Discussion in 'General Electronics Chat' started by zoom, Jul 11, 2014.

1. ### zoom Thread Starter New Member

Jul 11, 2014
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Since I have never used a transistor before, I am a bit unfamiliar with them.

I need to drive a 12Volt LED strip, which draws 0.4 mA for 1 Meter.

I found the tutorial below and understood almost everything but one thing;

http://teachmetomake.wordpress.com/how-to-use-a-transistor-as-a-switch/

When it comes to calculating the base resistor he checks the value of Vbe(sat) at 1 A even though the load requiers 1.25 A when stalled and 76mA when with no load.

I think he should select the least possible current, since less current corresponds to less Vbe(sat) value and this will eventually increase the current drawn from the base, right ?

5 - Vbe = R * I

where I is selected 20mA.
If the load current drops, Vbe will also drop and this will result in higher higher base current (I) for a given base resistor (R) . Am I wrong ? This point confuses me.

Also he claims that the selected base resistor is the maximum value. However if I decrease it, base current will increase and that may draw much more current than the arduino can handle (40mA).

Appreciate if you explain this phenomenon.

One more question;

Someone told me that you don't need to mess with base resistor if you're just intending to turn the transistor on and off as a switch. Just go ahead and select something around 220 or 330. Why is that so ?

And the last question

If I want to control the value of the voltage flowing from collector to emitter by base current, should I use a different type of transitor ? or what should I do ?

Thank you.

2. ### ericgibbs Senior Member

Jan 29, 2010
2,413
369
hi,
The Collector current, Ic is approx = Ibase * Beta [ Gain].

E

3. ### zoom Thread Starter New Member

Jul 11, 2014
29
0
However, it is still a bit unclear. The motor draws 76mA when it is not loaded. So, this is the least possible current that this motor will require. Why does he still uses 1A for the Vbe selection.

Okay, so let me ask the question that way; Assume there is a load attached to the emitter which requires 10 Amps. Would I still look at the 1A and the corresponding Vbe value and use it ?

4. ### ericgibbs Senior Member

Jan 29, 2010
2,413
369
hi,
If he used Ic= 76mA to calculate the required Base current, the transistor would not be fully switched ON [ have a low voltage across the transistor, say about 0.2v]when the motor is stalled or when the motor shaft is being 'loaded' .

This would mean the transistor will get very hot and possibly the motor would not run.

If Ic= 10A then Ib should be 1A to ensure that the transistor is saturated, ie: switched hard ON.
E

5. ### zoom Thread Starter New Member

Jul 11, 2014
29
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Okay then if the maximum current we might require is 1.25 why wouldn't we use 1.25 directly to find the Vbe ? Then we could be safer.

As for Ic = 10A case, you said 1A would be required by assuming Beta = 10 ?

How about the Vbe value that I have to pick up for Ic = 10A case to determine the base resistor ? This time I have to check for Ic = 10A on the graph and pick the corresponding Vbe value, let's say something around 3V according to this datasheet. Right or wrong ?

6. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,297
482
What Vbe got to do with it?

Vbe is built into the transistor. You, the user, don't find it, don't change it because you CAN NOT change it. That is why Vbe is given in the datasheet by the manufacturer.

7. ### zoom Thread Starter New Member

Jul 11, 2014
29
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Then why there is a Ic vs. Vbe(sat) graph in the datasheet ? This is the point confusing me.

8. ### ian field Distinguished Member

Oct 27, 2012
4,323
769
Sounds like you want to operate the transistor as a saturated switch - ie the transistor switches hard on and Vce-sat is the minimum it can be. If the transistor is not turned fully on, there will be a C/E volt drop - multiplied by collector current gives you unwanted dissipation.

Look at the transistor's data sheet for gain spread (max & min values) - assume the worst and take the lower figure and divide your target Ic by it. Now make sure the result of division is less than the current capability of your micro, if not you will need more stages - perhaps a Darlington if your micro outputs a high enough logic high.

The Ib value above has no margin for error such as resistor or PSU tolerance - you should reduce the calculated resistance anything from using the next preferred value down, to halving it depending how much drive you have to spare - you want to make sure the transistor is switched hard and not dissipating power in the linear region.

Much easier is to use a logic level MOSFET, but be careful to check Vgs-thr if its a 3.3V system - the only gate resistor you need is a "stopper" to prevent parasitic oscillation at the switching edges, about 22 - 47R is about typical.

Last edited: Jul 11, 2014
9. ### Brownout Well-Known Member

Jan 10, 2012
2,375
998
Not sure what shteii01 is talking about vbe certainly can be changed. You can calculate the value from: vbe = VTln(Ic/Iss) where VT=25mV at room temperature and Iss is a device parameter. Unfortunately, Iss isn't typically published in the datasheet.

10. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,297
482

Assume Beta of 10.
Ib=1.25/10=0.125 A

Assume Vbe of 0.7 volts.
Vinput-Rb*Ib-Vbe=0
Vinput-0.125*Rb-0.7=0
You have two variables here Vinput and Rb that you, the user/designer, can "play" with to get your circuit to work.

If you know/chose Vinput, you can find Rb.
If you know/chose Rb, you can find Vinput.

11. ### zoom Thread Starter New Member

Jul 11, 2014
29
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Okay but in opposition to your claim of constant Vbe, there is graph of Vbe changing by Ic in the datasheet.

12. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,297
482
Figure 2 on Page 3?

13. ### zoom Thread Starter New Member

Jul 11, 2014
29
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Exactly it is

14. ### shteii01 AAC Fanatic!

Feb 19, 2010
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Right.
Did you notice how the graph of Vbe(saturation) is gently sloping up until it gets to Ic=5 A? After it gets past Ic=5 A, it sharply goes up? Did you notice that?

In my education, such gentle slopes often approximated by a horizontal line. Meaning that at the interval from 0.1 to 5 amperes the Vbe(sat) is nearly constant. Notice that Vbe(sat) goes up from 1.2 to 1.7 volts, the change is .5 volts. So you change your Ic from .1 to 5 amperes, 5/0.1=50 meaning you increased you Ic 50 times! and what change did you get in Vbe? You got a measly 0.5 volts change in Vbe. Get it? A 50 times increase in Ic gave you a dinky half a volt increase in Vbe.

The Figure 2 on Page 3 is useful in that it shows you overall behavior of the part. It shows that you have a fairly steady LINEAR relationship between Ic and Vbe where up to Ic=5 A the Ic will change a lot (from 0.1 to 5 A), but Vbe will change very little. But once you get past Ic=5 A, Vbe is not steady anymore, it changes rapidly, a small change in Ic will increase Vbe more and more.

15. ### zoom Thread Starter New Member

Jul 11, 2014
29
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Ok thank you, got the drift

Well, not all the data sheets have that kind of graphs. Most of them have maximum Vbe(sat) values. For example for this transistor we consider, it has 2.5 V

Assume I don't have the graph. How come I determine what voltage value to use for Vbe ?

Thaks again.

Jan 10, 2012
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