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I am having trouble understanding the definitions for the transistor. I am checking many different transistors and unfortunately every company has different words/symbols. For example, in one company's transistor datasheet Io was the max output current, very easy to understand and Von was the voltage required to turn on/off the transistor.

Checking different transistors and datasheets I came through different terms. On this datasheet I understand that I can have up to 10A current flow through the transistor (the load which the transistor drive to be up to 10A), up to 60V but max 20W(V*I). enter image description here

Where is it written though how much voltage/current needs to be applied to base to turn on/off the transistor?

Why every company is using different terms?

EDITED: The other transistor I was referring to, wasn't mosfet but NPN transistor. Datasheet can be found here I am applying 3.3V from a microcontroller and transistor seems to work.. however from what I read here from you I understand that its not the voltage that enables this transistor but the current.. Is that correct?

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3 Answers 3

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You are (presumably) comparing a MOSFET with a BJT.

When you are using the transistor as a simple switch, the control current is what matters in the latter, and the voltage in the former.

In the case of the ancient 2N3055, you need to put 400mA into the base in order to have Vce drop to 1.1V (max) at 4A.

BJTs are different from MOSFETs are different from IGBTs and so on, so you need to get a basic understanding of how these parts work in order to interpret the datasheets.

For example, a MOSFET such as the PSMNR60-25YLH you can see that 4.5V or 10V will turn on the transistor to varying degrees with 25A drain current flowing. At 4A the 3055 will dissipate almost 5W, whereas the MOSFET linked above will dissipate 11mW with 10V drive or 16mW with 4.5V drive. That several hundred to one reduction in wasted heat/energy is why bipolar transistors are not very popular in modern times for switching low voltages at relatively high currents.

BJTs do not really have an Rds(on) like MOSFETs. The voltage drop is not linearly dependent on current as MOSFETs are (at least until Vds starts to approach Vgs(th)). So you find Rds(on) but not Rce(on) (Zetex marketing aside).


Edit: Here is the approximate circuit and a simulation:

schematic

simulate this circuit – Schematic created using CircuitLab

enter image description here

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  • \$\begingroup\$ I just edited the question. The other transistor i was referring to, seems to be also NPN BJT :/ \$\endgroup\$
    – Kris
    Mar 22, 2021 at 14:18
  • \$\begingroup\$ Well, that's in the "and so on" category I mentioned above - a small switching transistor with built-in resistors, aka pre-biased transistor or "digital transistor". It's not a simple BJT. With a voltage applied, a certain base current will flow to the actual BJT (depending on various factors) and they guarantee some useful characteristics under those conditions. \$\endgroup\$ Mar 22, 2021 at 14:19
  • \$\begingroup\$ on the last datasheet that I added(pre-biased one), regarding KRC406 VI(ON) has typ 0.9V and max 1.3V. If I understand correct, this is the voltage that needs to be applied to base to turn on the transistor? And if yes, then since I am applying 3.3V to the base I might destroy the transistor.. or VI(ON) is not that thing? \$\endgroup\$
    – Kris
    Mar 22, 2021 at 14:31
  • \$\begingroup\$ You're not applying that voltage directly to the base. The base has no external connection. You're applying to the input resistor relative to the emitter. The datasheet says that if you apply 3.3V to the input (Vi) it will drop less than 0.2V with 5mA collector current flowing at 25°C junction temperature. The absolute maximum voltage you may apply to the input is 20V and the minimum is -5V (but you should stay well away from abs max/min). Note that if your load will draw 100mA and you apply 1.3V you'll probably burn the transistor up because 5mA << 100mA. \$\endgroup\$ Mar 22, 2021 at 14:35
  • \$\begingroup\$ i.stack.imgur.com/nEQrP.png this is the schematic I don't understand why you say that I am applying to the input resistor relative to the emitter.to my understanding I am applying the voltage(IO18) to the resistor relative to the BASE. +12 to my +load and - of my load to the collector.. when voltage is applied to the Base(resistor of the base) then collector and emitter contact each other and its like the collector is grounded so the current flows on my load from + to -(where - is the collector connected to emitter which is connected to the ground).am i totally wrong? \$\endgroup\$
    – Kris
    Mar 22, 2021 at 14:55
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On this datasheet I understand that I can have up to 10A current flow through the transistor (the load which the transistor drive to be up to 10A), up to 60V but max 20W(VI).*

Here's why: -

enter image description here

At 10 amps, the 2N3055 will drop about 2 volts across collector to emitter and the current gain will be typically about 11. In other words, to get 10 amps flowing you need to supply about 0.9 amps into the base.

I am having trouble understanding the definitions for the transistor. I am checking many different transistors and unfortunately every company has different words/symbols.

I tend to look at the graphs in the data sheets because they are likely to be more consistent between devices and suppliers. Another important graph for the 2N3055 is this: -

enter image description here

And here you can see similar information but from a different perspective. However, it doesn't help much in these old data sheets when there are discrepancies; in this 2nd graph, when 10 amps collector current is flowing and base current is one-tenth i.e. 1 amp, the \$V_{CE(SAT)}\$ is graphed at about 0.85 volts when clearly, in the previous graph, \$V_{CE(SAT)}\$ is stated as being 2 volts.

If you are looking at these types of applications, it's probably preferable to use MOSFETs.

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  • \$\begingroup\$ Regarding the first graph, if Vce is 4V lets say, how do I calculate the Ic etc? \$\endgroup\$
    – Kris
    Mar 31, 2021 at 16:10
  • \$\begingroup\$ Try using the 2nd graph @Kris \$\endgroup\$
    – Andy aka
    Mar 31, 2021 at 16:12
  • \$\begingroup\$ on the left side voltage goes slightly above 1V only :/ \$\endgroup\$
    – Kris
    Mar 31, 2021 at 16:47
  • \$\begingroup\$ @Kris good point so, in the datasheet look at the electrical characteristics table and you'll see that the hFE is listed at a saturation voltage of 4 volts and that the collector current will be 4 amps when hFE is between 20 and 100. If you want 10 amps, hFE might be as low as 5. If you need a more typical kind of solution I would recommend using a simulator and they are free (micro-cap for one). \$\endgroup\$
    – Andy aka
    Mar 31, 2021 at 16:59
  • \$\begingroup\$ so for a load 4V and 4A (14W) I need to supply to base 4/0.02 A - 4/1 A ? \$\endgroup\$
    – Kris
    Mar 31, 2021 at 18:15
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Where is it written though how much voltage/current needs to be applied to base to turn on/off the transistor?

To turn on an NPN transistor, the Base-Emitter needs to be in forward mode. So look at \$V_{BE}(on)\$ which listed at the bottom of that table: 1.8 V maximum when \$V_{CE}\$ = 4 V and \$I_C\$ = 4 A.

Note how that 1.8 V is a maximum value, it can be much less than that.

Note how this is when \$I_C\$ = 4 A, now calculate the base current \$I_B\$ using the DC current gain \$h_{FE}\$. \$h_{FE}\$ is between 20 and 100. When \$I_C\$ = 4 A. So \$I_B\$ can be between 4 A / 20 = 0.2 A and 4 A / 100 = 40 mA.

Again: this is for \$I_C\$ = 4 A, when you're switching a much smaller current, the values change.

So there is no single voltage to turn on/off the transistor. It depends on how you're using the transistor.

Why every company is using different terms?

For the same type of transistor there should not be large differences. Maybe there might be different symbols used like \$\beta\$ or \$h_{FE}\$ for the DC current gain but these all refer to the same property.

If the differences are larger then make sure you're comparing similar types. MOSFETs and NPN transistors are both transistors but have some very different properties.

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  • \$\begingroup\$ so, for this kind of transistors(NPN BJT), I need to "feed" base with current instead of Voltage as in MOSFET. And the current has to be calculated as you did above but the Voltage on the base cannot be more than 1.8V due to the restriction as stated on the datasheet. So BJT I look for Vbe on how to turn it on/off. is that correct? \$\endgroup\$
    – Kris
    Mar 22, 2021 at 14:07

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