Troubleshooting a Current Divider Circuit for Dimmer Lights

In summary: The wattage rating of the pot should be enough to dissipate the power drawn by the dimmer. Unfortunately, the wattage rating of the pot may not be enough to dissipate the power drawn by the dimmer if the lamp's resistance is falling as the lamp heats up.
  • #1
GPT
11
0
Hello all,
I'm attempting to create a dimmer circuit for some lights Id like to install in my room. I've attached the diagram below.

Heres the problem: When I increase the resistance of potentiometer 2, for example, to decrease the current lamp 2 receives (thus dimming the lamp) the other lamps brighten. I want the other lamps to maintain their luminous output at a constant level dependant upon the amount of resistance of their respective pots.

Bear with me (this is my first circuit) but I don't understand why this is happening.

Two Questions
1) If the resistances of all the other pots remain the same, the same amount of current should flow through them, no? Why are they getting more current?
2) What are potential solutions to this problem?

Thanks for your help
Glenn

Edit:
Here are the parts I'm using on the breadboard:
-12v battery; 12v, 60mA mini indicator lamps; 10k ohm, .5W pot; 300ohm, .5W resistor in series with lamp and pot

These are the parts Id like to use in the finished circuit:
-12v transformer (wired to mains); 12v, 35W MR16 halogen lamps, appropriate pots and resistors
 

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  • #2
GPT said:
Hello all,
I'm attempting to create a dimmer circuit for some lights Id like to install in my room. I've attached the diagram below.

Heres the problem: When I increase the resistance of potentiometer 2, for example, to decrease the current lamp 2 receives (thus dimming the lamp) the other lamps brighten. I want the other lamps to maintain their luminous output at a constant level dependant upon the amount of resistance of their respective pots.

Bear with me (this is my first circuit) but I don't understand why this is happening.

Two Questions
1) If the resistances of all the other pots remain the same, the same amount of current should flow through them, no? Why are they getting more current?
2) What are potential solutions to this problem?

Thanks for your help
Glenn

Edit:
Here are the parts I'm using on the breadboard:
-12v battery; 12v, 60mA mini indicator lamps; 10k ohm, .5W pot; 300ohm, .5W resistor in series with lamp and pot

These are the parts Id like to use in the finished circuit:
-12v transformer (wired to mains); 12v, 35W MR16 halogen lamps, appropriate pots and resistors

If you measure the output voltage of the transformer as you vary a pot, does it vary? That would explain why the other bulbs change when you vary one pot. It would be because the output resistance of the transformer is not negligible in your circuit.
 
  • #3
Actually his trial version uses a battery, even more likely candidate for internal resistance I'd say, especially if the battery is "tired".

By the way, what about safety on the transformer version - need to be sure of the transformer rating, and with 35W lamps those "appropriate" pots may run a bit warm.
 
  • #4
By run a bit warm, do you mean warm enough to be a fire hazard, or too warm to touch?
Is there a formula I can use to calculate the heat dissipated by the pot and its accompanied fixed resistor?
 
  • #5
What I really meant was that resistive dimmers for 35W bulbs have to be fairly substantial, and mounted in suitable housings to deal with the heat to avoid burnt fingers. Resistive dimmers are generally pretty inefficient, and for this reason they are not so often used nowadays.

Unfortunately, filament lamps undergo a large increase in resistance as they heat up. Thus when you reduce the lamp voltage for dimming, the lamp resistance falls noticeably. Due to this effect, at some settings a dimming resistor may end up dissipating more than half of the lamp's nominal wattage. (If the lamp's resistance were constant, the dimmer could only dissipate up to half power). I'm sorry that I don't have a formula for this, perhaps it's safest to rate the pot for the full wattage.
 

Related to Troubleshooting a Current Divider Circuit for Dimmer Lights

1. What is a current divider?

A current divider is a basic circuit that divides a single current into two or more currents. It consists of resistors arranged in parallel, with each resistor allowing a portion of the total current to flow through it.

2. How does a current divider work?

A current divider works by utilizing the relationship between voltage and resistance in a parallel circuit. The total current flowing through the circuit is divided between the resistors based on their individual resistances, with the larger resistance receiving less current and the smaller resistance receiving more current.

3. What is the formula for calculating current divider assistance?

The formula for calculating current divider assistance is In = IT * (RT / Rn), where In is the current through a specific resistor, IT is the total current, RT is the total resistance, and Rn is the resistance of the specific resistor. This formula can be used to determine the current flowing through each resistor in a current divider circuit.

4. How is current divider assistance useful in circuit design?

Current divider assistance is useful in circuit design because it allows for the control and distribution of current flow in a circuit. By adjusting the resistances of the individual resistors in a current divider, the current can be divided and directed to different parts of the circuit as needed. This can help to ensure that each component in the circuit receives the appropriate amount of current.

5. Are there any limitations to current divider assistance?

While current divider assistance can be a useful tool in circuit design, there are some limitations to keep in mind. The total current in a current divider circuit cannot exceed the current supplied by the power source, and the total resistance of the circuit cannot be less than the smallest individual resistance. Additionally, current divider calculations assume ideal conditions and may not be accurate in real-world scenarios where factors such as temperature and component tolerances can affect the circuit.

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