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Topic: Weird problem when using a transistor (noob issue) (Read 251 times) previous topic - next topic

Dacha011

Nov 19, 2019, 08:34 pm Last Edit: Nov 19, 2019, 08:35 pm by Dacha011
Dear forum colleagues,

I am doing some basic tasks with and Arduino UNO and a NPN Tranistor - BC337-25 (a regular one, beta 160).

I am feeding the transistor base with current from my Arduino (+5V) and a 34K resistor.

There is a yellow LED in the collectors path. The collector current is limited with a resistor.

Here comes the problem :

If the collector current is provided from the Arduino +5v Rail - everything works perfectly fine.

As soon as I use some other current source for the collector (batteries or a breadboard power supply) , the transistor

becomes always ON ? :/ 

To me it looks like the the transistor base leaks when the current for the collector doesn't come from the Arduino +5v rail.

Can you help me with this problem ?

Thanks in advance <3


slipstick

As drawn that circuit can't possibly do anything because the battery -ve isn't connected anywhere. It must be connected to the Arduino GND.

Steve

Dacha011

Dear Slipstick,

You were right. My Arduino and my power source didn't have a common ground (instead, each had it's own GND).

Thank you for your fast reply !

You are the man !


Wawa

BC337-25 (a regular one, beta 160).
Beta is irrelevant for switching (lowest collector-emitter voltage).
If you use a transistor as switch, then you should provide a base current of 5-10% of the collector current.
Not relevant for a 20mA LED, but it is when you switch several hundred mA.
Leo..

Dacha011

#4
Nov 20, 2019, 11:56 am Last Edit: Nov 20, 2019, 02:01 pm by Dacha011
Hi Leo, thanks for the reply.

First time i read that Ib should be 5-10% of Ic. Interesting.

This is how I determine the how much Ib should be, in an common circuit where a transistor is acting as
a switch :

1) I choose the Ic. If i have a LED, then Ic should be 20mA. Than I do the following calculation
    First i find the resistor value
    Rc= (V supply - 1.83V (LED drop) - 0.3V) / 0.020A

2) Then I find the Ib
     Ic = Ib x Hfe(Beta)
     Ib = Ic / 160
     Ib = 0,000125 A or 125 µA
     Now the interesting part, i got this info in a bjt NPN transistor turorial
    Since I want the transistor to be in the saturated zone (I wrote active, my mistake), I multiply the Ib with a number between 4 and 10.
    Thats why  Ib = 4 * Ib
    And I supply the Transistor base with 500µA !

Do you thinks this calculation is right ?

Thanks and kind regards,


MarkT

When you use a (bipolar junction) transistor as a switch the device will either be in cutoff (off) or saturation (on).

Standard transistor action only applies to the active operating region where current is flowing but the emitter-collector voltage is several volts or more.  This is where the current gain figure applies, and the base-collector junction is reverse biased.

In full saturation the base current is about 5% to 15% of the collector current.  There is no normal transistor action in saturation because the base-collector junction is forward biased.  The amplification of current in saturation is an entropy effect due to the fact the emitter is doped about 10000 stronger than the collector in transistors, rather than due to charge carriers being injected into a reverse-biased BC junction and then being pulled across by the electric field.

To minimize power dissipation in the on-state you want full-saturation where the emitter-collector voltage is a fraction of a volt.   If the current is small and the voltage drop between emitter and collector can be greater you can use the active region of operation where beta is large, but if the load is high current this will cook the transistor.
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]

Dacha011

Hi MarkT,

Thanks for the lesson :)

Unfortunately I was not able to understand everything you wrote, just a portion.

But still, I made a screenshot, an I will read your reply once I have a greater understanding on how transistors work.

Thanks again !

Wawa

2) Then I find the Ib
     Ic = Ib x Hfe(Beta)
     Ib = Ic / 160
     Ib = 0,000125 A or 125 µA
No, Hfe only applies when there is some voltage left on the collector, say 4volt (transistor used as amplifier).
Hfe does not apply when the transistor is fully saturated (lowest voltage between CE).

If you want collector voltage to become lower than base voltage, then just pump 5% into the base.
Formula: 5% of 20mA is 1mA.

Collector voltage of a saturated transistor can become much lower (<0.2volt) than base voltage (0.65volt).
Important for a switch, because you don't want the transistor to get hot or 'steal' voltage from the load.

You can usually find switching specifications in a graph of the datasheet, with Ic:Ib ratios.
Leo..

Dacha011

#8
Nov 21, 2019, 08:28 pm Last Edit: Nov 21, 2019, 09:41 pm by Dacha011
Thanks Wawa, I will have this in mind next time I work with BJT transistors :)

For now I will just pump 5% of the Ic in the base when I'm using a BJT as a switch.

I understood the current flow of the transistor, but I have to read and learn more about the voltage drop that occurs in the BJT.

All i know is that the V drops approx. 0.7 when calculating the resistor value of Ib, and approx. 0.3 when calculating the resistor for Ic. But I am totally confused when we are talking about the relations between voltages on the C, B and E.

If you have a good tutorial at hand, I will be grateful if you share me the link.

:)

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