Interesting comments. Does it apply to the discussion?

Yes, it applies to the discussion because the Thread Starter is measuring the battery voltage at the transistor collector and can't seem to figure out where it's coming from. (It's coming from the battery.) For a real world circuit, see post #8

 

ok then it's just me being too dumb to understand how transistors work, I think I need to go to amazon and get a book on transistor theory.

thanks guys.

Neil.

No need to go to amazon, look at the education section, Vol111 semiconductors

ok then it's just me being too dumb to understand how transistors work, I think I need to go to amazon and get a book on transistor theory.

thanks guys.

Neil.

 

Yes, it applies to the discussion because the Thread Starter is measuring the battery voltage at the transistor collector and can't seem to figure out where it's coming from. (It's coming from the battery.) For a real world circuit, see post #8

Sorry, I had to read your answer twice before it made sense to me.

 

ok I just been staring blankley at this circuit for a while and I can't get my head around why when there is no current flow as in when the transistor is switched off why there would be a voltage across the transistor since no current flow means there cannot be a voltage present?

can someone give me a blow by blow account of what is going on here please?

sorry for being so dumb but this is very interesting to me and I want to understand.

That's kind of why I said measure from ground (Emitter). At least to me it makes more sense that way.
When the transistor is turned on all the way the collector and emitter are the same voltage (or real close) and since the base current is high the voltage drop from base to emitter is a little higher.
As the base current goes down the transistor starts to turn off and the voltage on the collector will rise.

 

I think what you are missing is the fact that this is an inverter. OFF means high voltage on the collector, ON means low voltage at the collector.

Bob

 

I think what you are missing is the fact that this is an inverter. OFF means high voltage on the collector, ON means low voltage at the collector.

Bob

Not if you measure from ground?

 

Yes I am measuring from ground and when the transistor is fully on I am getting high voltage on the collector which falls as I turn the transistor off via raising the base resistor value.

I am a relative noobe to this electronics lark not having dabbled in it since I was a teenager some 35 years ago and I am finding it very exciting and enthralling although also very confusing lmao

thankyou all for the many helpful replies....a fantastic forum.

Neil.

 

If the transistor is off (the term is cutoff), the voltage across the collector-emitter would be the supply voltage (which I didn't see mentioned anywhere). When you increase base current until the transistor is starting to conduct, that's called active mode. When you increase base current until increases no longer cause corresponding increases in collector current, the transistor is in saturation mode.

In active mode, the CB junction is reverse biased and the BE junction is forward biased. When the transistor is saturated, both junctions will be forward biased.

The fourth mode is called inverted, that's when the junctions are biased the opposite of active mode.

Supply voltage mentioned in post # 8 ;-)

 

Yes I am measuring from ground and when the transistor is fully on I am getting high voltage on the collector which falls as I turn the transistor off via raising the base resistor value.

I am a relative noobe to this electronics lark not having dabbled in it since I was a teenager some 35 years ago and I am finding it very exciting and enthralling although also very confusing lmao

thankyou all for the many helpful replies....a fantastic forum.

Neil.

QUOTE="Neil Groves 1, post: 999323, member: 344470"]Yes I am measuring from ground and when the transistor is fully on I am getting high voltage on the collector which falls as I turn the transistor off via raising the base resistor value.

Yep, that sounds backwards.

 

oh wait....lmao...

wait wait LMAO...I AM SO F****** STUPID....

I was measuring my collector voltage with respect to the positive rail....DUH!!!

now I have to do my experiment all over from scratch.

so sorry for wasting your time guys.

Neil.

 

oh wait....lmao...

wait wait LMAO...I AM SO F****** STUPID....

I was measuring my collector voltage with respect to the positive rail....DUH!!!

now I have to do my experiment all over from scratch.

so sorry for wasting your time guys.

Neil.

No, your chart is right. It lists the voltage across the collector resistor. When the transistor is in saturation VCC is dropped across the collector resistor and the transistor is all the way on. As the base resistor increases the transistor decreases conduction and voltage across the collector resistor decreases.
Your exercise is good.

Things to note:
How much base current causes how much collector current? (Current gain at a given collector current.) Change the 10K to 1K and compare the results.
Repeat with a different part number of transistor. Compare the results with a data sheet for the parts.

PNP transistors.. Emitter to +9 V. Base and collector resistors go to ground. (Nobody uses a negative power rail any more.)

Darlington NPN and PNP
N-MOSFET (Enhancement mode) transistors
P-MOSFET (Enhancement mode) transistors
SCR
Triac
J-FETs (not many people use them any more)

Then on to op amps and stuff.

 

ok one more question before moving on.....

I would have thought at some point, the VBE would start heading towards zero volts as the transistor shuts off, I was under the impression that the 0.7 volts is only present when the transistor is fully conducting (saturated)?

also It appears I cannot fully turn off the transistor in this simple circuit since my voltage across Rc will never reach zero and seems to have virtually levelled out at around 4v?

it needs to be zero volts when the transistor is fully shut off right?

 

Are you measuring the voltage across the base resistor and across the collector resistor at the same time (using two meters), or are you measuring one with a meter and then moving the meter over to measure the other?

What kind of meter are you using?

Try this -- remove the base resistor and measure the voltage across the collector resistor.

Then try this (if you have two meters) -- measure the voltage across the non-existent base resistor (i.e., measure the voltage between the two points where the base resistor was previously connected) and then measure the voltage across the collector resistor at the same time.

Report those results and then we will have quite a bit more to discuss.

 

I don't see how the voltage across your collector resistor could ever have been 9V (if the battery voltage was 9V) since you have to drop some voltage (usually two to three hundred millivolts) across the collector-emitter junctions of the transistor.

Other than that, with the 820 kΩ resistor your numbers yield a transistor beta of about 44, which seems reasonable (without knowing the exact transistor you are using).

 

You are measuring the voltage across the power supply, not the resistor.

To measure the voltage drop across a resistor, you must reverse the meter leads.

 

I would have thought at some point, the VBE would start heading towards zero volts as the transistor shuts off, I was under the impression that the 0.7 volts is only present when the transistor is fully conducting (saturated)?

It should, but for a typical silicon BJT it falls very slowly. At room temperature the Vbe drops by about 60 mV for every decade change in collector current. Your peak collector current was about 0.9 mA and your minimum was about 0.44 mA, or a change by a factor of 2. You typically expect to see a change in the Vbe of only about 20 mV for that about of change. Your base-emitter voltage changed by about 80 mV, though I think there is something hinky about that first column (and the second column, too). If we use the third column and the last column, then your collector current went from 733 uA to 444 uA with a change in Vbe of about 10 mV. You would expect the change in Vbe to be about 13 mV for that amount of collector current chance, which is well within the error bounds of your results.

 

You are measuring the voltage across the power supply, not the resistor.

To measure the voltage across a resistor, you must reverse meter leads.

Reversing the leads only changes the sign of the reading. It does not affect what the measurement is across.

He may not be measuring the voltage across the (collector) resistor, but he clearly isn't measuring it across the supply. He is measuring the highest voltage with the smallest base resistor and the smallest voltage with the largest base resistor. Since the collector resistor is 10 kΩ, even if it were placed directly across the 9 V battery the draw would only be less than 1 mA, which is hardly enough to drag down the terminal voltage very much (unless the battery is well and truly on its last legs).

 

Pardon me. I thought his last puzzlement was the voltage drop across the collector resistor when the transistor was off.

Red probe on bottom of resistor and black probe on top. Any voltage measured there, means current thru it.

That was my only point.

 

Usually measurements are made with respect to ground. Having them taken from the positive of the battery makes me want to stand on my head or do math in it - neither one is a good idea for me.
So I took the liberty of redoing your chart:

If you look at the transistor and think of the base emitter junction as a diode it will always be on if the base is more positive than the emitter by about a diode drop. So for the transistor to be off the base emitter voltage needs to be less than about .5 volts.

 

But why am I getting such a large change in the voltage across the collector resistor as I increase the base resistor value when nothing else is changing.....my VBE hasn't changed noticeably neither has the voltage across the base resistor.