# Circuit problem: Confused about results

• Xyius
In summary, Dave's problem was that he did not have an emitter resistor in his circuit. Without an emitter resistor, the transistor would be on all the time and the collector LED would not be affected. He has two circuits, one with an emitter resistor and one without, so he can choose which one to use.
Xyius
I am just starting to really get interesting in making circuits and I was doing a few experiments. One of my circuits I wanted to make was using a potentiometer to control the brightness of an LED. I also wanted to incorporate a transistor. So I made the following circuit. (Bear with me, I drew it in paint. )

http://img196.imageshack.us/img196/8831/circuitproblem.png

My logic was that by varying the potentiometer, both the LED on the base side and the collector side would change in brightness. To my surprise the LED on the collector side seemed unphased by the changes in the potentiometer and stayed at the same brightness reguardless. If I am changing the current to the base, shouldn't the output current to the LED (From the collector) change as well?

Here is a video showing what is happening and explaining my problem.

Thanks! :D

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I assume what's happening is that your pot doesn't go high enough to get the transistor out of saturation so it's just flat on all the time. Try a pot with a bigger resistance value.

Problem is your bias method. What you need is an emitter resistor.

Say if you want max of 10mA through the LED and you have a 12V supply, you use a 1K emitter resistor from E of the NPN to ground. Get rid of the resistor from the +12V to the LED at the collector, you don't need that. Then put a potentiometer with one side to the +12V, the other side to the ground. The wiper of the potentiometer to the base of the NPN, it will work.

The theory is the emitter resistor limit the current and it is called the emitter degeneration resistor. The potentiometer set up a voltage divider from 0 to 12V. Whatever voltage presented at the B, the E will follow with 0.7 drop from Vb. The voltage across the emitter resistor set up the current through the LED.

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OK, you haven't said what your battery voltage is. You haven't shown the negative side of your battery connected to anything. It should be connected to the Earth symbol on the emitter of the transistor. Nor have you stated the values of the fixed resistor in the collector circuit or the variable resistor in the base cct

Initially get rid of the LED in the base circuit...it will screw things up
If you are using a 9 to 12 Volt battery then the fixed resistor should be around 1k Ohms to ensure that the LED supply is correctly current limited, else you will burn out the LED in a brief burst of light.
You should have a minimum of ~ 470 Ohms resistor in series with the pot, so that when the pot is at minimum resistance the base of the transistor isn't getting "unlimited" current which will also cause its early demise.

Sort those things out and get back to us with some results... Assuming you haven't killed the LED in the collector cct or the transistorCheers
Dave

Keep in mind that the transistor will have a current gain of at least 100 so you may not even see the small current it takes to turn the transistor on in your base LED.

You could try modifying your circuit like this:
http://dl.dropbox.com/u/4222062/LED%20control.PNG

This will let you turn your collector LED on and off with the pot and protect the transistor and both LEDs from destruction.

The pot value is not very critical, but it should have at least 3 volts across it to give proper control. So anything above 5 K would be OK.

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I still think he should do the voltage divider bias and emitter resistor. Unless you get the base resistor just right, the transistor is just going to be on or off. And even he can get it just right, temperature drift of β is going to make the circuit very unstable.

And also, it is very important for beginners to get the concept that transistor is a current device and everything is control by the current of the emitter resistor unless you have matching current mirror.

He just wants to turn a light on and off.

Temperature drift and current mirrors are well in the future.

I left the input LED there so that the resistance at the bottom of the pot would not be enough to turn the transistor on. It will take the LED voltage plus Vbe to turn the transistor on and this should be well into the pot's normal range.

Anyway, he has both circuits, so he can choose.

Thanks a lot everyone! It is a little late at the moment but I am going to experiment with this more tomorrow. I will definitely let you know how it works out. :D

As of right now I employed the circuit presented by vk6kro and it worked fine. I did not have a ground after the potentiometer. I am still unsure as to exactly why that needs to be there. I realize putting it there allows me to apply a different voltage to the base, but I thought transistors were all about amplifying current. If the potentiometer went straight to the base without being grounded, why wouldn't it work? (The case I had originally)

But as yungman said, the LED behaves more like a switch of on or off. (Meaning it becomes bright extremely quickly.) The original intention was to be able to control the brightness with the potentiometer. I AM able to do this but as I said it becomes bright very quickly and there is hardly any control.

I will try what yungman said and see how that works. :]

If you want the LED to come on slowly, you can put a resistor in series with the base.

I have added it to the previous diagram to show how it is connected.

The value of 100 K I gave it will depend on the transistor so be ready to change it to get the effect you need.The reason the other pot arrangement didn't work was that the resistance wasn't high enough.
To make the LED turn on would require a collector current of 20 mA but a base current of (20 mA divided by the current gain of the transistor), say 200. So the base current would be 100 uA, and you would need a resistor of about 70 K.

Now, you would still be able to see the LED alight at 5 mA but that would require a resistor of about 280 K.

Good LEDs can still be seen at 1 mA which would require a resistor of 1.4 Megohms.

See why your pot used as a rheostat couldn't turn the light out?

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So I have been working on this more.

I wanted to try and design the circuit on my own, I took the advice here and eventually made a circuit that has the exact effect I am looking for.

While I tried to take as much advice as I could from this post, I am afraid I did not understand some of it :(

Below is a picture of my circuit (I am using a PNP transistor this time) that now works exactly how I want it to. I have some questions about it and I was wondering if anyone could answer them.

http://img3.imageshack.us/img3/2386/circuitproblem2.png​

1.) So this is a silicon transistor with a base conducting voltage of around 0.6 volts. Now, the LED is always turned on except for when the potentiometer gets close to 500K. (It dims gradually) My question is, how can I calculate the resistance needed from the potentiometer to get exactly 0.6 volts?

My first reaction was to do this..$$V=I(R+P)$$
Where P is the value of the potentiometer. But this has a variable current and resistor. I played around with different values of P until I got a P value of 400K ohms needed. This SEEMS to be consistent with experiment but I do not know if this is valid or not.

2.) What role does the 100K resistor play? Does it play any role at all? (EDIT: I believe I know this question, the 100K resistor allows the voltage divider to work. Without it, all the voltage would drop over the 100 ohm resistor on the base.)

Thanks for all your help :]

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Assuming the negative side of the battery is grounded, then the circuit would work.
I doubt if the 100 ohm resistor in the base is doing anything though.

There was a nice post in another thread that explained voltage dividers:

https://www.physicsforums.com/showpost.php?p=3832641&postcount=4

Could I suggest that you leave PNP transistors for later and go back to the previous diagram?
What didn't you understand about the previous discussion?

## 1. What is a circuit problem?

A circuit problem refers to an issue or complication that arises when trying to design, analyze, or troubleshoot an electrical circuit. It can involve unexpected results, incorrect measurements, or difficulty understanding the behavior of the circuit.

## 2. What are some common causes of circuit problems?

There are several potential causes of circuit problems, including faulty components, incorrect wiring or connections, inadequate power supply, and errors in calculations or circuit design. Environmental factors such as temperature and humidity can also affect circuit performance.

## 3. How can I troubleshoot a circuit problem?

The first step in troubleshooting a circuit problem is to carefully review the circuit diagram and check for any obvious errors or mistakes. Next, you can use a multimeter to measure voltage, current, and resistance at various points in the circuit and compare them to the expected values. You can also try replacing components one at a time to identify any faulty parts.

## 4. How can I prevent circuit problems?

To prevent circuit problems, it is important to carefully plan and design the circuit before building it. Double-check all connections and calculations before powering on the circuit. It is also helpful to use high-quality components and to regularly check for any signs of wear or damage. Following good electrical safety practices can also prevent potential circuit problems.

## 5. When should I seek help for a circuit problem?

If you have exhausted all troubleshooting methods and are still unable to resolve the circuit problem, it may be necessary to seek help from a more experienced engineer or technician. It is also important to seek help if you are unsure about the safety of a circuit or if you are dealing with high-voltage circuits.

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