# Transistor conduction

#### Neil Groves 1

Joined May 3, 2016
21
Not really homework although I guess it is as i'm doing this at home lol

I am playing with transistors building really simple circuits and taking measurements so as to get some idea of how tthey function in a circuit, I have a transistor which is fully turned on and I have the expected Base emitter at 0.67v, how can I turn off the transistor slowely whilst taking the voltage reading at this point so I can watch it fall as the transistor turns off please?

Neil.

#### #12

Joined Nov 30, 2010
18,210

#### wayneh

Joined Sep 9, 2010
16,399
Yeah, it's more useful to think of turning down the current in increments and letting the voltage fall where it likes. You will cover a wide range of current while seeing only a small voltage drop.

#### hp1729

Joined Nov 23, 2015
2,304
Not really homework although I guess it is as i'm doing this at home lol

I am playing with transistors building really simple circuits and taking measurements so as to get some idea of how tthey function in a circuit, I have a transistor which is fully turned on and I have the expected Base emitter at 0.67v, how can I turn off the transistor slowely whilst taking the voltage reading at this point so I can watch it fall as the transistor turns off please?

Neil.
Congrats on taking the time to learn the right way. A pot is good. Another method if you have a supply of resistors is to start with a base resistor of 10 M ohm, measuring base voltage, calculate base current, measure collector voltage, calculate collector current and note current gain. Then go to the next lower resistor and repeat until you get to saturation. Excel, or some other spreadsheet is very useful for this. Do it at a couple of different collector currents so you can see how things change.

A variable voltage works also if you have it available.

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#### #12

Joined Nov 30, 2010
18,210
Another method if you have a supply of resistors is to start with a base resistor of 10 M ohm,
May I remind you that a beginner might have difficulty measuring less than 1 microamp and he must consider the current lost through the meter when placing it in connection with a 10 meg resistor.

Simpler version: Watch out for calculation errors and measurement errors caused by the resistance of the meter you use.

#### HW-nut

Joined May 12, 2016
94
If you have access to a scope and a signal generator, you could set up a curve tracer.

#### #12

Joined Nov 30, 2010
18,210
If you have access to a scope and a signal generator, you could set up a curve tracer.
I think that's too crude of a method. There is no way you can see a 1% change by looking at a scope trace which is examining a range of 100 to 1 ratio, and my graph shows a million to 1 ratio.
That's why the true nerds (me and hp1729) did this with DC meters.

#### Neil Groves 1

Joined May 3, 2016
21
I don't get it!

here is the circuit I am playing with and my results don't make sense to me, as I increase the resistor going to the base, I am expecting the transistor to switch off being shown by the falling voltage across the collector resistor, this isn't happening, in fact little is happening at all except the voltage across the collector resistor falls as I increase the base resistor?.....my Vcc is 9v

#### #12

Joined Nov 30, 2010
18,210
as I increase the resistor going to the base, I am expecting the transistor to switch off being shown by the falling voltage across the collector resistor, this isn't happening,
Actually, that is exactly what is happening.

I would like to point out that using a 10 ohm resistor in the collector circuit looks like you are expecting the transistor to allow almost an amp of current. A square 9 volt battery doesn't have a chance of providing that much current, and you are probably using a small transistor that would melt if it did. You are all out of proportion trying to get an amp to flow when people who are good at this work in the range of a thousandth of that.

#### ronv

Joined Nov 12, 2008
3,770
Actually, that is exactly what is happening.

I would like to point out that using a 10 ohm resistor in the collector circuit looks like you are expecting the transistor to allow almost an amp of current. A square 9 volt battery doesn't have a chance of providing that much current, and you are probably using a small transistor that would melt if it did. You are all out of proportion trying to get an amp to flow when people who are good at this work in the range of a thousandth of that.
I think it is 10K in the collector.

#12

#### hp1729

Joined Nov 23, 2015
2,304
I don't get it!

here is the circuit I am playing with and my results don't make sense to me, as I increase the resistor going to the base, I am expecting the transistor to switch off being shown by the falling voltage across the collector resistor, this isn't happening, in fact little is happening at all except the voltage across the collector resistor falls as I increase the base resistor?.....my Vcc is 9v View attachment 106044 View attachment 106044
As the transistor turns off the voltage across the resistor decreases. The voltage across the transistor E-C increases. With the transistor off there is no voltage across the resistor, right?
Good exercise!!!!

#### #12

Joined Nov 30, 2010
18,210

#### wayneh

Joined Sep 9, 2010
16,399
Calculate the currents. I haven't done the math, but I think your results are entirely normal.

#12

#### ronv

Joined Nov 12, 2008
3,770
Normally you would measure the voltages from the emitter (ground).

#### Neil Groves 1

Joined May 3, 2016
21
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.

#### Neil Groves 1

Joined May 3, 2016
21
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.

#### dl324

Joined Mar 30, 2015
11,251
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?
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.

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#### wayneh

Joined Sep 9, 2010
16,399
Voltage with no current is like a mountain lake far above another lake. Or the live wire in your wall. There is potential but no current flows until a path is opened. You can definitely have a voltage difference with no current.

#### #12

Joined Nov 30, 2010
18,210
Voltage with no current is like a battery laying on a table. The fact that no current is flowing does not make the battery go dead then magically come back to life when you attach it to a circuit. Voltage does exist (in many places) without current flow.

#### hp1729

Joined Nov 23, 2015
2,304
Voltage with no current is like a battery laying on a table. The fact that no current is flowing does not make the battery go dead then magically come back to life when you attach it to a circuit. Voltage does exist (in many places) without current flow.
Interesting comments. Does it apply to the discussion? How about a real world circuit?