Current flowing through transistor even when base is shunted to ground.

[Th3_uN1Qu3]

Re: Current flowing through transistor even when base is shunted to ground.

NTE devices are designed to be replacements for other parts not the other way around. Hence why the NTE part has higher specs than the one it replaces.

[mrwebber35]

Re: Current flowing through transistor even when base is shunted to ground.

Old thread but I have an interesting comment about MMBT3906 mentioned here. NTE says NTE2407 is the equivilent but it has higher mA and Voltage ratings like 40v 200mA for 3906 and 60v 600mA for 2407. Close enough I suppose.

[EasyGoing1]

Re: Current flowing through transistor even when base is shunted to ground.

I just want to say that around 1990, I went through the electronics courses at my local junior college, starting from basic electronics and ending with two semesters of electrical calculus. At the end of the courses, we had the option of taking the CET exam which I did and I passed. But then I went on and built my career as a Network Engineer so I never really used my electronics education specifically ... but there were many times when that education served me well when I needed to troubleshoot servers and workstations...

Transistors were always something that I didn't feel like I fully understood internally, even though I could run the calculations and answer the questions.

This thread really helped me make more sense of transistors ... must have been something in the way you guys explained it. I know there is much more to know about them, especially when you start getting into oscillating current such as in audio applications etc., but in terms of the basics, I definitely feel like I have a much better grasp on them than I did before I started this post.

Thank you - all of you - who took the time to write what you did to explain things to me. I really do appreciate you making that effort.

[EasyGoing1]

Re: Current flowing through transistor even when base is shunted to ground.

with a bench supply it could be a well controlled destruction . throw in a video of it including infra red temp readings should spice things up a little

I don't have a bench supply, but I'm going to purchase one in about a month-ish. What I do have are those adjustable DC to DC buck converter circuits (you can get a hand full on Amazon for about $15) that also has the ability to adjust available current. I usually drive those with a 19V laptop power supply so I can set the buck to 5V and I'm sure I can push the amperage up to 5 amps with that power supply, but I'm sure anything over a couple of amps would do the job.

[EasyGoing1]

Re: Current flowing through transistor even when base is shunted to ground.

It is more fun for OP to do hands on!

I'm still gonna do the experiment ... it's fun letting the smoke out ... and I am curious to know how long it will take to destroy the junction... if it's as fast as an LED, I MIGHT see a number on the current meter when I light up the circuit. If it's a little more resilient than I will for sure.

If I could measure current with my scope, I'd set it to snapshot the turn-on event but low-end scopes don't come with leads for current.

What I find fascinating about electronics, in general (in this application anyway), is that all we need to do to prevent the damage is drop a 1k before the base and then the amount of current available at the source is irrelevant.

[petehall347]

Re: Current flowing through transistor even when base is shunted to ground.

with a bench supply it could be a well controlled destruction . throw in a video of it including infra red temp readings should spice things up a little .

[momaka]

Re: Current flowing through transistor even when base is shunted to ground.

It is more fun for OP to do hands on!

Yeah, I know.
Sorry for being the party pooper.

On the other hand, if O/P doesn't have a bench PSU with good or any current limiting, that test could end up like some of ElectroBOOM's videos

... and it would waste a good working part
... and the fumes smell bad

[petehall347]

Re: Current flowing through transistor even when base is shunted to ground.

i was wanting to do it with this but it wont let me ..maybe its my computer i dunno ..
http://www.falstad.com/circuit/e-npn.html

[budm]

Re: Current flowing through transistor even when base is shunted to ground.

Correct.
Keyword here being "less than".

And the best part is, you can test that by placing an Ammeter in series with the LED's path to measure the current and see if the circuit you built actually agrees with your theory (it probably will, or would be very close.) You can also move the Ammeter between the Emitter of the BJT and ground and measure that current. Then, if you compare Ie and Ic, you will see that they are almost identical, with Ie being only slightly higher due to the added Ib current.


It is possible, correct.
However, since h_FE can vary not only with current, but also with temperature and between even supposedly identical transistor parts, it's not practical to do.

That said, there is a way to "stabilize" the linear operation of the transistor and make it practically independent of h_FE by placing a resistor between Emitter and ground. With this, if any change occurs to the transistor h_FE so that it increases, the Emitter current would also try to increase for a given Base current. However, with the resistor between Emitter and ground, an increased current through it will cause a greater voltage drop, which would in turn raise the voltage observed at the Base of the transistor. And if a series resistor is used to limit Base current, then the Base current would decrease, which in turn would cause the Emitter current to decrease back to its original value, regardless of the increase in h_FE. In simpler words, the Emitter resistor in that case would provide negative feedback - a property extensively used in BJT amplifier circuits. But here, we are not doing that, so keeping the LED series resistor makes more sense.


Or just do the math.

Let's see here... 3904 BJT in standard TO-92 case is rated at maximum of 625 mWatts dissipation. No mention of the maximum Base current (though some datasheets do include that.)

Assuming a constant Vbe of 0.7V and dumping 5 Amps of current there would produce 0.7 x 5 = 3.5 Watts or 3500 mW - about 5.5 times more than what the whole transistor can handle.

So short answer: YES, it would fry the transistor.
... possible exception may exist if the Base current is pulsed, though. And in such case, the pulse width would have to be extremely narrow to no blow the BE junction. Also, the 625 mW rating is for the whole transistor die, so any current flowing through C-E multiplied by the Saturation voltage would add up to the dissipated power as well.

*EDIT*
Seems that CG beat me to it with the explanation.
But yeah, this is why you do need to limit current at the Base of a BJT. Otherwise, it can burn out the BE junction, or worse.

It is more fun for OP to do hands on!

[budm]

Re: Current flowing through transistor even when base is shunted to ground.

Do I need a complete transistor circuit to perform the test? In other words, do I need a circuit to complete C-E before B-E will saturate? I'm thinking not???

Nope, Just hook up the power +5V to the Base of NPN, Negative to the Emitter of a spare Transistor you have and then watch the current draw and see the effect.

[momaka]

Re: Current flowing through transistor even when base is shunted to ground.

I followed everything you said here, thank you. I get it now... my final circuit has a BC337 and its datasheet lists Hfe as a function of collector current. The range of Hfe being from 100 to 630. Since I'm using a resistor at the base of 1.7k and assuming a 5v output from the Arduino pin, my Ibe would be 2.5ma. The minimum gain is 100 so worst case, I can have 250ma at the collector, which would be limited by that 6.8Ω resistor in series with the LEDs, so since my actual source voltage for the LEDs is 9.8, and considering the voltage drop Vce, I am most likely giving those LEDs less than 150ma, which means they are being driven well within tolerance.

Correct.
Keyword here being "less than".

And the best part is, you can test that by placing an Ammeter in series with the LED's path to measure the current and see if the circuit you built actually agrees with your theory (it probably will, or would be very close.) You can also move the Ammeter between the Emitter of the BJT and ground and measure that current. Then, if you compare Ie and Ic, you will see that they are almost identical, with Ie being only slightly higher due to the added Ib current.

What I also realized reading your post, is that it would theoretically be possible to drive LEDs without a series resistor by properly selecting your base current and knowing exactly what your H value would be at that base current and voltage... which would take some experimentation to determine, but in theory it would be possible, would it not?

It is possible, correct.
However, since h_FE can vary not only with current, but also with temperature and between even supposedly identical transistor parts, it's not practical to do.

That said, there is a way to "stabilize" the linear operation of the transistor and make it practically independent of h_FE by placing a resistor between Emitter and ground. With this, if any change occurs to the transistor h_FE so that it increases, the Emitter current would also try to increase for a given Base current. However, with the resistor between Emitter and ground, an increased current through it will cause a greater voltage drop, which would in turn raise the voltage observed at the Base of the transistor. And if a series resistor is used to limit Base current, then the Base current would decrease, which in turn would cause the Emitter current to decrease back to its original value, regardless of the increase in h_FE. In simpler words, the Emitter resistor in that case would provide negative feedback - a property extensively used in BJT amplifier circuits. But here, we are not doing that, so keeping the LED series resistor makes more sense.

Do the experiment, set power supply to 5V with 5A limit then apply it to the E-B junction of spare Transistor you have and see what happen.

Or just do the math.

Let's see here... 3904 BJT in standard TO-92 case is rated at maximum of 625 mWatts dissipation. No mention of the maximum Base current (though some datasheets do include that.)

Assuming a constant Vbe of 0.7V and dumping 5 Amps of current there would produce 0.7 x 5 = 3.5 Watts or 3500 mW - about 5.5 times more than what the whole transistor can handle.

So short answer: YES, it would fry the transistor.
... possible exception may exist if the Base current is pulsed, though. And in such case, the pulse width would have to be extremely narrow to no blow the BE junction. Also, the 625 mW rating is for the whole transistor die, so any current flowing through C-E multiplied by the Saturation voltage would add up to the dissipated power as well.

*EDIT*
Seems that CG beat me to it with the explanation.
But yeah, this is why you do need to limit current at the Base of a BJT. Otherwise, it can burn out the BE junction, or worse.

[Curious.George]

Re: Current flowing through transistor even when base is shunted to ground.

Which is technically what I'm doing here. However, hypothetically, if your source voltage for the base of the transistor was ... let's say 5V but it was capable of supplying 5 amps of current, without that base resistor, wouldn't it fry the BE junction?

The V-I characteristics of the base-emitter junction are those of a diode. I.e., at low voltages (very close to 0), there is no current flowing. Once you get to about 0.6-0.7 volts, current flows. Pushing "extra" current through the diode will increase the voltage drop... but not very much! If you look at the curve, it heads ALMOST straight up beyond 0.7V.

So, imagine how high it will be (height = current) by the time the voltage has increased to equal that 5V that you are IMPOSING.

Ans: it will be so high that the diode will fry! :>

Interesting that simply adding the transistor can complicate the LED circuit that much.

You're using the transistor to give you current gain (and increased voltage compliance). You can't source (or sink) the amount of current that the LEDs want/need so use the transistor for "muscle".

It makes perfect sense of course. As it happens to be, when these LEDs turn on in this application, they are plenty bright enough to where normally, I would not look for any reason to change any of the parameters of the circuit. Based on what you're saying, I'm most likely not giving the LEDs their full 150ma, which is fine because that only means that I'm driving them at a safer current and hence they will last longer and the risk of any of them being damaged from overcurrent is a non-issue.

You can see what the ACTUAL current is by measuring the voltage drop across the 6R8 resistor and dividing by 6.8.

Based on what you're saying, since the output of pin 4 is roughly 5 volts (minus the .7 for the B-E junction), then when I divide that by 1.7k, I get a current of 2.5ma ... which shouldn't be enough to saturate the transistor.

At the minimum hfe of ~30, that would give you ~75mA of collector current. If 75mA flowing through the LEDs and 6R8 leaves you with ~0.2V to spare, then you're saturating the transistor.

If, OTOH, the LEDs only dropped 2.2V each, you might see a different result; the "excess/residual voltage" appearing across the transistor, increasing the power dissipated in it.

Note that there are other consequences to using a reduced drive... it also reduces the amount of conducted and radiated "noise" from the circuit (assuming you are flashing them).

[In your case, you wouldn't likely be flashing them fast enough to make a noticeable difference. But, imagine you had to drive 100 sets of LEDs. You'd likely drive some subset of them with circuitry SHARED among tham all -- multiplexing. If you drove 10 sets at a time, then moved on to the next set of ten, etc. until every set had been driven... lather, rinse, repeat... and did this fast enough that a person wouldn't see the fact that the LEDs are actually "flashing very very quickly", you start to generate noise that can upset other bits of nearby electronics.

Similarly, if you tried to pulse-width-modulate the drive to give you an "intensity" control, you'd be switching the *single* set of LEDs on and off very quickly to give the illusion of being lit continuously -- albeit at a varying intensity.

However, the LEDs are plenty bright enough (so much so, that you can't look at them directly without causing some discomfort - but they will be diffused in the final application of the circuit) ... so since I'm only driving the base at 2.5ma (the source is capable of 40ma), does that mean that the LEDs are not getting the full current that I originally intended?

Likely NOT getting 150mA -- simply because of the 6R8 (redo the math).

[EasyGoing1]

Re: Current flowing through transistor even when base is shunted to ground.

[I]"Do the experiment, set power supply to 5V with 5A limit then apply it to the E-B junction of spare Transistor you have and see what happen.

Do I need a complete transistor circuit to perform the test? In other words, do I need a circuit to complete C-E before B-E will saturate? I'm thinking not???

[EasyGoing1]

Re: Current flowing through transistor even when base is shunted to ground.

Now if the LEDs only actually draw, say for example, 100 mA of current, then you won't see 129 mA flowing out of the Emitter of your BJT and will just see 100 mA instead.
However, if the LED load does want to draw 200 mA, it will become limited by the 129 mA Ie we calculated and only 129 mA of current will be allowed through the LED load. If this is not enough, then reducing the value of the series Base resistor of the BJT will increase the Emitter current. Say you go with a 510 Ohm resistor instead. Now Ib will be 5 - 0.7 / 510 = 0.0084 mA (about 8.4 mA)... and with that you can do the rest of the calculations, I think, given the previous example above.

I followed everything you said here, thank you. I get it now... my final circuit has a BC337 and its datasheet lists Hfe as a function of collector current. The range of Hfe being from 100 to 630. Since I'm using a resistor at the base of 1.7k and assuming a 5v output from the Arduino pin, my Ibe would be 2.5ma. The minimum gain is 100 so worst case, I can have 250ma at the collector, which would be limited by that 6.8Ω resistor in series with the LEDs, so since my actual source voltage for the LEDs is 9.8, and considering the voltage drop Vce, I am most likely giving those LEDs less than 150ma, which means they are being driven well within tolerance.

What I also realized reading your post, is that it would theoretically be possible to drive LEDs without a series resistor by properly selecting your base current and knowing exactly what your H value would be at that base current and voltage... which would take some experimentation to determine, but in theory it would be possible, would it not?

[budm]

Re: Current flowing through transistor even when base is shunted to ground.

"Which is technically what I'm doing here. However, hypothetically, if your source voltage for the base of the transistor was ... let's say 5V but it was capable of supplying 5 amps of current, without that base resistor, wouldn't it fry the BE junction?"

Do the experiment, set power supply to 5V with 5A limit then apply it to the E-B junction of spare Transistor you have and see what happen.

" so since I'm only driving the base at 2.5ma (the source is capable of 40ma), does that mean that the LEDs are not getting the full current that I originally intended?"

Just check the Vdrops on the resistor and the Vdrops on E-C junction then you can calculate what the current flow is and if the Transistor is in saturation or not.

[EasyGoing1]

Re: Current flowing through transistor even when base is shunted to ground.

The other conversations that spun off here I'd like to keep going as I'm learning some things that I have forgotten over the years. However, as to the original problem, I replaced the transistor with a BC337 and I added a 1.7k resistor between the base and the output pin (schematic below). I also lowered the temp of my iron and used the full length of the transistors leads and got in and out quickly with the solder and the problem went away. When the circuit is idling, the LEDs are completely dark.

[EasyGoing1]

Re: Current flowing through transistor even when base is shunted to ground.

It's not important if you are SATURATING the transistor -- i.e., using it as an ON/OFF switch.

Which is technically what I'm doing here. However, hypothetically, if your source voltage for the base of the transistor was ... let's say 5V but it was capable of supplying 5 amps of current, without that base resistor, wouldn't it fry the BE junction?

If you fail to deliver enough base current, then the collector current will be reduced. This can result in the transistor operating in its linear region. There, it will dissipate power -- heat!

E.g., if you have 9V across the LEDs (at some particular operating current), the saturation voltage -- Vce(sat) -- of the transistor is probably around 0.2 - 0.3V (i.e., the transistor manifests as a 0.2V drop in much the same way the LEDs manifest as a 3 x 3V drop).

With 9.8V supply, less 9V of LED, less 0.3V across the SATURATED transistor means you have to drop 0.5V across your 6.8ohm ballast. 0.5/6.8=74mA flowing through the resistor (and, hence, the LEDs).

[Of course, the LEDs will likely have a lower voltage drop at that reduced current so there may be a higher voltage across the resistor -> higher current through it -> higher drop across LEDs, etc. You need to figure out where you REALLY are operating the LEDs in order to know for sure]

Interesting that simply adding the transistor can complicate the LED circuit that much. It makes perfect sense of course. As it happens to be, when these LEDs turn on in this application, they are plenty bright enough to where normally, I would not look for any reason to change any of the parameters of the circuit. Based on what you're saying, I'm most likely not giving the LEDs their full 150ma, which is fine because that only means that I'm driving them at a safer current and hence they will last longer and the risk of any of them being damaged from overcurrent is a non-issue.

Now, imagine the transistor didn't "conduct enough" to give you this 0.2-0.3V saturation. Imagine only 10mA flowed through the collector. Again, pretend 9V across the LEDs and 10mA through 6.8V drops just ~.1V across the resistor. So, the remaining 0.7V appears across the transistor -- it's operating in its linear region.

How do you get just 10mA through the transistor? If your base drive is too low ("light"), then the transistor doesn't turn on hard enough. In order to saturate the transistor, you want to ensure your base current is more than your desired collector current divided by your worst case (lowest) gain AT THAT COLLECTOR CURRENT.

In your case, if you are hoping for 150mA, then you need at least 1/30th of that into the base (because the minimum gain at 150mA Ic is about 30).

Then, you have to look to see how much current the MCU's output can source (cuz the base current has to come from somewhere!). You need to ensure you can drive the base to ~0.7V with ~5mA. E.g., old fashioned tonka toy logic could never do this because it can't source current very well. (as a result, you'd typically SINK current when driving a hefty load).

If your MCU output has a low output impedance, then you'd take the Voh (output high voltage) - 0.7V (the base voltage required to turn on the transistor) and figure out the resistance required to allow ~5mA to flow.

If, OTOH, your output isn't as "stiff", then it will sag as you put load on it -- because there is effectively an internal resistance in series with your load! So, you may have to tweak the resistance LOWER in order to get the 5mA.

I did change the circuit. Here is the updated schematic:


Based on what you're saying, since the output of pin 4 is roughly 5 volts (minus the .7 for the B-E junction), then when I divide that by 1.7k, I get a current of 2.5ma ... which shouldn't be enough to saturate the transistor. However, the LEDs are plenty bright enough (so much so, that you can't look at them directly without causing some discomfort - but they will be diffused in the final application of the circuit) ... so since I'm only driving the base at 2.5ma (the source is capable of 40ma), does that mean that the LEDs are not getting the full current that I originally intended?

[budm]

Re: Current flowing through transistor even when base is shunted to ground.

"Then, you have to look to see how much current the MCU's output can source (cuz the base current has to come from somewhere!)."

+10, always look at the spec sheet to know how much the device can Source and Sink.

[Curious.George]

Re: Current flowing through transistor even when base is shunted to ground.

Now THAT makes a lot of sense. I've noticed that resistor values make a huge difference when choosing one for LEDs. I was surprised when you said anything from 500 to 5k should work fine ... I guess bias current in a transistor isn't as sensitive in terms of results as a current limiting resistor is for LEDs?

It's not important if you are SATURATING the transistor -- i.e., using it as an ON/OFF switch.

If you fail to deliver enough base current, then the collector current will be reduced. This can result in the transistor operating in its linear region. There, it will dissipate power -- heat!

E.g., if you have 9V across the LEDs (at some particular operating current), the saturation voltage -- Vce(sat) -- of the transistor is probably around 0.2 - 0.3V (i.e., the transistor manifests as a 0.2V drop in much the same way the LEDs manifest as a 3 x 3V drop).

With 9.8V supply, less 9V of LED, less 0.3V across the SATURATED transistor means you have to drop 0.5V across your 6.8ohm ballast. 0.5/6.8=74mA flowing through the resistor (and, hence, the LEDs).

[Of course, the LEDs will likely have a lower voltage drop at that reduced current so there may be a higher voltage across the resistor -> higher current through it -> higher drop across LEDs, etc. You need to figure out where you REALLY are operating the LEDs in order to know for sure]

Now, imagine the transistor didn't "conduct enough" to give you this 0.2-0.3V saturation. Imagine only 10mA flowed through the collector. Again, pretend 9V across the LEDs and 10mA through 6.8V drops just ~.1V across the resistor. So, the remaining 0.7V appears across the transistor -- it's operating in its linear region.

How do you get just 10mA through the transistor? If your base drive is too low ("light"), then the transistor doesn't turn on hard enough. In order to saturate the transistor, you want to ensure your base current is more than your desired collector current divided by your worst case (lowest) gain AT THAT COLLECTOR CURRENT.

In your case, if you are hoping for 150mA, then you need at least 1/30th of that into the base (because the minimum gain at 150mA Ic is about 30).

Then, you have to look to see how much current the MCU's output can source (cuz the base current has to come from somewhere!). You need to ensure you can drive the base to ~0.7V with ~5mA. E.g., old fashioned tonka toy logic could never do this because it can't source current very well. (as a result, you'd typically SINK current when driving a hefty load).

If your MCU output has a low output impedance, then you'd take the Voh (output high voltage) - 0.7V (the base voltage required to turn on the transistor) and figure out the resistance required to allow ~5mA to flow.

If, OTOH, your output isn't as "stiff", then it will sag as you put load on it -- because there is effectively an internal resistance in series with your load! So, you may have to tweak the resistance LOWER in order to get the 5mA.

[momaka]

Re: Current flowing through transistor even when base is shunted to ground.

I was surprised when you said anything from 500 to 5k should work fine ... I guess bias current in a transistor isn't as sensitive in terms of results as a current limiting resistor is for LEDs?

Well, BJTs are more or less current amplifiers: a small current going into the Base (for NPN) will produce a larger current going out of the Emitter (supplied by a larger current going into the Collector.)

The relationship between Base, Collector, and Emitter currents is:
Ie = Ib + Ic
where Ic = collector current, Ib = base current, and Ie = emitter current.
What determines the Emitter current from the Base current is the transistor current gain, usually noted as h_FE in datasheets (a dimensionless number/ratio.)

h_FE is not a "solid" value and changes with voltage across the transistor C-E terminals, along with Emitter current. However, datasheets usually give you a pretty good idea with the expected range.

For that 3904, for example, the minimum current gain is stated as 30 and maximum as 300.

This means that in the worst case, if the current gain is indeed 30, then 0.01 Amps (10 mA) of current going into the base can make *up to* 0.3 Amps (300 mA) of current from the Emitter. Since Ib is relatively small in most cases compared to Ic and Ie, then you can approximate Ic and Ie as nearly equal, which means the Collector can also sink close to 0.30 Amps (300 mA) of current.

Of course with the LEDs and their series resistor in your above circuit, only as much current will flow through them as the LED series resistor allows. Thus, you really only need to bias the transistor as much as is needed to drive the load.

With that said and given your above circuit...
If the MCU is supplied by 5V and it can output up to 5V on D3 pin, then with a 1-KOhm resistor between Base and D3, the Base current will be approx. Ib = 5 - 0.7 / 1000 = 0.0043 Amps. Since you're not driving the transistor too hard, the gain (h_FE) will likely be around 100 (or more, but not more than 300). But even in the worst case when h_FE is 30, Ie can be up to 30 * 0.0043 = 0.129 Amps.

Now if the LEDs only actually draw, say for example, 100 mA of current, then you won't see 129 mA flowing out of the Emitter of your BJT and will just see 100 mA instead.
However, if the LED load does want to draw 200 mA, it will become limited by the 129 mA Ie we calculated and only 129 mA of current will be allowed through the LED load. If this is not enough, then reducing the value of the series Base resistor of the BJT will increase the Emitter current. Say you go with a 510 Ohm resistor instead. Now Ib will be 5 - 0.7 / 510 = 0.0084 mA (about 8.4 mA)... and with that you can do the rest of the calculations, I think, given the previous example above.