Try to built a simple transistor circut

[tefavolc]

i was try to built a simple transistor circut , i chose Ic=0.5 mA , i used s8050 transistor , by looking at its datasheet i found that i ve to make Vbe=0.56 V and Vce= 1 V in order to obtain Ic=0.5mA.i used power supply with output 12 volt and max current 9 A . by using 2 plots 10k and 200k i could establish the 0.56 and 1 volts for biasing the transistor .and when i connect the transistor to the voltage terminals guess what happened ? the transistor blow up ,, it toke my an hour to finish the calculation, the datasheet and the circut digram is attached pleasezzzzzzzzzzzz anyone tell my what's wrong?

 

[Bob S]

Try building the circuit in LTSpice circuit modeling application, available free for downloading at www.Linear.com;[/URL] type LTSpice in search box.

Bob S

 

[tefavolc]

bob s thnx for replaying but i really want to know that's wrong wiz my connection

 

[Bob S]

Insert a fixed emitter resistor. See thumbnail LTSpice simulation.

Bob S

 

[cabraham]

When you said that you "made Vbe 0.56V", how did you do that. Did you connect a power supply directly across the base-emitter junction. If so, that is why it blew up. In the forward direction, p-n junctions, such as bjt b-e junctions, LEDs, etc. must be current driven, never voltage driven. Have you had any formal education in electronics? If not, it would sure help. Please do not play with dangerous power, such as the ac mains.

Claude

 

[vk6kro]

What probably happened is that while you were adjusting the pots, the wipers were moved towards the + end, resulting in too much base current.

Try to work in terms of base current rather than base voltage. It is difficult to control base voltage to the necessary precision, but controlling the base current is fairly easy.

If you connect a resistor R from a + 10 V supply to the base and then have the emitter return to the - terminal of the supply, the base current is just the (supply voltage (10 V) minus 0.6 volts) / R

Suppose you want a collector current of 10 mA. The gain is about 100, so the base current is 100µA. So, the resistor reqired would be 10 volts - 0.6 volts / 0.0001 amps or 94 K. So, you could use a 100 K base resistor. You would add this as a fixed resistor, not a pot.

Now, it would be difficult for you to blow up the transistor.

Now, you need to use this as an amplifier.
There will be 10 mA flowing in the collector circuit and you want about half of the supply voltage across the transistor and the other half across a load resistor.
So, the size of the load resistor is 5 volts / 0.01 amps or 500 ohms.

Again you would add this as a fixed 470 ohm collector resistor, not a pot.

[PLAIN]http://dl.dropbox.com/u/4222062/NPN%20amp.PNG

 

[tefavolc]

thnx vk6kro really apprieciat your efforts , by the way I am studying electrical engineering but iam still a beginer

 

[skeptic2]

On the data sheet look at hFE. hFE is the current gain of the transistor. It gives a minimum value and a maximum value but no typical value. This is because the current gain of individual transistors varies widely. If by adjusting the pots you get a circuit to work, and then put a different transistor in the circuit, it probably won't work. In order for a circuit to work with a wide variety of current gains you must have something to stabilize the current gain. This is usually done with an emitter resistor.

As you adjust the pot connected to the base, you probably see the collector voltage switch from off to on very suddenly. This because the voltage change on the base to turn the transistor from off to on is only a few tens of millivolts.

You don't indicate the value of your voltage source nor the value of R3 but if the amount of resistance in the collector circuit is low enough, as soon as the transistor turns on it could heat up and destroy itself. To prevent that from happening and make the circuit work with other transistors, negative feedback is used in the form of an emitter resistor.

An emitter resistor is typically between 100 and 1000 ohms and inserted between the emitter and ground. The voltage across the emitter resistor is about 0.6 V less than the base voltage but not lower than zero.

As you turn up the pot on the base and base current begins to flow, some value of collector current that is a multiple of the base current by a factor between the minimum and maximum values given for the transistor. Both the base current and the collector current flow through the emitter resistor. If the gain of the transistor is high, more current will flow into the emitter resistor and tend to bias the transistor off until equilibrium is reached. If the gain of the transistor is low, less collector current will flow, less voltage will be developed across the emitter resistor and the transistor will be biased on more resulting in more collector current until equilibrium is reached.

The use of emitter resistors does reduce gain but that is not really a disadvantage. You are really trading gain for stability and gain also becomes much easier to calculate. It is approximately equal to the collector resistance divided by the emitter resistance.

Try it again with an emitter resistor and report back.

 

[tefavolc]

skepitic u said that by putting emitter resistor we can establish stable gain , would you show me that by equations ? and i ve another question ,, i know that current gain hfe is dependent on Ic, then as iam going to built a circuit i ve to chose Ic and look at the graph for Ic vs hfe and find the corresponding hfe , then i use hfe and Ic to get Ib,then i calculate the collector and the base resistance which will alow the currents to pass with Vbe=0.7 and Vce>0.3,, is it right ?

 

[vk6kro]

skepitic u said that by putting emitter resistor we can establish stable gain , would you show me that by equations ? and i ve another question ,, i know that current gain hfe is dependent on Ic, then as iam going to built a circuit i ve to chose Ic and look at the graph for Ic vs hfe and find the corresponding hfe , then i use hfe and Ic to get Ib,then i calculate the collector and the base resistance which will alow the currents to pass with Vbe=0.7 and Vce>0.3,, is it right ?

skepitic u said that by putting emitter resistor we can establish stable gain , would you show me that by equations ?

You can estimate the gain by the equation Vgain = RC / RE within limits.

i know that current gain hfe is dependent on Ic

Current gain is a property of the transistor but it does vary over a small range with collector current.

You choose the Ic you want then divide by the current gain (about 100 for your transistor) to get the base current.

That gives you a base current and almost any base current will give a base-emitter voltage of about 0.6 volts. This varies with temperature and to some extent with the base current. However, the base current will not vary much if the base resistor is fed from a much higher voltage than Vbe. After that, you don't need to worry about Vbe.

then i calculate the collector and the base resistance which will alow the currents to pass with Vbe=0.7 and Vce>0.3,, is it right ?

Not the Vce part. Can you see why you want Vce to be about half the supply voltage?

 

[tefavolc]

thnx again vk6kro for making thing clearer ,

Not the Vce part. Can you see why you want Vce to be about half the supply voltage?

i want Vce to be greater than 0.3 coz i want to operate the transistor in the active mode not the saturation one.

 

[vk6kro]

Yes, that is right, but if you want the output to follow an input signal that swings positive and negative, you need to have the output positioned in the middle of the possible output range.

This allows maximum swing in each direction.

 

[Bob S]

tefavolc-
Please very carefully look at the npn circuit simulation in my post #4.

1) The base is nominally set at ~1/3 of Vcc by two resistors; a 10 k and a 20 k.
2) The 500 ohm emitter resistor determines the emitter and collector current.
3) The 500-ohm collector resistor cleverly sets the collector voltage ~half way between the emitter voltage and Vcc. This operates the transistor in the middle of the linear region.

Write down these equations, including both the Vbe voltage drop and the base current.

Bob S

 

[tefavolc]

Write down these equations, including both the Vbe voltage drop and the base current.

Bob S

Here u are