Trouble shooting basic transistor test circuit

Thread Starter

mahela007

Joined Jul 25, 2008
45
Hi.. I tried to build a basic circuit which uses a transistor as a switch to turn an LED on or off.
I tried several transistors thinking the first one was broken but I always get the same result..
The LED stays ON even when I disconnect the base. The supply voltage is +9 volts.

The emitter was connected to ground and collector to +9 volts. The base was connected to a voltage divider which produced 4.5 V (9/2 = 4.5).

What could the problem be?
 

Audioguru

Joined Dec 20, 2007
11,248
The emitter is connected to ground.
The collector is connected to +9V.
The base is connected to a voltage divider that produces +4.5V.
And the LED is not connected to the transistor.

Then the transistor is shorting the battery.
Maybe the LED is also shorting the battery.
There is no current-limiting resistor for the LED.
 

Thread Starter

mahela007

Joined Jul 25, 2008
45
Then the transistor is shorting the battery.
Maybe the LED is also shorting the battery.
There is no current-limiting resistor for the LED.
I forgot to mention that there was a resistor in series with the LED. BTW: I'm using a 'wall-wart' .. not a battery. (I'm certain that it produces 9V DC)

The schematic is very similar to this one
http://www.kpsec.freeuk.com/images/trancurr.gif

However, I do not have LED B. The 10k resistor is replaced with a 1K resistor and the LED (C) is connected to the emitter (with the proper polarity).
 

elec_mech

Joined Nov 12, 2008
1,500
Hi mahela007,

We need to know the part number(s) of the transistor(s) you are using. The schematic shows an NPN transistor, but you may be using a PNP transistor which will act differently. Also, the part number will allow us to better determine if your resistor values need to change.

Also, it's always a good idea to add a pull-down or pull-up resistor on the base depending on what you're doing. In this way, the transistor is forced into a known state when you aren't pushing the switch (in this example).

Based on this circuit, I'd suggest adding a 10kΩ resistor between base and ground so the transistor is forced off whenever the switch is not pressed (this assumes you are using a NPN transistor, not a PNP).
 

Thread Starter

mahela007

Joined Jul 25, 2008
45
One of the transistors I tried was "BC550" and the other was labelled C828, which I assumed to be 2SC828.
By the way, in the data sheets, in the diagram showing the pin orientation, are the pins facing towards or away from the reader ?
 

SgtWookie

Joined Jul 17, 2007
22,230
You need to post links to the datasheets that you are viewing, otherwise we may very well be looking at different datasheets.
 

flat5

Joined Nov 13, 2008
403
I think you have the transistor pins wrong.

A very clear picture would be nice.
What voltage do you see on the base to emitter legs when switch is closed?
 

Audioguru

Joined Dec 20, 2007
11,248
Usually the datasheet shows the pins from a bottom view.
Why don't you user a transistor that has a datasheet with a very clear photo of how the pins are arranged?
 

Attachments

Thread Starter

mahela007

Joined Jul 25, 2008
45
This transistor appears to have widely varying beta values.. ( DC current gain).
With reference to the link posted by SgtWookie, I think my transistor falls into the R category (there is a number beinning with R after the model number).
HOwever, the R category shots beta values from 180 to 360..
Which value should I use in biasing calculations?
 

SgtWookie

Joined Jul 17, 2007
22,230
You're not using the transistor in the linear region; you're using it as a saturated switch.

Use Ib=required Ic/10
For a common emitter circuit used as a saturated switch, to calculate the base current limiting resistor, use:
Rbase = (Vin - Vbe) / (Ic/10)
where:
Vin is the voltage supply to the end of the base resistor that is away from the base, referenced to the emitter.
Vbe is the voltage on the base referenced to the emitter when applying the required current. This is usually in a range of 0.63v for very low Ic to ~.8v when the collector is carrying about 1/2 of its' maximum rated current (if you need more than 1/2 the rated current, you should look for a more capable transistor)
Ic is the required collector current.

So, if your required collector current is 25mA, your voltage supply is 9v, and assuming 0.7v for Vbe:
Rbase = (9-0.7) / (25mA/10) = 8.3/.0025 = 3320 Ohms. 3.3k Ohms is the nearest standard value.
 

SgtWookie

Joined Jul 17, 2007
22,230
No.
In post #4 the OP said the LED is connected to the emitter and the base is biased at half the supply voltage.
Please re-read the OP's statements and links.
In the original post, they said:
The emitter was connected to ground and collector to +9 volts. The base was connected to a voltage divider which produced 4.5 V (9/2 = 4.5)
And in post #4, nothing was mentioned about the connections, but they linked to this schematic:



...which I see as a common emitter configuration where the transistor is used as a switch.

Since they have not posted a schematic of their actual circuit, I'll have to go by what they've posted already.
 

Audioguru

Joined Dec 20, 2007
11,248
And in post #4, nothing was mentioned about the connections
As said in post #4:
"The schematic is very similar to this one ....
.... the LED (C) is connected to the emitter"

I don't know why the OP does not post the exact schematic he used since now we don't know where the 470 ohm current-limiting resistor is connected.
 

SgtWookie

Joined Jul 17, 2007
22,230
OK, I missed that - thanks.

I wish people would provide schematics that actually matched what they have, rather than simply saying "It's like this, but I changed...."

Providing an accurate schematic takes care of just about all the questions.
 

Thread Starter

mahela007

Joined Jul 25, 2008
45
OK.. sorry about the confusion. The reason I didn't post a exact schematic was because I wasn't sure how to draw one.
Anyway, my main problem is solved. Thanks for the help.
 
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