About Using transistor BJT as a switch

In summary, the transistor will still work in saturation mode even if the connection between the base and the collector is reversed.
  • #1
ajack
6
0
In order to use BJT as a switch, it's very popular to do as the picture.
A Switch.JPG

In the picture, V_B > V_E, so there's a current flow I_BE control the I_CE. In the other way, V_B > V_E generates the current I_CE, and the lamp will light on. But, I change the connection a little:
B Switch.JPG

I'll let the V_B > V_C, it means, i change the position of C and E in that picture. And I found out that the lamp still light on. That means V_B > V_C also generates the current I_CE to make the light on. But in theory, only V_B > V_E make the light on
My questions are:
1. in saturated mode, there's no difference between C and E, you could use V_B > V_E or V_B > V_C and the lamp still light on, is it right?
2. Which way is better?
 
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  • #2
When biasing a transistor circuit, the base and the collector are usually connected to the same battery terminal. For an npn transistor (as with your first diagram), the base and the collector should have a positive voltage, and for a pnp transistor the base and the collector should have a negative voltage.

You will see that the many methods of biasing a transistor follows what I stated above.

WHen used in a switching circuit and the switch is open, the transistor is said to be cutoff. If the switch is closed, this means that electrons flows from the emitter through to the base of the transistor. The base current will enable a much larger electron flow from emitter to collector, thus lighting and LED for example. In this state of maximum circuit current, the transistor is said to be saturated.
 
  • #3
I'll let the V_B > V_C, it means, i change the position of C and E in that picture. And I found out that the lamp still light on. That means V_B > V_C also generates the current I_CE to make the light on. But in theory, only V_B > V_E make the light on
If the LED is lighting in the 2nd diagram, it's probably because you are putting the B-E juntion of the NPN transistor into reverse Zener breakdown, and that current is then flowing B-C. Depending on the current flowing, reverse Zener breakdown of a B-E junction on a BJT may not be destructive.

The second diagram is not the way you connect an NPN transistor -- not unless you want to reverse Zener the B-E junction on purpose (there are some times when you might want to do that -- advanced quiz question... When?).

For the second diagram, you should use a PNP transistor and pull the base down through a resistor as Ranger has said.
 
  • #4
When using transistor NPN in switch mode as 2nd diagram, I use the 6V source, and I use Q2SC1815. But the Q2SC1815's breakdown Volt of B-E Junction is 15V. That means the transistor still work in staturation mode. After doing as 2nd diagram, although what Ranger said is right, but I wonder the reason people follow that way is habit, or there's another reason?
 
  • #5
Well if it's not Zener breakdown of the B-E junction, then you are using the NPN transistor in reverse mode, where the beta is much lower than in forward mode. Don't use a transistor upside-down like that -- it's a mistake.
 
  • #6
berkeman said:
Well if it's not Zener breakdown of the B-E junction, then you are using the NPN transistor in reverse mode, where the beta is much lower than in forward mode. Don't use a transistor upside-down like that -- it's a mistake.
Yes, that's right. But the matters are switching the light on. In switching mode (saturation mode), the current I_C is constant (I_C doesn't increase although you increase I_B). We only care about how to light the LED on because we use the transistor as a switch in this case. So, in this case, I wonder how the LED still light on :(
 
  • #7
It is working because the NPN transistor is somewhat symmetric, with diode structures in both the B-E and B-C directions. *BUT* the transistor silicon structure of the transistor is optimized for forward operation, not reverse operation. If you did some tests of beta and frequency response, you would find that the forward (regular) mode of operation is much better than the reverse mode. That is why the C and E are labelled on the dang part. So you don't get the part put in backwards and get less performance than the datasheet specs. Your LED drive circuit is not very challenging for the transistor, so it is able to work even when installed incorrectly. If you draw a transistor like that on a test, however, you will get the problem wrong. And if you draw a transistor like that in an interview, you will not get the job. Do you understand that part?
 
  • #8
Yeah, thank for your advice. And now I understand why my teacher is really angry when I do as the 2nd diagram :P. Anyway, thank you very much!
 
  • #9
ajack said:
When using transistor NPN in switch mode as 2nd diagram, I use the 6V source, and I use Q2SC1815. But the Q2SC1815's breakdown Volt of B-E Junction is 15V. That means the transistor still work in staturation mode. After doing as 2nd diagram, although what Ranger said is right, but I wonder the reason people follow that way is habit, or there's another reason?
Using the collector as the emitter is not habit! While it is not the normal arrangement, it is not unknown. There are drawbacks, including the need to pump in a lot more base current (for the same load) because Beta is so much lower. From memory, you can get a slightly better performance as a switch in one or two parameters. I think for small collector currents, using the transistor inverted results in lower ON voltage. Or maybe a lower small signal resistance, I just forget exactly.

But certainly, as you have been told, it should not be used for normal logic switching. In any case, OP is probably in middle management by now, and no longer interested in design detail. :)
 

1. How does a BJT transistor work as a switch?

BJT transistors can work as switches by using their three layers of doped semiconductor material to control the flow of current between two terminals: the collector and the emitter. When a small current is applied to the base terminal, it allows a larger current to flow from the collector to the emitter, effectively switching the transistor "on". When no current is applied to the base, the transistor is in the "off" state.

2. What are the advantages of using a BJT transistor as a switch?

One of the main advantages of using a BJT transistor as a switch is its ability to handle high power and high voltage applications. It also has a fast switching speed and can operate in a wide range of temperatures. Additionally, BJT transistors are relatively inexpensive and easy to use compared to other types of switches.

3. What is the difference between a BJT transistor and a MOSFET transistor?

While both BJT and MOSFET transistors can be used as switches, they operate in different ways. BJT transistors use a small current to control a larger current, while MOSFET transistors use an electric field to control the flow of current. MOSFET transistors also have a higher input impedance, making them more energy efficient.

4. Can a BJT transistor be damaged if used as a switch?

Yes, it is possible for a BJT transistor to be damaged if it is used as a switch without proper precautions. If the current flowing through the transistor exceeds its maximum rating, it can overheat and potentially fail. It is important to calculate the necessary current and voltage ratings for the transistor and use proper circuit protection to prevent damage.

5. Are there any limitations to using a BJT transistor as a switch?

One limitation of using a BJT transistor as a switch is that it requires a small amount of current to turn it on, which can lead to power loss and reduced efficiency in high-power applications. Additionally, BJT transistors have a higher saturation voltage compared to MOSFET transistors, which can limit their use in low voltage applications.

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