Transistors: Connecting Bases

[Falcon69]

I am having trouble with a circuit, and I suspect it could be because the way I have the transistors set up.

Just ignore the inputs/outputs, only look at how the bases are connected together in the picture.

Which is the correct way to do it? The top or the one below?

Also, can the emitter of an NPN be connected to the emitter of a PNP, or should there be a diode/resistor in between them?

Thanks in advance for your insight.

Jason

[Simon]

Both would work in theory although if you use one base resistor it needs to be 1/3 the value for 3 transistors. I'm no BJT expert and I'd go with caution and use individual base resistors because unless the Vbe is identical then they won't all come on by the same amount or at all.

[Falcon69]

Ya, I have them wired like the top.

A single 10k to run 3-4 transistors.  It doesn't seem as if they are all turning on.

So removing that 10k at the power input, and then placing a 10k between power and each base, should allow them all to turn on at same time when switch is hit?

[Simon]

it sounds like 10K is only enough for 1 transistor so the best wins, you need then 3.3K or better 10K for each

[Falcon69]

okay, i will try that Simon, thank you.

So what s happening, the 10k is eating up too much voltage, and not leaving enough to turn on each transistor. But reducing that to 3.3k should leave enough voltage to switch each one on at same time?  That makes sense I guess.  But, it sounds like they still may not switch on, unless the vbe is the same, which I doubt they are.

I will wire a 10k to each one, and see how that works.

Thank you Simon.

[Simon]

you are not using voltage to turn on the transistors but current, you need to know your load current, then your base current (Ib) is Iload/beta of transistor, your base resistor will then be (Vsupply-0.7)/Ib.

The problem you can get with paralleled bases is that if one transitor heats up the base voltage drops allowing more current through so it will take up all of the current and starve the other 2 as they are a less easy path.

[Andy Watson]

Either circuit should work. But if you are using different colour LEDs expect a lot of variation in light output - due to the different operating voltage of the LEDs.

[janoc]

okay, i will try that Simon, thank you.

So what s happening, the 10k is eating up too much voltage, and not leaving enough to turn on each transistor. But reducing that to 3.3k should leave enough voltage to switch each one on at same time?  That makes sense I guess.  But, it sounds like they still may not switch on, unless the vbe is the same, which I doubt they are.

I will wire a 10k to each one, and see how that works.

Thank you Simon.

Strictly speaking, a BJT turns on with base current, not voltage (unlike a FET). So if you use a single resistor for all 3, the resistor needs to pass 3x as much current in order to open all three transistors (Kirchhoff's law). If your BJT works with a 10k resistor solo (with about 500uA base current), then using 3k3 (or even 2k7) should do it.

However, putting a separate resistor for each transistor is better - the transistors don't have the same beta, so they may need different base currents.

[Simon]

Could the base/emitter volt,age vary with temperature of the working BJT? like i said i'm no expert on BJT's and I'm wondering myself ?

[jucole]

This is a really good video which might be useful.

[mikerj]

Are you simply trying to switch LEDs on and off, or are you controlling the brightness with some external circuit (i.e. instead of the switch in the schematic)?  If you only want to switch them on and off, then why have you chosen an emitter follower configuration rather than common emitter? 

[mij59]

Hi,

This is a very expensive circuit just to turn 3 led's on, wondering why you want 3 transistors.

The 10k resistors are not needed, it's better to connect a 10k resistor between the bases and ground, so the bases don't float.

The function you need can also made with just one transistor.

[Simon]

You talking about mosfets not bjt's, remove the resistor and it all goes bang

[mij59]

You talking about mosfets not bjt's, remove the resistor and it all goes bang

Why should it go bang ?

[rs20]

You talking about mosfets not bjt's, remove the resistor and it all goes bang

Woah, woah woah. The BJT does *not* go bang if you remove the resistor, not even close. See attached sim. The top circuit would work perfectly if the LEDs were simple 2V red LEDs, but from context I suspect these are actually LED arrays designed to run at 12V which is why it's not working.  I'm going to continue under the former assumption though.

I'm not advocating this design, but let's actually model what we would expect to happen with the top circuit. Firstly, the LED/resistor pairs will never see more than 5V, so the fact that they're being powered off 12V is weird, they may was well be powered off 5V as well. If you want to fix this, the LEDs should be connected to 12V and switched from below (still with the resistors down to ground for the reason explained below). So from +12V, it goes LED, then BJT, then resistor.

That point aside, the bases of all the transistors are going to have the same voltage, because they're shorted together -- let's say 4V for the sake of argument (probably much lower if the resistor is too large, but that's not my point). Since the VBE of a transistor will stick to a narrow range around 0.7V more-or-less regardless of base current, we expect that the emitters of all the transistors will all be around 3.3V. If the LED resistors are appropriately chosen, this is a perfect outcome, they'll all shine equally brightly regardless of transistor hFE and so on. You do not need the 10k resistor at all, the BJTs will not blow up because the emitter voltages will rise.

In short, get rid of the base resistors entirely (just short them out) and that's actually the best thing to do, not a recipe for a bang. Assuming those LEDs are just single, naked LEDs as opposed to 12V modules.

[LvW]

Strictly speaking, a BJT turns on with base current, not voltage (unlike a FET).

Perhaps not too important within the context of THIS thread - however, this assertion is not true.
It is the base-emitter VOLTAGE which controls the collector current and NOT the base current (which cannot be avoided).
Physically - it is, of course, the changing voltage across the depletion layer which allows the current to become larger or smaller.
A larger current (like Ic) never can be controlled by a smaller current (like Ib). 
This was discussed already several times in various electronic forums.

[janoc]

Perhaps not too important within the context of THIS thread - however, this assertion is not true.
It is the base-emitter VOLTAGE which controls the collector current and NOT the base current (which cannot be avoided).
Physically - it is, of course, the changing voltage across the depletion layer which allows the current to become larger or smaller.
A larger current (like Ic) never can be controlled by a smaller current (like Ib). 
This was discussed already several times in various electronic forums.

Care to explain (or point me to some place discussing this), please?

AFAIK, for BJTs, beta = Ic/Ib, thus you are controlling the collector current by the base current (current amplification). Of course, in order to get the current flow you need to apply voltages with the correct polarity, but if the device was voltage controlled as you seem to imply, then the OPs circuit would have been working as he had sufficient voltage applied to the bases. It didn't work, because the transistors split up the available 5V/10k = 500uA of base current among themselves and that portion was not enough to keep them open.

You could say that the base voltage was too low because of the large voltage drop over the resistor, but that voltage drop happens because there is current flowing through the resistor -> no current, no voltage drop.

It is quite well explained for ex. here, including what is happening with the minority/majority carriers and the depletion regions: http://ecee.colorado.edu/~bart/book/book/chapter5/ch5_2.htm

[David Hess]

10k is too high a value to drive the three transistors in parallel; with a worst case hfe of 50, the total base current would be almost 200 microamps so the 10k base resistance would have a voltage drop of about 1.8 volts limiting the LED current significantly.

A resistance below 3k would work to drive all three transistors with their bases tied in parallel but there is no reason to use one that high.  I would probably just use another 680 ohm resistor; the value is not critical.

Maximum LED current in this circuit is about 3 milliamps.  Is that what you intended?  One transistor would certainly work to drive all three LEDs.

[rs20]

Maximum LED current in this circuit is about 3 milliamps.  Is that what you intended?  One transistor would certainly work to drive all three LEDs.

Good point.

10k is too high a value to drive the three transistors in parallel; with a worst case hfe of 50, the total base current would be almost 200 microamps so the 10k base resistance would have a voltage drop of about 1.8 volts limiting the LED current significantly.

A resistance below 3k would work to drive all three transistors with their bases tied in parallel but there is no reason to use one that high.  I would probably just use another 680 ohm resistor; the value is not critical.

Why not zero ohms?

[David Hess]

10k is too high a value to drive the three transistors in parallel; with a worst case hfe of 50, the total base current would be almost 200 microamps so the 10k base resistance would have a voltage drop of about 1.8 volts limiting the LED current significantly.

A resistance below 3k would work to drive all three transistors with their bases tied in parallel but there is no reason to use one that high.  I would probably just use another 680 ohm resistor; the value is not critical.

Why not zero ohms?

Adding the single resistor may make the board layout easier.  Other than that I do not see any problem with zero ohms.  Sometimes the base resistor and a shunt capacitor are added for ESD protection if the LEDs are exposed.

In other circuits the base resistor might be added to isolate the output of the stage driving the transistor or as a snubber to prevent the transistor from oscillating if it is driving a low impedance load but neither consideration applies here.

[LvW]

Care to explain (or point me to some place discussing this), please?
.....
It is quite well explained for ex. here, including what is happening with the minority/majority carriers and the depletion regions: http://ecee.colorado.edu/~bart/book/book/chapter5/ch5_2.htm

Hi janoc,
I think, in general we have to distinguish between a simple formula (like Ic=B*Ib) which can be used during the design and other formulas which describe the real physical effects. Of course, Ib is a small (nearly fixed) percentage of Ic - but does this automatically mean that Ib controls Ic?
More than that - for my opinion, even the document referenced by you supports the voltage-control approach. See the following excerpt:

We consider here only the forward active bias mode of operation, obtained by forward biasing the base-emitter junction and reverse biasing the base-collector junction. .......
Once the electrons arrive at the base-collector depletion region, they are swept through the depletion layer due to the electric field. These electrons contribute to the collector current. In addition, there are two more currents, the base recombination current, .....and the base-emitter depletion layer recombination current.

For my opinion. the base current is clearly considered as secondary (not as the cause of Ic).

Here are some other references:

1.) Winfield Hill (Co-author of the popular Horowitz/Hill book), see his post#2 and post#10:
http://cr4.globalspec.com/thread/68055/voltage-vs-current

2.) Berkeley University (see chapt.
www.eecs.berkeley.edu/~hu/Chenming-Hu_ch8.pdf

Excerpt: An undesirable but unavoidable sideeffect of the application of Vbe is a hole
current flowing from the base, mostly into the emitter. This base (input) current, Ib,
is related to Ic by the common-emitter current gain.


3.) Book from Sedra/Smith (chapt. 5)

Excerpt: Basically, the forward-bias voltage Vbe causes an exponentially related current Ic to flow in the collector terminal. The collector current Ic is independent of the value of the collector voltage as long as the collector–base junction remains reverse-biased; Thus in the active mode the collector terminal behaves as an ideal constant-current source where the value of the current is determined by Vbe.

[Falcon69]

Hi guys, and thanks for your input

The input voltages and outputs (led's in this case) are not important. I was merely wondering about what is needed to successfully drive the transistors when the bases are tied together.  As in my circuit, they are not turning on.  I will post my circuit in the original thread I had started early, as that is the correct place for it.

So it seems that the consensus of everyone is saying to place a resistor at each base of each transistor.

There may be something else wrong with my circuit, but I think I have my answer here. I just need to add a resistor at the base of each transistor or (for layout purposes) divide the 10k by how many transistors (10k/3 = 3K3) and use the 3K3.

Thanks everyone!

[Dinsdale]

Your transistors have no reliable bias.  Your bias current is highly dependent on the transistor gain.  Plus the voltage at each emitter is probably highly variable.  You need a resistor divider on EACH base.  Try 5k up and 1k down and adjust from there.

[c4757p]

Your transistors have no reliable bias.  Your bias current is highly dependent on the transistor gain.  Plus the voltage at each emitter is probably highly variable.  You need a resistor divider on EACH base.  Try 5k up and 1k down and adjust from there.

Seriously, dude? Did you look at the schematic?

They're common collector for god's sake. They don't need "bias resistors". They don't need base resistors at all.

[janoc]

Hi janoc,
I think, in general we have to distinguish between a simple formula (like Ic=B*Ib) which can be used during the design and other formulas which describe the real physical effects. Of course, Ib is a small (nearly fixed) percentage of Ic - but does this automatically mean that Ib controls Ic?

OK, thanks for the references, I am chewing through it (I am a software guy by trade, not EE, but physics and electronics was always my interest). However, as you have said - that's really an enormous overkill for what the OP was asking. Also most literature shows the current-based analysis, using load lines, characteristics, beta and such, so that's where I am coming from.

[rs20]

Hi guys, and thanks for your input

The input voltages and outputs (led's in this case) are not important. I was merely wondering about what is needed to successfully drive the transistors when the bases are tied together.  As in my circuit, they are not turning on.  I will post my circuit in the original thread I had started early, as that is the correct place for it.

If you try to drive a 12V led module with 5V, it's not going to turn on. That's why your circuit isn't working, and that's why input and output voltages are critically important.

So it seems that the consensus of everyone is saying to place a resistor at each base of each transistor.

Most of the more recent posts are agree to have no resistors at all, or to only put resistors for a bit of a laugh. No base resistors are called for at all.

[Falcon69]

Actually, the circuit I am using, the collectors are not common at all, in the schematic for this purpose, it is.

Also, I have it drawn with 5v for power for the bases.  It is actually 12 volts, the whole circuit is 12 volts.  If there is no resistor at the base of the transistors, They Blow. I've fried several already because I forgot to place them on the breadboard.

[c4757p]

Actually, the circuit I am using, the collectors are not common at all, in the schematic for this purpose, it is.

What? Then there is a huge, fundamental difference between what you're doing and what you're showing us. Please fix the schematic so we know what you are actually doing.

[Dinsdale]

It is actually 12 volts, the whole circuit is 12 volts.

In that case, try 10K up and 620 down.
I guess you are just playing with transistors for fun, because you don't need any transistors at all if everything is 12V.

[c4757p]

you don't need any transistors at all if everything is 12V.

 

How can you say that, knowing nothing about the rest of his circuit? 12V != high current...

[IanB]

Actually, the circuit I am using, the collectors are not common at all, in the schematic for this purpose, it is.

Also, I have it drawn with 5v for power for the bases.  It is actually 12 volts, the whole circuit is 12 volts.  If there is no resistor at the base of the transistors, They Blow. I've fried several already because I forgot to place them on the breadboard.

It is not possible to guess why the transistors are failing unless you post the actual circuit you have built on the breadboard, with actual voltages and component values.

The behavior of a circuit depends on what it actually is, and not at all on what it isn't...

[Falcon69]

Okay, I understand.  I thought that there was just a general standard for connecting the bases of multiple transistors to get them all to activate correctly.

I have posted the schematic in the other post about sequential LED's.

[LvW]

OK, thanks for the references, I am chewing through it (I am a software guy by trade, not EE, but physics and electronics was always my interest). However, as you have said - that's really an enormous overkill for what the OP was asking. Also most literature shows the current-based analysis, using load lines, characteristics, beta and such, so that's where I am coming from.

Yes - you are right. It is really surprising (funny?) that in some documents/papers - even in some textbooks - the simple formula Ic=beta*Ib (which certainly is correct) is used to describe the physical control mechanism of a BJT. Just because it is so simple? On the other hand, there are many application-oriented properties of transistor circuits which clearly show that the BJT is voltage controlled. For example, take two simple gain stages having the same DC operating point but two different beta values.
The voltage gain will be equal - independent on beta.     

[T3sl4co1l]

In the example exactly as shown, each individual emitter current is controlled by a separate resistor.  Therefore, no current sharing is necessary; the bases will each draw as much current as they need, and no base resistance is required -- anywhere, not even by the switch.

Possible gotchas: emitter followers can have a negative input impedance, which can cause oscillation.  A series "base stopper" resistor can be helpful to limit this, especially if bypass capacitors (on any combinations of two terminals, B-C-E), or long wires, are present.

If the emitters are common, however, you are paralleling transistors and must observe correct use:emitter resistors, since the effect is hFE times more reliable than base resistors (which however might still be required for the above reasons).  This is fundamentally because transistors are transconductance devices (voltage in, current out), the base current is only a convenient side effect of their construction (e.g., hFE varies all over the place, Gm follows a well defined function).

Tim

[David Hess]

Possible gotchas: emitter followers can have a negative input impedance, which can cause oscillation.  A series "base stopper" resistor can be helpful to limit this, especially if bypass capacitors (on any combinations of two terminals, B-C-E), or long wires, are present.

An emitter resistor is even more effective than a base resistor for preventing this problem and the circuit as drawn has one for each transistor although we admittedly do not know how close to the emitter it is.

[Falcon69]

Thank you everyone!

Placing a 10k resistor at the base of each transistor for the same trigger/input worked!  they all turn on now.

Thank You Thank You Thank You!

I still have some problems with the circuit, but at least the transistor turning on is solved.      

Jason

[WarSim]

LOL you actually posted for help with a cct that wasn't even the cct in question and got two pages of help. 
Even funnier that it wasn't until the second page somebody saw the floating problem of the original fake cct. 

This is a perfect example why I wait before consider helping. 


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[Falcon69]

Untrue, I believe one of the first posts said to place  10k resistor at the base of each transistor. That was what was needed. I will apply that logic to any other circuit i try and work on from now on.

[WarSim]

Yeah I know you have no idea what I am referring to and that is ok.    it is a moot point now that we know the cct given was wrong, as stated in response 27. 


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