less efficient than a simple resistor.
Only if you include the current through the ground or adjust pin. For some parts that is uA.

ak

#### stspringer

May 10, 2019
116
Well, I didn’t do the power calculation.

P = I * V

The current is 0.3 and the voltage across the resistor is 9- 1.5 = 7.5

So the power is 2.25 Watts. You need a resistor rated at 3W or more.

Bob
Thank you Bob I will get a 3 watt resistor. I will study your calculations. Much appreciated

#### Bluejets

Oct 5, 2014
6,583
When you use these so called "smoke alarm batteries" you have to remember the amount of current they can deliver is only small.

Not suitable at all for what you want to do as others have pointed out.

You want to drive a load at 300mA then get something decent like a plug pack for 9v or / if only for a short time test, a battery box with an inbuilt switch that will take 6 * AA batteries.

As Bob said back in #18, a dc dc converter is much more efficient than resistors or linear regulators.

The one I show below ( buck converter) will handle a couple of amps, one can have voltages input from 5 to 36V and an output from 1.25V up to around a couple of volts below your input supply. (remember this)
It is cheap, should be ok for years on your test bench and will give your battery maximum life.
Cost about $6.00 delivered. https://www.ebay.com.au/itm/DC-DC-S...f8a3:m:mGzXuIOgAerpD-jIG6w2tXw&frcectupt=true If you need one to go into a small project then the one below, similar specs just watch the minimum output voltage and the maximum output current levels. https://www.ebay.com.au/itm/DC-Buck...163635?hash=item5903d1b233:g:h3sAAOSwUoNaPIJX There are also project size units with pot adjustable output rather than the "step output" in this one. These can be really tiny. Im presently using one in a r/c altitude meter I'm building. 17*10mm. OLd type were twice the size for same specs. Cost of the one below maybe$1.20 delivered.

Last edited:

#### stspringer

May 10, 2019
116
What you need is a resistor in series with the bulb.

If the bulb draws 300mA at 1.5V we can use Ohms law to find the resistance of the bulb when it is lit. This is different than the resistance when it is not lit because, as the filament heats up, its resistance increases. So, using Ohms law:

V = I * R
R = V / I

R = 1.5 / 0.3 = 5 Ohms.

Now, we want to make a voltage divider using the bulb itself as the lower resistor so that the voltage across the bulb is 1.5V.

Using the equation of a voltage divider:

Vout = Vin * R2 / (R1 + R2)

replacing R2 with the bulb resistance, and the two voltages we get:

1.5 = 9 * 5 (R1 + 5)

1,5 * (R1 + 5) = 9 * 5

1.5 * R1 + 1.5 * 5 = 9 * 5

1.5 * R1 = 9 * 5 - 1.5 * 5

1.5 * R1 = 35

R1 = 35 / 1.5 = 25

Use a 27 Ohm resistor in series with your bulb and your bulb will light. With a fresh battery it might light for 20 minutes.

Edit: If you have successfully used LED, you should have already known how to calculate this.

The way it is usually expressed for LEDs is

R = (Vin - Vf) / If

in this case:

R = (9 - 1.5) / 0.3 = 25

Edited to fix original calculation and show alternate calculation.

Bob
Hello Bob,
You gave me the best answer but I am a little confused on the math. What is the value of R1? Could you please explain your math for me I am not good at math. I know you are substituting here but I am confused. Thank you

John

#### BobK

Jan 5, 2010
7,682
Basically you are making a voltage divider using a series resistor (R1) and the bulb (R2.)

The first equation is the equation for a voltage divider. You know all of the variables except R1, so you substitute all of the other values and solve for R1 using basic algebra.

Bob

#### hevans1944

##### Hop - AC8NS
Jun 21, 2012
4,855
@stspringer: I think it is very admirable that you are taking a "hands on" experimental approach to learning electricity and electronics. You are in the process of learning some valuable lessons about voltage dividers and practical values of the resistance associated with them. Take careful notes, and try to keep a journal or a notebook of what you measure, lest you forget what you did before and keep repeating the same thing while expecting different results. Burning up resistors is all part of the learning experience. We have all done it at one time or another. Just determine what went wrong, correct it, and move on.

Voltage dividers as used today are very seldom used to provide significant levels of power to a load, such as your 1.5 V incandescent lamp for example. The main reason is they are terribly inefficient for that purpose. Instead, most applications of voltage dividers use them as a low-level signal attenuator, the volume control in a radio for example, or to create a low-level, relatively high impedance, control signal for some other circuit, such as a user-adjustable power supply output voltage for a programmable power supply.

As you can imagine, while the basic voltage divider equation hasn't changed, the actual values selected for R1 and R2 can be all over the map while still providing the same attenuation ratio. The only thing that changes is how the voltage divider appears electrically to circuit with which it is connected.

In days of old, when incandescent lighting was first used to illuminate theater stage lighting, physically very large variable resistances (called rheostats) were used in series with the lights to control their brightness. These variable resistances were typically capable of dissipating hundreds, or even thousands, of watts of power. The entire back-stage, behind-the-scenes area, was kept toasty warm in the winter from the heat given up by immense banks (often hundreds in a Broadway production) of these rheostats. Of course the heat during a summer production was pretty much unbearable.

That all changed, practically overnight, during the middle of the twentieth century with the advent of solid-state lighting controls. Some prior efforts had used mercury-vapor thyratrons to control lighting intensity, which had the advantage of allowing separation between the control electronics and the thyratron, but it was all pretty much a kludge.

Good hunting on your electronic adventure! Make sure to nail the basics before moving further into the jungle.

#### stspringer

May 10, 2019
116
@stspringer: I think it is very admirable that you are taking a "hands on" experimental approach to learning electricity and electronics. You are in the process of learning some valuable lessons about voltage dividers and practical values of the resistance associated with them. Take careful notes, and try to keep a journal or a notebook of what you measure, lest you forget what you did before and keep repeating the same thing while expecting different results. Burning up resistors is all part of the learning experience. We have all done it at one time or another. Just determine what went wrong, correct it, and move on.

Voltage dividers as used today are very seldom used to provide significant levels of power to a load, such as your 1.5 V incandescent lamp for example. The main reason is they are terribly inefficient for that purpose. Instead, most applications of voltage dividers use them as a low-level signal attenuator, the volume control in a radio for example, or to create a low-level, relatively high impedance, control signal for some other circuit, such as a user-adjustable power supply output voltage for a programmable power supply.

As you can imagine, while the basic voltage divider equation hasn't changed, the actual values selected for R1 and R2 can be all over the map while still providing the same attenuation ratio. The only thing that changes is how the voltage divider appears electrically to circuit with which it is connected.

In days of old, when incandescent lighting was first used to illuminate theater stage lighting, physically very large variable resistances (called rheostats) were used in series with the lights to control their brightness. These variable resistances were typically capable of dissipating hundreds, or even thousands, of watts of power. The entire back-stage, behind-the-scenes area, was kept toasty warm in the winter from the heat given up by immense banks (often hundreds in a Broadway production) of these rheostats. Of course the heat during a summer production was pretty much unbearable.

That all changed, practically overnight, during the middle of the twentieth century with the advent of solid-state lighting controls. Some prior efforts had used mercury-vapor thyratrons to control lighting intensity, which had the advantage of allowing separation between the control electronics and the thyratron, but it was all pretty much a kludge.

Good hunting on your electronic adventure! Make sure to nail the basics before moving further into the jungle.

Thank you Sir, that was helpful.
I am just a hobbyist, and a gadget man, and enjoy learning new things.

John

#### stspringer

May 10, 2019
116
If the 9V battery kept its voltage at 9V (it won't) when overloaded with the high current, then the total resistance is 27 ohms + 5 ohms= 32 ohms and the current is 9V/32 ohms= 281mA.
The heating is called Power Dissipation and the formula is (P= I squared x R) so the 27 ohms resistor heated with 2.13W.
You need a huge 5W resistor. What is the power rating of your smoking resistor?
You are killing the 9V battery with such a high current so it also got hot.

Thank you,
Can you help me by explaining Bobs post,on how he found R1 value using algebra? Post #22

#### stspringer

May 10, 2019
116
Thank you to all of you for your help.

I am just a hobbyist and a gadget man. I like to learn new things. I started using breadboards with led's and got them working from tutorials.

I had no idea what was in store for me when I decided to try a 1.5 volt bulb to light up with a 9 volt battery, I just assumed resistors were the way to go, just like for led's. I am glad I tried it because it turned out to be a great learning experience in electronics.

Some of your answers are way over my head. I am now finding that I will have to dive into algebra. I was never good at math,but I will try. I am doing the google for algebra on finding resistor values. See Bob's post #22

As you know some google tutorials are for people who already know their stuff and not for guys like me. I have to hunt for a tutorial that explains it to a person like myself who knows next to nothing about electronics or algebra.

As always any help is much appreciated.

Thanks again to all
John

Last edited:

#### Audioguru

Sep 24, 2016
3,650
Thank you,
Can you help me by explaining Bobs post,on how he found R1 value using algebra? Post #22
I don't use Algebra, instead I use simple Ohm's Law and simple Math Fractions.

1) What is the resistance of the white-hot bulb? It is spec'd at 1.5V/0.3A and OHM'S LAW is R= V/I. Then 1.5V/0.3A= 5 ohms. .

2) What value resistor will allow the bulb to work from a very strong 9V battery? This resistor will have 9V - 1.5V= 7.5V across it and will have 0.3A in it and OHM'S LAW is R= V/I. Then 7.5V/0.3A= 25 ohms. But 25 ohms is not a standard value so use 27 ohms.

3) Heating power? P= V x I or I squared x R. Then the heating is a little less than 2.25W. A 3W resistor will get hot enough to melt nearby plastic so I would use a 5W resistor.

#### stspringer

May 10, 2019
116
I don't use Algebra, instead I use simple Ohm's Law and simple Math Fractions.

1) What is the resistance of the white-hot bulb? It is spec'd at 1.5V/0.3A and OHM'S LAW is R= V/I. Then 1.5V/0.3A= 5 ohms. .

2) What value resistor will allow the bulb to work from a very strong 9V battery? This resistor will have 9V - 1.5V= 7.5V across it and will have 0.3A in it and OHM'S LAW is R= V/I. Then 7.5V/0.3A= 25 ohms. But 25 ohms is not a standard value so use 27 ohms.

3) Heating power? P= V x I or I squared x R. Then the heating is a little less than 2.25W. A 3W resistor will get hot enough to melt nearby plastic so I would use a 5W resistor.

Thank you you explained it well

#### BobK

Jan 5, 2010
7,682
I don't use Algebra, instead I use simple Ohm's Law and simple Math Fractions.
No, you are using algebra, you just let someone else do the work.

Ohms law is usually stated as V = I * R

So how did you get the equation R = V / I?

Perhaps you found all 3 forms of Ohms law written out somewhere.

Myself, I prefer to remember one form and get the other forms as needed by algebra, which I learned in second grade (that is a whole 'nother story), so I have been using it for 60 years now.

For the TS:

The equation:

V = I * R

Can be rearranged to isolate R on one side. This is called solving for R.
This is how you do it:

Since both sides of an equation are equal you can apply the same operation to both sides and they remain equal.

So take:

V = I * R

and divide both sides by I, getting:

V / I = (I * R) / I

Then you can cancel the I in the numerator and the denominator, getting:

V / I = R

That's all there is too it. For more complex equations, you keep applying the same kind of rules until you have isolated the variable you are trying to solve for.

You can, among other things,

Add the same quantity to both sides
Subtract the same quantity from both sides
Multiply both sides by the same quantity
Divide both sides by the same quantity. (though it must not be zero)
Take the square root of both sides

And so on.

Bob

#### Audioguru

Sep 24, 2016
3,650
I remember Ohm's law the same way I remember how many fingers I have on each hand without calculating it using Algebra.
1) The total of my fingers is 10 due to luck (I played with chemical explosives when I was young but they all behaved like missiles. Guess what my missiles did?).
2) I count 2 hands.
3) 10 fingers divided by 2 hands= 5 fingers on each hand.
I wonder how to calculate how many toes I have on each foot? With Algebra?

Ohm's Law is not Voltage Law or Current Law but they are all combined.

#### stspringer

May 10, 2019
116
Hi all,
I found a great link on this subject of lighting high powered led's with a 12 volt car battery

Curious if I understand this video correctly
In my circuit , if I put 3 of my 1.5 volt bulbs in series, I could then use a smaller resistor right?

Thanks

Last edited:

#### Audioguru

Sep 24, 2016
3,650
In my circuit , if I put 3 of my 1.5 volt bulbs in series, I could then use a smaller resistor right?

Thanks
Yes, a smaller resistance and smaller physical size. But the high current of 0.3A will quickly kill a little 9V battery, you will see the lights dim more and more as they kill the battery..

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