# Need help in a small project please

Discussion in 'The Projects Forum' started by shubham123, Nov 11, 2012.

1. ### shubham123 Thread Starter New Member

Nov 11, 2012
5
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I have to make a project on 'Transistor As A Switch' and draw a graph between the input & output voltage showing the cut-off region,saturation & the active region.
I have attached the circuit diagram.The problem is that I dont know what values to take for the various components.

I tried making it with BC547 or BC 548 transistor,100k and 1k as Rb & Rc respectively,3volt for Vbb and Vcc(varied the voltage using a 10k potentiometer),Multimeter as Vi and Vo to measure the input and output voltage.Could not get any significant result,the votage kept varying a lot.

From what I understand,when Vbb will be less than 0.7V then multimeter Vo reading is equal to Vcc.As Vbb increases above 1V,Vo decreases.But as I said there was irregular voltage in the multimeter Vi.
Thanks a lot

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2. ### tubeguy Well-Known Member

Nov 3, 2012
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When using a transistor as a switch you would normally have a constant voltage on the load - Rc. So connect that directly to 3v not through a pot. The ratio of Rb to Rc is 100, should be less for more reliable switch operation. Make Rb 10k instead of 100k.

If still having troubles, double check your wiring. Also measure 3v directly with your meter to verify the meter and test leads are OK.

3. ### shubham123 Thread Starter New Member

Nov 11, 2012
5
1
The basic idea is to measure the input and output voltage and draw a graph marking the active region,stauration region & the cut off region.

I made the circuit again using the following components:
BC547 transistor
Vbb = 1.5volt cell (100k potentiometer for varying Vbb)
Vcc = 3volt cell
Rb = 680k+680k+120k resistors in parallel
Rc = 10k resistor

I got some reading at Vo & Vi.As I increased Vi,Vo decreased by a small amount(approx. 0.1volt decrease in Vo with every 0.2volt increase in Vi).I think this lies in the active region.
Now I want to know how do I increase the range of Vi so as to get 0.5volt as min. voltage(currently getting 0.9volt to 1.6volt using the above components)

I was given to understand that Vo would near zero volt as Vi becomes greater than 1volt
and Vo will be equal to Vcc when Vi is equal to or less than 0.6volt.
I have attached my reading of Vi & Vo.
Thanks a lot

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Nov 3, 2012
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5. ### tubeguy Well-Known Member

Nov 3, 2012
1,157
197
Here's a tip:

With the pot wired properly you can get 0-1.5v with 1.5v battery.

Pot has 3 terminals. Connect battery V+ to one end, V- to other end.
DO NOT connect either end of battery to center terminal of pot or you will
damage the pot or short out the battery as you rotate the pot.

Now experiment with rotating the pot as you measure voltages between the center and outside terminals.

Then reverse the battery connections to the outside terminals and measure again.

6. ### Audioguru New Member

Dec 20, 2007
9,411
896
If you wired the circuit like I show here then your transistor has its pins connected backwards or it is defective.

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7. ### takao21203 Distinguished Member

Apr 28, 2012
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A circuit like that indeed is highly useful as transistor tester.

I did not draw a schematic.

However, I added DIL switches, LEDs, and a jouletheif as power supply.

So I am able to figure out the pinout, PNP or NPN, and get an idea about the hFE. I can vary the base voltage, as well the base current!

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8. ### shubham123 Thread Starter New Member

Nov 11, 2012
5
1
Just got what I think is the correct readings.I had connected the potentiometer
& transistor wrongly.I have attached the new readings.Let me know if they are correct.Thanks a lot to everyone who answered.

Last edited: Nov 14, 2012
9. ### Audioguru New Member

Dec 20, 2007
9,411
896
Now the transistor is working correctly but the circuit is so simple that it is sensitive to temperature change.
Heat the transistor by squeezing it with your fingers while it shows an output of 1V to 2V and see the output voltage drop a little.