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So I have been researching on how a bipolar NPN transistor work. Here are my understandings:

  1. In order to turn on a transistor, you need to apply a voltage (from base to emitter) to overcome the depletion region, that voltage is usually 0.7Vdc
  2. Once the transistor is "on", as the base current increases, the current flowing from the collector to emitter will also increase proportionally by a hfe constant

Armed with this knowledge, what I should expect to see in a data sheet should be the following:

  1. The minimum current/voltage to turn the transistor on (voltage needed to overcome the barrier)
  2. The maximum current allowed at the base to avoid damaging the transistor

How do I navigate the data sheet to obtain this data?

Finally, there is a DC current gain chart in the data sheet, here are my questions

  1. Why they used the collector current instead of the base current? shouldn't this makes more sense to use the base current? because you would think "if I have x amount of current in the base, I will get x amount of current from collector to emitter?
  2. What if your Vce is larger than 1.0Vdc? say 5Vdc? what would the gain be then?

Datasheet:

transistor https://www.onsemi.com/pub/Collateral/2N3903-D.PDF

Relay https://www.circuitbasics.com/wp-content/uploads/2015/11/SRD-05VDC-SL-C-Datasheet.pdf

Circuit I am trying to solve

enter image description here

Thank you

As you can see from the diagram, the relay will only turn on is if there is 5Vdc and 71.4mA across it. Knowing this fact, how do I determine the current I need at the to achieve this?

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    \$\begingroup\$ In my opinion, trying to understand an NPN transistor by studying the datasheet is a complete waste of time and you will be confused (as I can see from the questions you have). You're much better of looking at simple circuits using an NPN transistor and trying to figure out how these work. The datasheet is really only intended for engineers that already know how a transistor works and how to use it. \$\endgroup\$ – Bimpelrekkie Aug 16 '20 at 13:26
  • \$\begingroup\$ A better source of information would be: learn.sparkfun.com/tutorials/transistors/…. and after you have advanced further: electronics-notes.com/articles/electronic_components/transistor/… \$\endgroup\$ – Bimpelrekkie Aug 16 '20 at 13:28
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Vbe/Ice is logarithmic when Vbe=594mV or 0.6V/1.00mA then rBE adds ohmic drop >0.6. So 0.7 is closer to nominal currents >10mA then rBE dominates the Vbe rise above this so it becomes quasi-linear.

hFE varies widely between batches and device doping on different types due to differences in heavy doping of B-E and light doping of B-C so Ic output is easier to work with than Ib, yet both dependant on hFE and Ic.

hFE drops rapidly when Vce<1 because when Vce~0V, BC junction is now forward biased shunting the current source which is high impedance. therefor all Vce(sat) specs are done with hFE=10 @ some currents to measure saturation voltage. It depends on rBE and hFE Max because in general we learn to expect this is a smooth transition as hFE drops with Vce so hFE can drop towards 10 in most devices. But if hFE >>500 to 1500 it might be rated for Ic/Ib=50. But these are more expensive from Diodes Inc.

So we consider as a switch the gain drops towards 10% of Max hFE. yet as a linear amplifier we know hFE increases with current then declines and at max current it can affect linear sine gain with Vce<2V. This one reason why Op Amps using BJT’s do not work well within 2V of supply rail due to lack of gain and hFE effects.

In general with an emitter R base impedance becomes hFE*(rBE+RE) and since Ie=Ib+Ic because hFE is so nonlinear and generally high in linear mode we simplify design with small Re to make it more linear thus voltage gain becomes Rc/Re with linear input current Ib=Ve/Re/hFE=Ie/hFE.

Then after you try H bias you realize it is only linear with small signal output then you decide what distortion is acceptable and learn to use negative feedback with R ratios to regulate gain using multiple stages such as Op Amps.

shorter answer

  • You can think of transistors as logarithmic Vbe controlled currents sinks acting as a voltage controlled resistance with current regulated by Vbe=0.6V at Ic=1mA or Vbe =0.7V at Ic = TBD near 10mA or Vbe=0.8V at Ic =TBD near 100mA to 1A depending on bulk size, and many other factors.

  • or you can add Re to linearize it and choose Ib more linear Ic/Ib controlled emitter depending on Ve drop relative to diode drop alone.

  • then you choose type of transistor based on speed GBW, fT, rise time, power dissipation with heatsink or none, saturation voltage or high linear current gain like 2N5088/5089 (old gold added to doping for audio small current transistor with very high hFE). For binned parts choose Rohm who specialize in ABC binned hFE parts, and specialized hybrid combinations or Diodes Inc for very high hFE, low Vsat=Vce(sat) and low Rce switches rated in milliOhms.

  • the base-collector junction reducing from negative voltages to 0 then towards +0.7 affects hFE slightly Vce <2V and greatly Vce <1 with more reduction at higher currents thus for best linearity Vce>1V vs 2V depends on max used/rated current but above this it is a good high Z current source/sink for PNP/NPN’s.

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  • \$\begingroup\$ Thank you, I have added extra in my question, basically I am try to get a 5v relay to work on an Arduino. From the relay data sheet, it needs 71.4mA and 5V to work, knowing this, how to I determine what current at the base should be? \$\endgroup\$ – Brendon Cheung Aug 16 '20 at 21:46
  • \$\begingroup\$ 3mA minimum 7mA with margin. for 10%Ic \$\endgroup\$ – Tony Stewart EE75 Aug 17 '20 at 1:00
  • \$\begingroup\$ I use 10% hFE max Rule of Thumb or Ic/Ib=(sat)spec(10) whichever is greater, as long as Coil armature speed and current are adequate and Pd in driver is cool. \$\endgroup\$ – Tony Stewart EE75 Aug 17 '20 at 8:01
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"Why they used the collector current instead of the base current? shouldn't this makes more sense to use the base current? because you would think "if I have x amount of current in the base, I will get x amount of current from collector to emitter?"

Start by looking at figure 11 of your first data sheet.

Since collector current will be greater than the base current (hfe > 1), it is the current used to power a load. Since you typically work backwards from the required load (collector) current to get the base current, it makes much more sense to characterize the transistor in terms of collector current rather than base current. What you normally think is, "I need x amount of current in my load, so how much base current do I need to provide the base?"

"What if your Vce is larger than 1.0Vdc? say 5Vdc? what would the gain be then?"

You have it backwards. If Vce < 1.0 volts, your transistor is in saturation, and the gain will be quite low, typically 10 to 20 or so. Again, it will change with current levels.

Refer once more to Figure 11 of the first data sheet. Notice that hfe is characterized for Vce held constant at 10 volts.

Furthermore, your statement that "Once the transistor is "on", as the base current increases, the current flowing from the collector to emitter will also increase proportionally by a hfe constant" is only approximately true over fairly small changes in collector current. Again, look at Figure 11. hfe changes with collector current.

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  • \$\begingroup\$ This makes so much sense, you would need to know what current you need and then from there you work backwards. But one obstacle that I can't wrap my head around is the gain. If all I know what the required current is 71.4mA to drive my relay, how do I determine what is my gain so I can determine my base current? And, once I get that base current, how do I know if that base current is below the maximum rated base current before it destroys the transistor? \$\endgroup\$ – Brendon Cheung Aug 17 '20 at 2:46
  • \$\begingroup\$ Also, from the Figure 11, why is it significant to know that Vce is held at 10vdc? how does this help me to design the circuit? \$\endgroup\$ – Brendon Cheung Aug 17 '20 at 2:51
  • \$\begingroup\$ @BrendonCheung - "If all I know what the required current is 71.4mA to drive my relay, how do I determine what is my gain so I can determine my base current?" You want the voltage drop across the transistor to be as low as possible - less than 1 volt. This means that the transistor needs to be in saturation. This means that you should assume a gain of about 10, so you need 7 mA of base drive. \$\endgroup\$ – WhatRoughBeast Aug 17 '20 at 3:16
  • \$\begingroup\$ @BrendonCheung - "Also, from the Figure 11, why is it significant to know that Vce is held at 10vdc? how does this help me to design the circuit?" It doesn't. You want the transistor to be in saturation. But that is not what you asked. You asked about the gain if Vce is greater than 1 volt. I pointed out that hfe is specified for a greater voltage. To a first approximation, hfe will increase with Vce, but not by much. \$\endgroup\$ – WhatRoughBeast Aug 17 '20 at 3:18

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