BJT transistor DC analysis

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komay

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I want to calculate the (IE, IB, IC, VBE, VCE) for the following transistor can any one help me
i have β=100
 

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I can see you have a +10 & a -10 volt supplies. I can not see what voltage the Base is connected to.
 
Since the base of the transistor is not used then the transistor is turned off and the currents are zero.
 
using the drawing as shown:
Ie=0
Ib=0
Ic=0
Vbe=0
Vce=-20V
was this a "bonus" question on a test? i've seen quite a few in my time. i've used some of my own when asked to make a tech test for companies i've worked for. always good to see if the person is paying attention. such as putting a 7400 gate in a drawing with a 5.0V supply, and a 470 ohm pullup to 50V on one of the inputs, and the question " what does this circuit do?" answer i'm looking for: "smokes", "burns", or anything to that effect (it's a WAS gate).
 
Ut is a homwo
It is a homewor and this transistor is just a one question of it
I have solved all the other 20 question and thes wase the only one left
Thank you verey much for the help
 
Hi,

I dont know what level of detail you are looking for here, but in some problems when the base is open then the collector base current becomes a deciding factor on what the transistor is doing. The collector base leakage is always present, but you would have to look at the data sheet and temperature to determine if the current is turning on the transistor a little. To compensate for this, most transistor circuits have a resistor from emitter to base so that this leakage current is swamped out. With the resistor, the leakage current is not high enough to reach the turn on threshold voltage of the base emitter.

If the base is connected directly to +10v in this circuit, then all the leakage current is forced into the +10v supply line and the transistor would be considered completely 'off' in most cases.

This could be an exercise to show the difference between having an open base and a base that is connected so as to shunt all leakage current bypassing the emitter base.
Look up the data sheet, then decide what you think about the leakage current specification. If you dont have a part number, trying using a common PNP transistor part number.
 
Thank you for this valuable information. This circuit is part of my homework and I could not solve it, becuse they did not connect the base to any source of voltage. When I asked my teacher, he told me that the base is connected to the positive voltage.can you help my to solve the question
 
ronsimpson said:
Same answer(s) as not connected. because there is no base current in both options.
Click to expand...

Hi,

Sorry to disagree just a little but the collector base leakage current can turn the transistor slightly on. That's a known problem in some circuits so a base emitter resistor is used to swamp out any small current. It changes with temperature too. Check out a data sheet on one of the common transistors you know about.

And dont forget to have a Happy New Year
 
i dont need that much of detils in my study the transistor in this case is off and we ignor the verey small currents .and happy new for you too
 
Hello again,

Thanks for the 'like'

Just so you know what i was talking about, i looked up a common transistor the 2N2222 transistor.
The leakage current is 10ua max at max temperature, and with a Beta of 100, that would mean approximately 1ma could flow collector to emitter. In the circuit here that would mean that the upper resistor drops 2 volts meaning the collector voltage would be around 8 volts, and the emitter resistor would drop about 1 volt, meaning that at the emitter we would have about -9v. So that case would be different than being fully off.
That's with max temperature though so we might only see around half of that in most applications.
The specification to look for is often called "Collector Leakage Current".
The design goal is to keep the base emitter voltage less than 0.4v which is often referred to as the start of the turn on region.

I realize this does not matter for your particular case in this thread, but thought you might want to know for the future when you start making real designs for real companies.
 

It's a PNP transistor (The arrow always points to the 'N' material) If the base is connected to the positive rail, then the transistor is cut off.

So there's no current flow (apart from whatever leakage current there is; this differs from transistor to transistor)

Take a look at the spec sheet for a 2N3906.
 
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