I think my fundamentals in circuits is pretty poor so I have to ask some questions over Ohm's law

I'll use a BJT (npn) for my question
If a voltage source is connected to the base of a BJT, emitter connected to ground and collector connected to another voltage source.

If there are no resistors is there no current flow?

Conversely I also ask myself if there was no resistor and Vin was lets say 0.5V it would still be 0.5 V whether or not the resistor was there.

I seem to have alot of the same problems this person was having
http://www.physicsforums.com/showthread.php?t=470212&page=2

 

If a voltage source is connected to the base of a BJT, emitter connected to ground and collector connected to another voltage source.

Assuming that your voltage source on the base is big enough to turn your transistor on, you'll have a voltage drop from base to emitter, which is about 0,6V, the pn-junction potential barrier voltage, and you'll have a voltage drop from collector to emitter, saturation voltage of your transistor.
There will be current, almost only limited by the internal resistance of your voltage sources. (or are you talking about an ideal voltage source with zero internal resistance?)

 

You need resistance in order to restrict the flow of current. When the resistance is zero, you will have maximum current flowing, that is, your power source will attempt to pump as much current as it can supply, usually only limited by its own internal resistance (which is non-zero).

So when you connect your power source across the base-emitter junction of your transistor you will fry your transistor, i.e. say bye-bye.

The same thing happens when you connect a power source greater that 2V directly to an LED. That is why you must have a current-limiting resistor in series with the power source.

 

OK I get all of that so far

If I have a 10k resistor at the base and I put in a voltage into the base

What would be the voltage after the 10k resistor?
Is it Vin - Vbe

also is this Vsat the same thing as Vbe and turn on voltage?

 

There were a couple of recent threads that I responded to that you may want to check up on. One on "voltage drop" and the second on "why 0.7V drop across a Si pn junction".

But I will try and answer your questions here. The voltage across a forward biased pn junction (e.g. Vbe) will increase until it becomes saturated at about 0.7V.

Let me put in differently, the current will increase exponentially as the voltage applied is increased from zero. When the voltage reaches 0.7V the current will sky rocket and only your series resistance will prevent damage to the junction.

So what is the voltage at the resistor-base junction? The voltage drop across the base-emitter will be 0.7V. If the emitter is at 0V, the base will be at 0.7V, i.e. Vbe, not (Vin - Vbe).

(The circuit in the link you provided in your original post is an emitter follower circuit - the emitter is NOT at 0V.)

 

Thanks for the help so far

I'll switch over to the thread you recommended but let me ask one last question

on wikipedia im reading that for
cutoff: Vb<Ve
saturation:vb>vc
active: vc>vb

but on
http://web.engr.oregonstate.edu/~traylor/ece112/lectures/bjt_reg_of_op.pdf

They use Vbe>0.7
and Vce>Vsat as the thresholds for region of operation

They can't be the same because Vbe>0 and Vce>0 would be the equivalent statement

 

There were a couple of recent threads that I responded to that you may want to check up on. One on "voltage drop" and the second on "why 0.7V drop across a Si pn junction".

But I will try and answer your questions here. The voltage across a forward biased pn junction (e.g. Vbe) will increase until it becomes saturated at about 0.7V.

Let me put in differently, the current will increase exponentially as the voltage applied is increased from zero. When the voltage reaches 0.7V the current will sky rocket and only your series resistance will prevent damage to the junction.

So what is the voltage at the resistor-base junction? The voltage drop across the base-emitter will be 0.7V. If the emitter is at 0V, the base will be at 0.7V, i.e. Vbe, not (Vin - Vbe).

(The circuit in the link you provided in your original post is an emitter follower circuit - the emitter is NOT at 0V.)

Quick side note, there are lot of transistor numberw for what is basically the same kind of device. Personally most of my transistor projects use very low currents so I use 0.6V when calculating transistor BE (Base Emitter) drop. As you increase the base emitter current this voltage goes up, and) it is pretty logarithmic. A 2N3055 (a TO3 case style NPN power trasistor) can drop 1.0V BE voltage before you hit the region that will damage the transistor, but then the current is approaching or exceeding 1A. This transistor is meant to be a driver, which translates into high wattages and low gain.

The general rules you stated are valid, but the exceptions are killers.

Pretty much all solid state electronics needs resistors to control their current. One of the common beginner mistakes is trying to use LED without resistors as if they were light bulbs.

They can't be the same because Vbe>0 and Vce>0 would be the equivalent statement

BJTs have some really nice characteristics. When they are on the CE (Collector Emitter) drop is well under 0.1 volts (I generally just call it 0V, but there are exceptions) even though the BE is dropping 0.7V. They also make really nice constant current sources for many applications with a minimum of parts. Many people make the mistake of linking the BE drop with the CE drop, they are independent. There are math models where a BJT are voltage controlled, but the better math models use current to control these devices. The math is much simplier, and is accurate.

 

There is no contradiction between the two references. I think you are reading too much into what they are saying.

Firstly, here are the two extreme cases when the transistor is operating as a switch. It is either OFF or ON.

OFF - when Vbe is below 0.6V. If Vb<Ve this condition is satisfied. The base current is too low to turn on the transistor.

ON - when Vbe>0.7 . If Vb>Vc the same thing will occur. The transistor is fully ON and acts almost like a short circuit between the collector and emitter. Make sure you have a current-limiting resistor at the collector source otherwise once again you will blow the transistor.

The third case is where we find a sweet spot between the above two extremes and this becomes a linear amplifier. The base current will be amplified by the current gain beta. The emitter current = base current x (1 + beta).

Yeh, Bill is right, it's more like 0.6V for a small signal BJT.

 

No, I don't think you are confused. What you say is correct. Vbe = Vb - Ve.
But we are not discussing Vb>Ve. The condition was Vb<Ve, i.e. there is essentially ( I was going to say basically) zero base current.


Here is the thread on "voltage drop"

http://forum.allaboutcircuits.com/showthread.php?t=56372

And the thread on "pn junction voltage"

http://forum.allaboutcircuits.com/showthread.php?t=57044

No more from me for now. I gotta go to bed.

 

K thanks a lot for the help

 

So what is the voltage at the resistor-base junction? The voltage drop across the base-emitter will be 0.7V. If the emitter is at 0V, the base will be at 0.7V, i.e. Vbe, not (Vin - Vbe).

(The circuit in the link you provided in your original post is an emitter follower circuit - the emitter is NOT at 0V.)

So if I want to turn on the transistor where I need base to be higher than emitter I have to make sure than Vb>0.7 (in this example). How am I suppose to raise the voltage at the base when My Vin didn't change the base voltage

 

So the base is at 0.7V regardless of my Vin? Then how would I increase the base to ever turn it on.

No. If Vin is less than 0.6V the transistor is turned off. The p-n junction of the base-emitter is a highly non-linear device. As you increase Vin above 0.6V the base-emitter current will begin to increase but will be limited by the series resistance from your source to the base. You will measure only a marginal change in Vb but the base current will increase dramatically because of the high non-linearity. (the p-n junction is not a resistor).

The transistor will then start to conduct. In the linear region:

Ic = beta x Ib
Ie = Ib + Ic

The base-emitter p-n junction will act like a diode clamp. Vbe = Vb - Ve will not rise much beyond 0.7V (don't forget that series resistance is going to limit the current and save the p-n junction from self destruction.)

Another thing to remember: a BJT (bipolar junction transistor) amplifier is a current amplifier, not a voltage amplifier.

 

I think I'm beginning to understand what my problem is

I'm confusing voltage at a point with voltage drop.

When you said Vb is 0.7V that's true because that's what the "barrier" voltage is and my Vin can't change that.

To turn it on I have to have Vb>Ve and if I put a voltmeter at base to emitter it reads 0.
But when I put Lets say 1V in at Vin then I get 0.3V reading from base to emitter which is greater than 0 and it turns on.

I tried to imagine a voltage source in series with a resistor that was in series with another voltage source.

I think have been trying to change voltages with voltages
Not sure if all of what I have said is correct but I feel like I'm honing in on my problem

 

Are you actually testing this out on a real transistor? If so, what is the part number on your transistor? Germanium (Ge) transistors will turn on with Vbe > 0.3V while Silicon (Si) transistors need Vbe > 0.6V

Think of the base-emitter junction of the npn Si transistor as in the circuit above. You can increase the input source V1 to much above 3V. The "voltage drop" across the diode will not rise much more than 0.7V

 

If we had a transistor now with a Vin at base with no base side resistor
and only a resistor on emitter and we put in lets say 1V and the Vbe_on is 0.7V
and I took a voltmeter from B to ground it would be 0.3 correct?
and at the top of the resistor it would also be 0.3?

 

Are you actually testing this out on a real transistor? If so, what is the part number on your transistor? Germanium (Ge) transistors will turn on with Vbe > 0.3V while Silicon (Si) transistors need Vbe > 0.6V

Oh I'm just thinking about an example to just help me understand BJTs I guess we can talk about it generally as Vsat and Vbe_on

EDIT: Yup I am confused again...I don't know why but I just have trouble when I see voltages. Like Vb > Ve to be on

and if Vb is 0.7V and Ve is 0V its always on (I know this is wrong but I dont know why)
And I for some reason expect my Vin to change the base voltage because the voltage at Vin "drops" and thus Vb is another voltage.

All of those things I said are wrong, but like I said I just don't know why. I feel like I don't understand some aspect of ohm's law like I said in my first post. Voltages just generally confuse me...

 

If we had a transistor now with a Vin at base with no base side resistor
and only a resistor on emitter and we put in lets say 1V and the Vbe_on is 0.7V
and I took a voltmeter from B to ground it would be 0.3 correct?
and at the top of the resistor it would also be 0.3?

No. I don't know where you are coming up with 0.3V. Can you give me some background with this so I know where you are coming from?

Let us use the circuit above for our discussion. This is quite different from what we have been discussing before. The base-emitter voltage drop Vbe is determined by the current flowing through the base. We will assume for the sake of this discussion that the transistor is turned on and Vbe = 0.6V

We still do not know the current flowing through the emitter resistor R1 because this is the emitter current:

Ie = Ib + Ic = Ib + (beta x Ib) = Ib x (1 + beta)

Hence at this stage we do not know the voltage drop across R1. To find out we would have to do the complete circuit analysis with the rest of the circuit drawn in. That is why I pointed out earlier that the first circuit that was troubling you in the OP was called an "emitter follower" circuit.

I'll end here for today.

 

Some comments that might help clarify things.

If you are going to use the expression Vb > Ve to turn on, to be more precise we would have to rewrite it as (Vb - Ve) > 0.6

Secondly a p-n junction is highly non-linear. We cannot apply Ohms Law because the effective resistance keeps changing as current changes. That is why the Vbe voltage stays around 0.6 to 0.7 volts when the transistor is turned on.

Finally, all voltage measurements are "voltage drops". Voltage is ALWAYS the voltage of Point A with respect to Point B.
When we say "voltage drop" we really mean the "voltage difference" or "voltage across" or "voltage with respect to...".

 

You are mentally connecting the base emitter to the collector emitter. Don't. The relationship is current, not voltage, and you would be better thinking of a transistor as a specialized relay than as two diodes.

The AAC book goes through this, I did when I was a kid (it seemed so logical). Transistors have quantum effects that isolate voltage drops between BE and CE. CE can go down to almost 0 volts, the BE never can (and still be conducting). If the BE is 0V the transistor is off.

http://www.allaboutcircuits.com/vol_3/chpt_4/3.html

 

Actually I read through the thread myself and learned quite a bit. Dah!