# What's the Logic behind selecting a resistor for NPN base?

Discussion in 'General Electronics Chat' started by strantor, Oct 3, 2010.

1. ### strantor Thread Starter AAC Fanatic!

Oct 3, 2010
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I'm want to use a TIP120 transistor (the only type I have laying around) to switch 5V (low amps) to my IC. This is for isolation from a higher voltage range (5-24V) which will be doing the switching. I realize that this is backwards from the way you normally use a transistor (normally to Amplify) but I want it to work anyways; please don't break my heart. I am trying to select a resistor for the base that will work with voltages as low as 5 volts and as high as 24V, but I really don't understand the logic behind the value of resistor. When I used a transistor before in the opposite configuration (5v to switch 24v) I was told to use a 1kohm (I did; it worked; but it was just PFM to me; didn't understand the purpose). I have been pouring over the data sheet and came to the conclusion that I need to keep the current across the base-emitter <120mA (IB Base Current (DC) 120 mA max), but >2mA (IEBO | Emitter Cut-off Current | VBE = 5V, IC = 0 | 2 mA) at a voltage of < 2.5v (VBE(on) * Base-Emitter ON Voltage VCE = 3V, IC = 3A | 2.5 V). I don't see how I can do any of the ohm's law equations for this unless I know the resistance of the Base-Emitter, which as I understand, varies depending on the current across the base-emitter (which seems quite circular to me). I am throughly confused. If someone just gave a resistance value to use, I would be grateful, but what I am really after is an explanation.
Thanks!

2. ### tom66 Senior Member

May 9, 2009
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Usually you'd use the following equation:

R = (Vs - 0.7) / Ib

0.7 = approximate base-emitter voltage drop - this does change with base current, temperature and a hundred other factors.

For your question I'll take Ib = 60mA and Vs as 5V.

R = (5 - 0.7) / 0.06 = 71.6 ohms.

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3. ### strantor Thread Starter AAC Fanatic!

Oct 3, 2010
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Found out that I CAN switch a lower voltage with a higher voltage. Tested switching 5V from collector to emitter with a 2.2Kohm resistor at base @ 5,12, and 24V.
Ibe @ 5V = 2.32mA
Ibe @ 12V = 5.62mA
Ibe @ 24V = 11.18mA

all well below the 120mA max, however at 5V I am near the cutoff of of 2mA (IEBO Emitter Cut-off Current VBE = 5V, IC = 0 2 mA). So, if changes in temp effect this, would you reccomend using a lower value base resistor in case it were to get hot inside the enclosure? if it were to get hot, would it no longer switch at 5V?

Last edited: Oct 3, 2010
4. ### marshallf3 Well-Known Member

Jul 26, 2010
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Generally the Vbe decreases as the temperature rises so you should be fine.

5. ### Audioguru New Member

Dec 20, 2007
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Everybody is confusing you because they are all WRONG!
The TIP120 is not a transistor, it is a darlington pair of transistors with lots of gain.
Its base-emitter voltage is not 0.7V, it is typically 1.8V when its collector current is 3A.

It turns on pretty well when its base current is 1/250th its collector current.
You are switching 3A so the base current should not be less than 12mA.

The IEBO is with the base-emitter backwards with 5V (its max allowed) reverse-bias because normally it has 0V or forward bias.

6. ### Jaguarjoe Active Member

Apr 7, 2010
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If you look at the "collector-emitter saturation voltage" graph that is part of a transistor's data sheet, look in the upper left corner and you'll see a value for beta or hfe. This is the gain used to fully turn on the transistor. Some could be 10, some 100, or like your TIP120, 250. The required base current will then be your collector current divided by that beta.
Subtract the Vbe of the transistor from the driving voltage for the base and use that result along with the Ib previously calculated to calculate your resistor value. R=E/I

7. ### Wendy Moderator

Mar 24, 2008
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I have to go with Audioguru on this one. The parameters for a Darlington are a bit different than a single transistors, that and you also have a built in resistor between the emitter to base to mess things up. Many of the rules of thumb do not transfer very well from a single transistor.

There is a rule of thumb to reliably drive a single transistor into saturation base emitter current should be 1/10 collector emitter current. For a Darlington I translate that into 1/100, but I'm sure you can get by with less base current. This just sets a rough way to do the job.

Just be aware that an off the shelf Darlington is a different animal than a conventional transistor. Because of the built in resistor it is different that two conventional transistors wired in a Darlington configuration.

In any case a single transistor starts with 0.6V BE drop and goes up, a Darlington starts with a 1.3V BE drop and goes up with current.

8. ### SgtWookie Expert

Jul 17, 2007
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Just to expound a bit more on AG and Bill Marsden's posts;

With a single bipolar junction transistor (aka BJT), the typical Vbe used for Ic (collector current) up to about 1/2 the maximum rated collector current is 0.7v. If you go much higher than 1/2 the maximum rated current, then start using 0.8v. If you're nearing 2/3 of the rated current, you really need to look for a more capable transistor. The majority of switching transistors use the formula:
Ib=Ic/10
for the basis of calculating the required base current.

You can think of the BE junction as a forward-biased diode. At very low current levels, it will have a low forward voltage.

A single transistor is generally considered "in cutoff" when Ic is less than 250uA; the collector current is negligible. This usually occurs when Vbe is <= 500mV.

Darlington transistors are a special case; there are two diode drops between the "base" terminal and the "emitter" terminal. The gain of the input BJT and the output BJT are multiplied. However, Vce in a Darlington configuration seldom goes below 0.7v; so power dissipation in the Darlington is rather high, particularly when compared to a modern power MOSFET, which is why they are used predominantly nowadays.

9. ### Wendy Moderator

Mar 24, 2008
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To restate what Wookie said,

A single transistor when saturated drops less than 0.1V between collector emitter. I've seen textbooks quote 0.2V for saturation, but my experience for low power applications is it is a lot less, for all intense and purposes a dead short or closed switch.

A Darlington, any Darlington, will drop a minimum of 0.6V on the collector emitter when saturated, and it could be more. This means the transistor gets hot (again, restating what Wookie said), sometimes very hot.

Sziklai Pairs is also a dual transistor configuration that resemble single transistors, and they also have the 0.6V collector emitter saturation problem. Their difference is they still drop 0.6V on the base emitter.

10. ### marshallf3 Well-Known Member

Jul 26, 2010
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This was the original question I was responding to, the answer is he's probably fine.

11. ### strantor Thread Starter AAC Fanatic!

Oct 3, 2010
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Sorry for the delay in responding to my own post, duty calls! Thanks everybody for the info, You've all been really helpful. I am straightened out now.