Maker Pro
Maker Pro

1.5 volt bulb won't light in my circuit newbie

stspringer

May 10, 2019
116
Joined
May 10, 2019
Messages
116
Hello all,
Here is what I have:

9 volt battery - VCC
breadboard
1 - 10k resistor
1 - 2k resistor
1 - 1.5 volt light bulb


I read 1.5 volts output with my digital meter, but for some reason the light bulb will not light.
The light bulb is good, I can light it with a 1.5 volt battery.
I get continuity on the light bulb socket too.

Here are some pics. I don't understand why it doesn't work. Please advise.

Thank You

John
 

Attachments

  • 20190602_092619_1.jpg
    20190602_092619_1.jpg
    32.7 KB · Views: 18
  • 20190602_092619_2.jpg
    20190602_092619_2.jpg
    32.1 KB · Views: 18
  • 20190602_092619_3.jpg
    20190602_092619_3.jpg
    25.7 KB · Views: 16

AnalogKid

Jun 10, 2015
2,927
Joined
Jun 10, 2015
Messages
2,927
A.l.w.a.y.s post a schematic.

This is one of the things that is not intuitively obvious to a beginner, because it involves the consequences of how a battery actually functions, and it appears to the rest of the circuit.

My GUESS is that you are trying to step down the 9 V to 1.5 V with a two-resistor voltage divider. If so, the problem is output impedance, or Thevenin equivalent resistance.

Assuming you are connecting the bulb across the 2K resistor, that means the two are in parallel. I measured a 1.5 V penlight bulb, and its cold resistance is around 1.0 ohm. So the equivalent resistance at the bottom of your divider isn't 2,000 ohms, it is 1 ohm. If you re-calculate the output voltage, you will see it is very low.

That is a quickie estimate. There is a right way to put the impedance of a voltage divider in context with the load attached, called the Thevenin equivalent resistance. for your circuit; it goes like this:

The battery has a (theoretically) zero ohm output impedance, and between its two terminals are two resistors in series. The *effective* impedance at the center point isn't 10K or 2 K, it is 10K and 2K in parallel, or 1.67 K. This is because the zero ohm output impedance of the battery effectively "shorts together" the 10K and 2K resistor ends. The voltage at the center point is 1.5 V as you've calculated. So what the light bulb "sees" as its power source is a 1.5 V battery in series with a 1.67K resistor. That 1.67K equivalent resistance in series with a 1 ohm bulb makes for a huge attenuator (0.06%), which is why the bulb doesn't light.

This is a bit math-heavy, but covers the topic: https://en.wikipedia.org/wiki/Thevenin_circuit

A.l.w.a.y.s post a schematic.

ak
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Joined
Jan 21, 2010
Messages
25,510
Try reading the voltage when the bulb is connected.

Do you know how much current is needed to light the bulb? Try measuring it using a 1.5V battery (make sure you know how to measure current first).

Then measure the current drawn from the 9 Volt battery when your voltage divider is connected.

A rule of thumb is that the load on a voltage divider should be less than about 10% of the current through it.

How does the current required by the bulb compare to the current through the voltage divider alone?

Imagine you simply replaced the 2k resistor with your bulb? What is the maximum current that would flow through it? (Assuming the resistance of the bulb is zero is reasonable for a tough calculation). How does this current compare with the current required by the bulb?
 

stspringer

May 10, 2019
116
Joined
May 10, 2019
Messages
116
A.l.w.a.y.s post a schematic.

This is one of the things that is not intuitively obvious to a beginner, because it involves the consequences of how a battery actually functions, and it appears to the rest of the circuit.

My GUESS is that you are trying to step down the 9 V to 1.5 V with a two-resistor voltage divider. If so, the problem is output impedance, or Thevenin equivalent resistance.

Assuming you are connecting the bulb across the 2K resistor, that means the two are in parallel. I measured a 1.5 V penlight bulb, and its cold resistance is around 1.0 ohm. So the equivalent resistance at the bottom of your divider isn't 2,000 ohms, it is 1 ohm. If you re-calculate the output voltage, you will see it is very low.

That is a quickie estimate. There is a right way to put the impedance of a voltage divider in context with the load attached, called the Thevenin equivalent resistance. for your circuit; it goes like this:

The battery has a (theoretically) zero ohm output impedance, and between its two terminals are two resistors in series. The *effective* impedance at the center point isn't 10K or 2 K, it is 10K and 2K in parallel, or 1.67 K. This is because the zero ohm output impedance of the battery effectively "shorts together" the 10K and 2K resistor ends. The voltage at the center point is 1.5 V as you've calculated. So what the light bulb "sees" as its power source is a 1.5 V battery in series with a 1.67K resistor. That 1.67K equivalent resistance in series with a 1 ohm bulb makes for a huge attenuator (0.06%), which is why the bulb doesn't light.

This is a bit math-heavy, but covers the topic: https://en.wikipedia.org/wiki/Thevenin_circuit

A.l.w.a.y.s post a schematic.

ak
Hi AnalogKid,
Thanks for your reply. Here is the schematic. I will now try to digest what you posted. LOL
 

Attachments

  • 20190602_092619_4.jpg
    20190602_092619_4.jpg
    56.9 KB · Views: 23

stspringer

May 10, 2019
116
Joined
May 10, 2019
Messages
116
Try reading the voltage when the bulb is connected.

Do you know how much current is needed to light the bulb? Try measuring it using a 1.5V battery (make sure you know how to measure current first).

Then measure the current drawn from the 9 Volt battery when your voltage divider is connected.

A rule of thumb is that the load on a voltage divider should be less than about 10% of the current through it.

How does the current required by the bulb compare to the current through the voltage divider alone?

Imagine you simply replaced the 2k resistor with your bulb? What is the maximum current that would flow through it? (Assuming the resistance of the bulb is zero is reasonable for a tough calculation). How does this current compare with the current required by the bulb?

Hello Steve,
The bulb is 1.5 volts. I can light it directly with a 1.5 volt battery.
"Then measure the current drawn from the 9 Volt battery when your voltage divider is connected"
I will try this.
 

stspringer

May 10, 2019
116
Joined
May 10, 2019
Messages
116
A.l.w.a.y.s post a schematic.

This is one of the things that is not intuitively obvious to a beginner, because it involves the consequences of how a battery actually functions, and it appears to the rest of the circuit.

My GUESS is that you are trying to step down the 9 V to 1.5 V with a two-resistor voltage divider. If so, the problem is output impedance, or Thevenin equivalent resistance.

Assuming you are connecting the bulb across the 2K resistor, that means the two are in parallel. I measured a 1.5 V penlight bulb, and its cold resistance is around 1.0 ohm. So the equivalent resistance at the bottom of your divider isn't 2,000 ohms, it is 1 ohm. If you re-calculate the output voltage, you will see it is very low.

That is a quickie estimate. There is a right way to put the impedance of a voltage divider in context with the load attached, called the Thevenin equivalent resistance. for your circuit; it goes like this:

The battery has a (theoretically) zero ohm output impedance, and between its two terminals are two resistors in series. The *effective* impedance at the center point isn't 10K or 2 K, it is 10K and 2K in parallel, or 1.67 K. This is because the zero ohm output impedance of the battery effectively "shorts together" the 10K and 2K resistor ends. The voltage at the center point is 1.5 V as you've calculated. So what the light bulb "sees" as its power source is a 1.5 V battery in series with a 1.67K resistor. That 1.67K equivalent resistance in series with a 1 ohm bulb makes for a huge attenuator (0.06%), which is why the bulb doesn't light.

This is a bit math-heavy, but covers the topic: https://en.wikipedia.org/wiki/Thevenin_circuit

A.l.w.a.y.s post a schematic.

ak

I just measured my resistance on my 1.5 v 0.3A bulb and it was 00.8
 

stspringer

May 10, 2019
116
Joined
May 10, 2019
Messages
116
Try reading the voltage when the bulb is connected.

Do you know how much current is needed to light the bulb? Try measuring it using a 1.5V battery (make sure you know how to measure current first).

Then measure the current drawn from the 9 Volt battery when your voltage divider is connected.

A rule of thumb is that the load on a voltage divider should be less than about 10% of the current through it.

How does the current required by the bulb compare to the current through the voltage divider alone?

Imagine you simply replaced the 2k resistor with your bulb? What is the maximum current that would flow through it? (Assuming the resistance of the bulb is zero is reasonable for a tough calculation). How does this current compare with the current required by the bulb?
Steve my amps are .306 on the light bulb with 1.5 volt battery
volts with voltage divider 1.5v
current draw on 9 volt battery with voltage divider 1.8 a
not sure I did it right testing current draw on 9 volt battery, battery was getting warm and amps kept going down
 
Last edited:

stspringer

May 10, 2019
116
Joined
May 10, 2019
Messages
116
Try reading the voltage when the bulb is connected.

Do you know how much current is needed to light the bulb? Try measuring it using a 1.5V battery (make sure you know how to measure current first).

Then measure the current drawn from the 9 Volt battery when your voltage divider is connected.

A rule of thumb is that the load on a voltage divider should be less than about 10% of the current through it.

How does the current required by the bulb compare to the current through the voltage divider alone?

Imagine you simply replaced the 2k resistor with your bulb? What is the maximum current that would flow through it? (Assuming the resistance of the bulb is zero is reasonable for a tough calculation). How does this current compare with the current required by the bulb?


current on bulb .306 a
volts on voltage divider 1.5 v
 

stspringer

May 10, 2019
116
Joined
May 10, 2019
Messages
116
A.l.w.a.y.s post a schematic.

This is one of the things that is not intuitively obvious to a beginner, because it involves the consequences of how a battery actually functions, and it appears to the rest of the circuit.

My GUESS is that you are trying to step down the 9 V to 1.5 V with a two-resistor voltage divider. If so, the problem is output impedance, or Thevenin equivalent resistance.

Assuming you are connecting the bulb across the 2K resistor, that means the two are in parallel. I measured a 1.5 V penlight bulb, and its cold resistance is around 1.0 ohm. So the equivalent resistance at the bottom of your divider isn't 2,000 ohms, it is 1 ohm. If you re-calculate the output voltage, you will see it is very low.

That is a quickie estimate. There is a right way to put the impedance of a voltage divider in context with the load attached, called the Thevenin equivalent resistance. for your circuit; it goes like this:

The battery has a (theoretically) zero ohm output impedance, and between its two terminals are two resistors in series. The *effective* impedance at the center point isn't 10K or 2 K, it is 10K and 2K in parallel, or 1.67 K. This is because the zero ohm output impedance of the battery effectively "shorts together" the 10K and 2K resistor ends. The voltage at the center point is 1.5 V as you've calculated. So what the light bulb "sees" as its power source is a 1.5 V battery in series with a 1.67K resistor. That 1.67K equivalent resistance in series with a 1 ohm bulb makes for a huge attenuator (0.06%), which is why the bulb doesn't light.

This is a bit math-heavy, but covers the topic: https://en.wikipedia.org/wiki/Thevenin_circuit

A.l.w.a.y.s post a schematic.

ak
My bulb reads .7 ohms sound right? its a miniature lamp 1.5 volt 0.3a bulb
 

Audioguru

Sep 24, 2016
3,656
Joined
Sep 24, 2016
Messages
3,656
A little 9V alkaline battery is designed for a 20mA load, not 15 times more that you are trying to do!
Also the contacts on a solderless breadboard probably have trouble passing such a high current.

I think you measured the current draw of the multimeter (not of the light bulb) because you probably used it to short circuit the battery so you killed most of the life of the battery.

A light bulb is a heater. You never use a little battery to power a heater.
To measure a current then the current meter must be in series with the load.
 

Alec_t

Jul 7, 2015
3,673
Joined
Jul 7, 2015
Messages
3,673
With your circuit there won't be 1.5V across the bulb..... much less.
Think of it this way. The 2kΩ in parallel with the 0.7Ω bulb is virtually short-circuited by the bulb, so can be ignored. That means the voltage divider is now just 10kΩ and 0.7Ω. Virtually all the circuit current passes through the bulb, but as it has to pass through the 10k resistor it will be very small (9V/10,000.7 Ohms ~= 0.9mA). The voltage across the bulb will be 0.9mA x 0.7Ω = 0.63mV.
 

stspringer

May 10, 2019
116
Joined
May 10, 2019
Messages
116
With your circuit there won't be 1.5V across the bulb..... much less.
Think of it this way. The 2kΩ in parallel with the 0.7Ω bulb is virtually short-circuited by the bulb, so can be ignored. That means the voltage divider is now just 10kΩ and 0.7Ω. Virtually all the circuit current passes through the bulb, but as it has to pass through the 10k resistor it will be very small (9V/10,000.7 Ohms ~= 0.9mA). The voltage across the bulb will be 0.9mA x 0.7Ω = 0.63mV.

Thank you for your reply. So can you tell me how to get it to work so I can learn. What do I need to do? What resistors should I be using? Thank you .
 

stspringer

May 10, 2019
116
Joined
May 10, 2019
Messages
116
With your circuit there won't be 1.5V across the bulb..... much less.
Think of it this way. The 2kΩ in parallel with the 0.7Ω bulb is virtually short-circuited by the bulb, so can be ignored. That means the voltage divider is now just 10kΩ and 0.7Ω. Virtually all the circuit current passes through the bulb, but as it has to pass through the 10k resistor it will be very small (9V/10,000.7 Ohms ~= 0.9mA). The voltage across the bulb will be 0.9mA x 0.7Ω = 0.63mV.

Thank you for your reply. So I need a much smaller resistor than 10 k right? What do you suggest to make it work?
 

Alec_t

Jul 7, 2015
3,673
Joined
Jul 7, 2015
Messages
3,673
It won't work (at least, not for many seconds) with a little 9V battery if the bulb needs 0.3A to light up. See post #11.
Have you considered replacing the bulb with a white LED? That will be pretty bright with much less current.
 

stspringer

May 10, 2019
116
Joined
May 10, 2019
Messages
116
It won't work (at least, not for many seconds) with a little 9V battery if the bulb needs 0.3A to light up. See post #11.
Have you considered replacing the bulb with a white LED? That will be pretty bright with much less current.


I have all kinds of breadboards with led's working. I just thought I would try something with a miniature bulb and never knew it could not be done. It amazes me that it can't be done.
 

AnalogKid

Jun 10, 2015
2,927
Joined
Jun 10, 2015
Messages
2,927
That 1 ohm resistance measurement is the clue. Incandescent bulbs are very hungry.

ak
 

BobK

Jan 5, 2010
7,682
Joined
Jan 5, 2010
Messages
7,682
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
 
Last edited:

Nanren888

Nov 8, 2015
622
Joined
Nov 8, 2015
Messages
622
If "it" is light a bulb from a battery of greater voltage, then, it can be done. A resistive divider is just not an efficient way to do it.
.
Look up "voltage regulator".
Voltage regulators usually look like a variable resistance in series with the bulb, as mentioned above, staying just right to get the voltage you need. if chosen well.
.
Also as mentioned above, that little 9 volt battery is designed for low current, so it is designed with a relatively high source impedance. Your bulb is low resistance (impedance). Driving a low impedance load from a high impedance source is going to be inefficient. LOts of the power is going to be disappated in the source.
Bulbs are a pain like that. Need lots of current. LEDs are a great invention.
.
It's a great learning experience. Just decide which way you want to go next.
Flashing?
 

BobK

Jan 5, 2010
7,682
Joined
Jan 5, 2010
Messages
7,682
If "it" is light a bulb from a battery of greater voltage, then, it can be done. A resistive divider is just not an efficient way to do it.
.
Look up "voltage regulator".
Voltage regulators usually look like a variable resistance in series with the bulb, as mentioned above, staying just right to get the voltage you need. if chosen well.
.
Also as mentioned above, that little 9 volt battery is designed for low current, so it is designed with a relatively high source impedance. Your bulb is low resistance (impedance). Driving a low impedance load from a high impedance source is going to be inefficient. LOts of the power is going to be disappated in the source.
Bulbs are a pain like that. Need lots of current. LEDs are a great invention.
.
It's a great learning experience. Just decide which way you want to go next.
Flashing?
A linear voltage regulator is less efficient than a simple resistor. A DC to DC converter can be more efficient.

Bob
 
Top