Super new to reading Datasheets :(

[JacobEdward]

I just want someone to confirm for me that this transistor is turned on with a minimum voltage of .65V and a maximum of .85V when there's 1mA of current on the base pin...  Not sure why its rated in negative voltage though... does anybody know of a good tutorial for going through a datasheet step by step for beginners?

https://www.fairchildsemi.com/datasheets/2n/2n3906.pdf

[TimFox]

Better to state that with 1.0 mA of base current and 10 mA of collector current (both established with external voltages and resistors), the base voltage is guaranteed to be between 0.65 and 0.85 V.  One should not drive the transistor from a "stiff" voltage source, since the base current is not well-determined at a given voltage.
This is measured with the device saturated, i.e. Vce < 0.25 V,  the maximum value shown in the block above.  The spec is relevant to switching operation, not linear amplification.
That these voltages are negative means that the device is PNP.  When turned on, the voltages at the base and collecter are both negative with respect to the emitter for PNP devices.

[JacobEdward]

Better to state that with 1.0 mA of base current and 10 mA of collector current (both established with external voltages and resistors), the base voltage is guaranteed to be between 0.65 and 0.85 V.  One should not drive the transistor from a "stiff" voltage source, since the base current is not well-determined at a given voltage.
This is measured with the device saturated, i.e. Vce < 0.25 V,  the maximum value shown in the block above.  The spec is relevant to switching operation, not linear amplification.
That these voltages are negative means that the device is PNP.  When turned on, the voltages at the base and collecter are both negative with respect to the emitter for PNP devices.

Just to check, you apply 1mA to the base and 10mA to the collector to turn it on, but you also apply over 3V to the Base and over 30V to the Collector to turn it off?  Where would I find the proper current ranges to shoot for if it's current I should be looking at and not Voltage?

[TimFox]

In the OFF case, you apply reverse bias to both the collector and the base.  In the reverse-bias case, you apply a fixed voltage and measure a small current, which the manufacturer guarantees to be less than the spec of 50 nA.  Both collector-base and base-emitter diodes will act as Zeners for too large a voltage, which is why the manufacturer measures the leakage currents at 3.0 V for the base, and 30 V for the collector, which is his measurement to verify a sufficiently high breakdown (Zener) voltage.
A proper specification is one that can be measured on each device with automatic test equipment.  In the ON case, the stimuli are currents and the responses are voltages;  in the OFF case, the stimuli are voltages and the responses are currents.
You do not need to apply 3.0 V of reverse DC bias to the base to turn the device OFF, you just need to reduce the base current to zero.  Driving the base further negative will speed up the switching by sweeping out the stored charge, however.  Exceeding the base-emitter breakdown voltage (minimum 5 V here, which is very low compared with the other voltage ratings of a normal transistor) will probably damage the device.

[JacobEdward]

Better to state that with 1.0 mA of base current and 10 mA of collector current (both established with external voltages and resistors), the base voltage is guaranteed to be between 0.65 and 0.85 V

So just to be clear, if you apply a 1mA to the base and a 10mA to the collector, but the voltage applied to the base is 5V and the voltage applied to the collector is 40V, it will still work?  The only thing that determines if the transistor is on or not is the current?

I read your other reply, but I don't think it was really addressing the question I was trying to ask... so you are suppose to apply 1mA to the base, is there a minimum and a maximum range I could work with?  Like maybe .8mA to 1.2mA or does it have to be a perfect 1mA to activate the base?  Same with turning the transistor off?  I'm not really interested in any other mode apart from saturation and cut-off.

[TimFox]

When using a transistor as a switch, the requirement is to supply enough base current to saturate the collector current, achieving minimum collector voltage.  Applying more current to the base has little effect on the saturated collector, once the collector voltage falls below the base voltage, reaching the saturation voltage.  The transistor is no longer operating linearly.
If you apply 1 mA to the base, the base-emitter voltage will be less than 0.85 V.  With a resistor in series with the collector, the voltage from collector to emitter will be less than 0.25 V if the collector current is less than or equal to 10 mA when the base current is 1 mA.  These are the direct conclusions from the manufacturer's specification.
Applying 5 V (forward) directly (no series resistance) to the base-emitter diode will smoke the device.  Applying 5 V (reverse) to the diode may break down the diode (manufacturer only guarantees > 5 V breakdown voltage) and smoke the device.  These statements assume negligible resistance in series with the 5 V.
Similarly, applying 40 V to the collector without a series resistor may smoke the device with 1 mA base current, since the saturated collector current is not well-determined. 
With 4000 ohms from the 40 V to the collector, the collector current will be limited to slightly less than 10 mA which will work.  However, since the specification for collector breakdown is > 40 V, this would be marginal.  This could be done with 4000 ohms from the 5 V supply drive to the transistor base to obtain slightly more than 1 mA.
There is a large variation from unit to unit in many parameters of a transistor.  The ratio between collector current and base current can be specified as a minimum, but may have a 4 to 1 ratio from maximum to minimum over production units.  The manufacturer chooses minimum values he can guarantee.  In this case, if you want to turn on the transistor to 10 mA collector current, the manufacturer states that 1 mA into the base is sufficient to guarantee that collector current.

[JacobEdward]

If you apply 1 mA to the base, the base-emitter voltage will be less than 0.85 V. 

When you say the base-emitter voltage will be .85V, is that what is suppose to be going on inside the transistor?  I'm not sure why that would be the case, or why any other scenario would be the case since the voltage should be determined by whatever voltage/current is being applied to the base pin... lol I think I'm getting more confused.

With a resistor in series with the collector, the voltage from collector to emitter will be less than 0.25 V if the collector current is less than or equal to 10 mA when the base current is 1 mA. 

Is the voltage from collector to emitter not effected by resistor values of the connecting circuits?

Applying 5 V (forward) directly (no series resistance) to the base-emitter diode will smoke the device.  Applying 5 V (reverse) to the diode may break down the diode (manufacturer only guarantees > 5 V breakdown voltage) and smoke the device.  These statements assume negligible resistance in series with the 5 V.

What do you mean by applying reverse voltage?  I'm assuming you mean 5V forward is just 5V to the base pin, right?

Similarly, applying 40 V to the collector without a series resistor may smoke the device with 1 mA base current, since the saturated collector current is not well-determined. 
With 4000 ohms from the 40 V to the collector, the collector current will be limited to slightly less than 10 mA which will work.  However, since the specification for collector breakdown is > 40 V, this would be marginal.  This could be done with 4000 ohms from the 5 V supply drive to the transistor base to obtain slightly more than 1 mA.

Doesn't the series resistance value for the collector really get determined by the series resistance value of the emitter since while the transistor is on, it would be one continuous series circuit, which would then effect the overall current of the circuit...

There is a large variation from unit to unit in many parameters of a transistor.  The ratio between collector current and base current can be specified as a minimum, but may have a 4 to 1 ratio from maximum to minimum over production units.  The manufacturer chooses minimum values he can guarantee.  In this case, if you want to turn on the transistor to 10 mA collector current, the manufacturer states that 1 mA into the base is sufficient to guarantee that collector current.

Does that mean that in this situation, it's a 10 to 1 ratio between collector current and base current?  If that's true, why do manufacturers give specific values instead of ratio's with acceptable ranges?

[TimFox]

(Answers in blue)  In general, each line of the data sheet gives the measurement conditions (e.g., Ib = -1 mA, Ic = -10 mA) and the range of measured value (e.g., Vbe between -0.65 and -0.85 V) under those conditions that is acceptable.  A good device must pass all of these measurements.  Sometimes the range of measurements is from min to max, sometimes only min, sometimes only max, and sometimes includes a typical value which is within the range.

If you apply 1 mA to the base, the base-emitter voltage will be less than 0.85 V. 

When you say the base-emitter voltage will be .85V, is that what is suppose to be going on inside the transistor?  I'm not sure why that would be the case, or why any other scenario would be the case since the voltage should be determined by whatever voltage/current is being applied to the base pin... lol I think I'm getting more confused.

You cannot force both the voltage and current through a two-terminal device to arbitrary values.  In the case of a PN junction (e.g., base-emitter), there is a monotonic function of voltage vs. current.  Since the inverse function of current vs. voltage is very steep, the voltage at a specified current is a better-determined value.  The manufacturer specifies the maximum voltage at specified current values.

With a resistor in series with the collector, the voltage from collector to emitter will be less than 0.25 V if the collector current is less than or equal to 10 mA when the base current is 1 mA. 

Is the voltage from collector to emitter not effected by resistor values of the connecting circuits?
  The (low) saturation voltage from collector to emitter is a very weak function of the collector current.  The collector resistor limits the current when the transistor saturates to approximately (Vcc / Rc), since Vce(sat) << Vcc (supply voltage).

Applying 5 V (forward) directly (no series resistance) to the base-emitter diode will smoke the device.  Applying 5 V (reverse) to the diode may break down the diode (manufacturer only guarantees > 5 V breakdown voltage) and smoke the device.  These statements assume negligible resistance in series with the 5 V.

What do you mean by applying reverse voltage?  I'm assuming you mean 5V forward is just 5V to the base pin, right?
When I said "5 V forward", I meant in the forward-bias direction for the diode.  That is a huge voltage for a silicon diode, and would require a huge current that would smoke the device.  If you apply 5 V in the reverse-bias direction (base to emitter voltage positive for PNP device), you are getting too close to the manufacturer's guaranteed minimum value for breakdown voltage on that device.

Similarly, applying 40 V to the collector without a series resistor may smoke the device with 1 mA base current, since the saturated collector current is not well-determined. 
With 4000 ohms from the 40 V to the collector, the collector current will be limited to slightly less than 10 mA which will work.  However, since the specification for collector breakdown is > 40 V, this would be marginal.  This could be done with 4000 ohms from the 5 V supply drive to the transistor base to obtain slightly more than 1 mA.

Doesn't the series resistance value for the collector really get determined by the series resistance value of the emitter since while the transistor is on, it would be one continuous series circuit, which would then effect the overall current of the circuit...
No.  The transistor is an active device, which is very non-linear in switching circuits.  When you saturate the transistor with high base current, the collector to emitter is almost shorted, with only a small Vce(sat) voltage that is roughly constant.  Therefore, the collector current is determined by the collector supply voltage Vcc and the collector resistor.  The designer can choose the collector current at will (subject to manufacturer's maximum values) and then choose the collector resistor to obtain that current.

There is a large variation from unit to unit in many parameters of a transistor.  The ratio between collector current and base current can be specified as a minimum, but may have a 4 to 1 ratio from maximum to minimum over production units.  The manufacturer chooses minimum values he can guarantee.  In this case, if you want to turn on the transistor to 10 mA collector current, the manufacturer states that 1 mA into the base is sufficient to guarantee that collector current.

Does that mean that in this situation, it's a 10 to 1 ratio between collector current and base current?  If that's true, why do manufacturers give specific values instead of ratio's with acceptable ranges?

For that transistor, the manufacturer guarantees that a base current greater than 1/10 of the collector current will saturate the transistor.  The specific values quoted are guaranteed minimum values.  If you are using the transistor as an amplifier, not a switch, with a collector-emitter voltage substantially larger than the saturation voltage, the manufacturer will quote a range of "beta = hfe", which is the ratio of (change in collector current)/(change in base current), or, the "current gain".  On another page of the 2N3906 data sheet, we see this gain quoted at different collector currents.  For Ic = -10 mA and Vce = -1.0 V, they quote hfe between 100 and 300.  For other currents, they only quote a minimum value of hfe.  This is a linear parameter, and does not apply directly to the non-linear switching case.  The ratio 10:1 for the switching case is sometimes called hFE (note the capitals).  As with all active devices, the linear parameters are approximations to the detailed behavior shown in I-V curves.

[JacobEdward]

(Answers in blue)

I'm still confused, but I found this video tutorial of how to read a datasheet and would like to switch to that transistor just to stay with the tutorial.

This is the datasheet https://www.fairchildsemi.com/datasheets/TI/TIP120.pdf

According to the tutorial, this transistor only required 120mA on the base pin to turn it out and it can handle as high as 5A across the Collector to Emitter.  Does that mean I can use a signal anywhere from 120mA to 5A on the base pin?

I'm having a hard time understanding the "Electrical Characteristics" section of this transistor datasheet...

The guy doing the tutorial said that for the Collector Cut-Off Current with the base at 0, "using at 30V, it's going to take .5mA before the current cuts off, that's how far it's going to drop down"

What does he mean by the current cuts off?  Is that just the transistor acting like an open switch?

For the Icbo section, I'm confused as to how there could ever be a situation where 60V would be going from Collector onto the Base... wouldn't that be the transistor being fried?

When a transistor is referred to as an "amplifier", does that just mean it's a switch that can be toggled with a much lower current/voltage than what is actually going through the collector/emitter?  I'm not sure why there would be a "DC Current Gain" if it's just a switch being toggled by a smaller signal... is that just referring to what the collector/emitter can handle, not what is actually "gained" through amplifying some signal?

What does it mean to have saturated the transistor?  On this datasheet, it says that the transistor can "sustain" 60V, but then is saturated at 2V when the collector is 3A and the base is 12mA...

The Base-Emitter On Voltage is the voltage it would take to turn the transistor even though it would still only take 120mA?  So the example they gave was that when there was 3V and 3A being applied to the collector, it would take 2.5V (and 120mA) to turn the transistor on?

[TimFox]

(Answers in blue)

I've answered most of these questions, so I will only deal with a few below.  Again, read the data sheet literally so that each line has a measurement condition and a measured result.  I have attached a basic graph of the voltage-current curve for a diode to indicate the forward-biased current (exponential in voltage) and the reverse-biased current (low leakage followed by avalanche).
Remember that the data sheet has two important sections:  ratings (maximum stresses that the device can withstand) and characteristics (measured results from specific operating conditions).

I'm still confused, but I found this video tutorial of how to read a datasheet and would like to switch to that transistor just to stay with the tutorial.

This is the datasheet https://www.fairchildsemi.com/datasheets/TI/TIP120.pdf

According to the tutorial, this transistor only required 120mA on the base pin to turn it out and it can handle as high as 5A across the Collector to Emitter.  Does that mean I can use a signal anywhere from 120mA to 5A on the base pin?
This is a power transistor, meant to operate at much higher current than the small-signal 2N3904/2N3906 series.  The 5A value is a collector current, not a base current.  You could put 5 A into the base, but the voltage would be higher than the low collector saturation voltage and the resulting power would be higher, possibly damaging.
I'm having a hard time understanding the "Electrical Characteristics" section of this transistor datasheet...

The guy doing the tutorial said that for the Collector Cut-Off Current with the base at 0, "using at 30V, it's going to take .5mA before the current cuts off, that's how far it's going to drop down"

What does he mean by the current cuts off?  Is that just the transistor acting like an open switch?
With the base-emitter shorted, and 30 V applied to collector-emitter, the leakage current is 0.5 mA.  That is the meaning of the parameter Ices:  collector-emitter current with base shorted.  A related parameter is Iceo:  collector-emitter current with base open-circuited.

For the Icbo section, I'm confused as to how there could ever be a situation where 60V would be going from Collector onto the Base... wouldn't that be the transistor being fried?
If you apply 60 V from collector to base, the normal polarity applies reverse bias to that diode.  There will be a small leakage current, but the manufacturer guarantees that the diode will not break down at that voltage.  The small power from this voltage and current is safe for the transistor.  Icbo is the diode current of the collector-base diode with the emitter open circuit.
When a transistor is referred to as an "amplifier", does that just mean it's a switch that can be toggled with a much lower current/voltage than what is actually going through the collector/emitter?  I'm not sure why there would be a "DC Current Gain" if it's just a switch being toggled by a smaller signal... is that just referring to what the collector/emitter can handle, not what is actually "gained" through amplifying some signal?
When the transistor is used an "amplifier" circuit, the collector voltage is much higher than the "saturation" voltage, and the operation is considered "active".  Varying the input current to the base results in a larger change in collector current and a large swing in the collector voltage, due to an appropriate value of load resistance from collector to power supply.  In a "switch" circuit, the transistor changes quickly from "Off" (very low current, high voltage) to "On" (high current, saturated voltage)":  in both states, the power dissipated in the transistor is low, while the power dissipated by the amplifier circuit is relatively large.  Again, in a bipolar junction transistor (BJT), when the transistor is saturated, the low Vce value is a very weak function of the current.  "Gain" is a factor:  the output (collector) current is proportional to and higher than the input (base) current.  A constant ratio ("current gain") is a linear approximation to a smooth function.  /color]
What does it mean to have saturated the transistor?  On this datasheet, it says that the transistor can "sustain" 60V, but then is saturated at 2V when the collector is 3A and the base is 12mA...
"Saturation" is a very important concept for BJT switches.  I refer you to any textbook on solid-state devices.
The Base-Emitter On Voltage is the voltage it would take to turn the transistor even though it would still only take 120mA?  So the example they gave was that when there was 3V and 3A being applied to the collector, it would take 2.5V (and 120mA) to turn the transistor on?
The "On" voltage is the result of applying the 120 mA base current   The manufacturer guarantees that the voltage measured at the transistor will not exceed that value.  For BJTs, one normally treats the input as a current-activated connection with a medium impedance, and the voltage is a result.   Vacuum tubes and FETs are normally considered voltage-activated, since they have very high input impedance.

[JacobEdward]

With the base-emitter shorted, and 30 V applied to collector-emitter, the leakage current is 0.5 mA.  That is the meaning of the parameter Ices:  collector-emitter current with base shorted.  A related parameter is Iceo:  collector-emitter current with base open-circuited.

What do you mean by "leakage current"?  Is that the current that can be measured at the emitter pin or the base pin?

[JacobEdward]

For the Icbo section, I'm confused as to how there could ever be a situation where 60V would be going from Collector onto the Base... wouldn't that be the transistor being fried?

If you apply 60 V from collector to base, the normal polarity applies reverse bias to that diode.  There will be a small leakage current, but the manufacturer guarantees that the diode will not break down at that voltage.  The small power from this voltage and current is safe for the transistor.  Icbo is the diode current of the collector-base diode with the emitter open circuit.

How do you apply 60V from collector to base?  Doesn't anything connected to the collector naturally travel to the emitter (and if it can't travel to the emitter because the base hasn't connected the collector and emitter, the circuit is open)?

[JacobEdward]

(Answers in blue)

If the maximum voltage for the Collector-Emitter is 60V and it is 5A for the maximum current for the Collector... and the saturation rate is 5A and 4V on the collector-emitter with 20mA on the Base, what Voltage/Current would be needed for the base to allow the maximum 60V and 5A from the collector to emitter?

Sorry this is taking so long... and thank you for helping me.

[flynwill]

I suggest you might want to go find any of a number of basic EE text books and carefully read the section on Bipolar Transistors (aka BJTs).  There is quite a lot to understand if you are coming from no knowledge at all more than can be practically explained in a web forum like this one.  You can also look a the wikipedia article, but it is written at a pretty high level with a lot detail you probably don't need.

As I think was explained already...  You will never see 5V Base to Emitter voltage on a BJT.  If you try the current through the base will rise exponentially and the device will go up in a puff of smoke.  Practical BJT design will always have this current limited in some way, typically by the use of resistors in series with the base, the emitter or both.  The 5V number on the datasheet refers to the base being more than 5 volts negative with respect to the emitter (once again for an NPN device).  With the base voltage negative there will be almost no base current, and the device will be "off".  If you exceed that value in the negative direction the junction will "breakdown" and the current will rise rapidly with increasing voltage.  This is a condition your circuit should avoid by design.

Likewise you won't see 5 A collector-emitter current and 60V collector-emitter voltage at the same time in a device like this.... that's 300W and it will once again go up in puff of smoke if you try.  In fact for a small package 5A for the fully saturated device (that is "on") is 2 - 3 watts and it's going to get pretty hot pretty quickly.

"Saturation" by definition is the condition where the collector voltage is less than the base voltage for an NPN device (or visa-versa for a PNP device).  One -- only slightly flawed -- way to look at it is that the Collector-Base is a diode that is normally reverse-biased, but when the device goes into saturation some of the base current is sapped into the collector, through the now forward-biased diode, and no longer contribute to the gain mechanism.  In practical terms the collector-emitter voltage drops to something in the 0.3-0.6 V range (depending on the device and the load) and stays there no matter how much more base current you apply.

So the main thing to understand is that it is a current device.  Generally the collector current equals the base current times the hfe (aka "beta") UNLESS whatever is supplying the collector current is limits the current to a lower value or the device goes into saturation.    Note that for the device you linked hfe can range for 30-300... Your circuit has to be designed to work over this range of conditions.

Even with no base current if there is a voltage between the collector and emitter there will be a tiny current flowing "leakage". 

If you exceed the rated Collector-emitter voltage (or related collector-base voltage) with the device "off" (ie no base current) at some point the the device will "break down" and a large collector current will flow, possibly with catastrophic results depending on the circuit design.

[TimFox]

With the base-emitter shorted, and 30 V applied to collector-emitter, the leakage current is 0.5 mA.  That is the meaning of the parameter Ices:  collector-emitter current with base shorted.  A related parameter is Iceo:  collector-emitter current with base open-circuited.

What do you mean by "leakage current"?  Is that the current that can be measured at the emitter pin or the base pin?

Please read the above sentence for the measurement condition:  base-emitter shorted, 30 V applied to collector emitter.  There are only two terminals for that condition.  You can measure the current at either end.  The "leakage current" corresponds to the label on the graph of diode current vs. voltage that I attached.

[TimFox]

For the Icbo section, I'm confused as to how there could ever be a situation where 60V would be going from Collector onto the Base... wouldn't that be the transistor being fried?

If you apply 60 V from collector to base, the normal polarity applies reverse bias to that diode.  There will be a small leakage current, but the manufacturer guarantees that the diode will not break down at that voltage.  The small power from this voltage and current is safe for the transistor.  Icbo is the diode current of the collector-base diode with the emitter open circuit.

How do you apply 60V from collector to base?  Doesn't anything connected to the collector naturally travel to the emitter (and if it can't travel to the emitter because the base hasn't connected the collector and emitter, the circuit is open)?

With the emitter open circuited (dangling in mid air), there are only two terminals left:  collector and base.  The transistor between those two terminals is a diode, and one applies voltage and measures current in the usual fashion.

[TimFox]

(Answers in blue)

If the maximum voltage for the Collector-Emitter is 60V and it is 5A for the maximum current for the Collector... and the saturation rate is 5A and 4V on the collector-emitter with 20mA on the Base, what Voltage/Current would be needed for the base to allow the maximum 60V and 5A from the collector to emitter?

Sorry this is taking so long... and thank you for helping me.

60 V is the maximum rating on the collector voltage and 5 A is the maximum rating on the collector current.  That does not mean that the transistor can sustain both of those values simultaneously:  that would be 300 watts of power dissipation in the transistor and that device cannot dissipate that much power.  When the manufacturer specified the base current for saturation, the collector voltage is much lower.