# Transistor questions...

Discussion in 'General Electronics Chat' started by alank2, Jul 24, 2009.

1. ### alank2 Thread Starter Member

Jul 14, 2009
26
0
Hi,

I'm playing around with a 7 segment numeric LED. I've got resistors on the anode side that allow for 25ma to pass with my 5v supply. If I ground the cathode all 7 segments come up to 179ma or about 25.6ma for each one.

I have a package of NPN switching transistors laying around I got from radio shack. I measured the hfe of each one and found them to be all over the map. Here are the way they came out, 73, 85, 91, 93, 99, 101, 102, 108, 182, 185, 185, 226. The package says 200 hfe typical. Question # 1 : Is this range normal or are these not very consistent???

Question # 2, I tried a 101 hfe in my circuit. I have a 1K resistor connecting the base up to the 5V, so 5/1000 = 5ma. Shouldn't it allow 5 * 101 hfe = >500 ma to flow ??? In my case it only allows about half or around 102 ma to flow instead of he full 179ma. Am I missing something here??

Thanks,

Alan

2. ### Audioguru New Member

Dec 20, 2007
9,411
896
hFE is used when the transistor is a linear amplifier with voltage across it, not as a saturated switch. When a transistor is saturated as a switch then its HFE is not used, instead the base current should usually be 1/10th the collector current.

Also, the HFE of a transistor is measured at a certain low current. Your current is higher so the hFE is lower.

3. ### millwood Guest

that's fairly normal.

beta is useful for linear applications. in this case, your transistors are used for switches so don't think about beta (with one exception).

with 5ma base current, you would have driven the transistor into deep saturation (Vce is very low) so the current going through the LEDs are defined by its current limiting resistors and the supply voltage there.

now, if you had used a much higher base resistor, to the point where the transistor is NOT in saturation, you will then need to consider its base. But that would be a poor design for a bjt switch.

4. ### millwood Guest

the 10:1 ratio is used to define Vce(sat).

5. ### alank2 Thread Starter Member

Jul 14, 2009
26
0
Hi,

So if I want 180ma, I should allow 18ma to make sure it is saturated?

Thanks,

Alan

6. ### millwood Guest

it will depend on the transistor. take a look at its datasheet and it will give you some sense of its "on characteristics" and its beta at the desired collector current. all you need to do is to drive more base current into it for the device to stay in saturation.

most modern power transistors will have beta > 25 under heavy load and typically > 50. medium power transistors are usually above 75.

so if you want to be conservative, you need 180/50=3ma of base current.

Jul 14, 2009
26
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8. ### millwood Guest

which part are you referring to?

the basic principle is always the same: figure out how much (collector) current you need, check out the datasheet to see how much base current you need, and then work out the resistor values.

9. ### alank2 Thread Starter Member

Jul 14, 2009
26
0
Hi,

This would be the part:

The most commonly used transistor switch is the PNP variety shown in Figure 1. The secret to making a transistor switch work properly is to get the transistor in a saturation state. For this to happen we need to know the maximum load current for the device to be turned on and the minimum HFE of the transistor. For example, if we have a load that requires 100MA of current and a transistor with a minimum HFE of 100, we can then calculate the minimum base current required to saturate the transistor as follows:

Minimum base current = 100 MA / 100
Minimum base current = 1 MA

Thanks,

Alan

10. ### millwood Guest

that section is correct.

you can use the same math on your design but may not be the same numbers.

11. ### alank2 Thread Starter Member

Jul 14, 2009
26
0
Hi,

I think where I am getting lost is that the HFE I am getting is from a tool measuring it. It is a peak atlas dca55 and it says:

npn silicon transistor
current gain Hfe=103
test current Ic=2.50ma
base-emitter v Vbe=0.74v
test current Ib=4.66ma
leakage current Ic=0.00ma

So since I am using a higher current, the hfe is a lower value than what I see reported on the test tool?

How do I measure HFE in circuit? By measuring the current flowing into the base vs through the collector/emitter?

Thanks,

Alan

12. ### millwood Guest

hFE typically goes up with Ic and then goes down with Ic: transistors have little amplification capability at extremely low Ic (for small signal transistors it is usually <100ua), and the same at very high Ic (for power transistors usually > 5 - 10amp).

your test seems to suggest that hFE is 103 at Ic=2.5ma? that hFE may not hold if you run the transistor at a higher Ic level (180ma in your case).

so you either test the transistor at working Ic to be sure, or over drive the transistor.

you measure hFE in circuit by measuring its Ib and Ic.

13. ### Audioguru New Member

Dec 20, 2007
9,411
896
The datasheet for the 2N3906 shows that its minimum hFE is only 30 when its collector current is 100mA and there is at least 1v from collector to emitter. Then the base current should be at least 3.3mA but the transistor might not be saturated.

The saturation voltage loss is shown with the base current 1/10th the collector current.

Therefore your article is wrong. Only some of the transistors might saturate, maybe none.

14. ### millwood Guest

the article he quoted is based on a "minimum" hFE of 100.

the article is giving the correct approach / thought process, and you just need to apply that approach to your particular situation and recalculate.

you cannot blame the article for not applying its approach correctly.

15. ### millwood Guest

I ran a sim on 2n3906's Ic vs. Ib chart, at varying Vce.

as you can see, with Vce = 1, the max Ic you can get is about 30ma, regardless of Ib.

with Vce=5.5v, you reach 180ma Ic at about 6ma Ib. at that point, the transistor is dissipating about 1w, way bigger than its spec.

I also attached the same sim for TIP42. As you can see, it reaches 200ma at about 6ma of Ib and 1v of Vce (for a dissipation of 200mw).

so you will need a bigger transistors, like TIP to deliver more current more efficiently.

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16. ### SgtWookie Expert

Jul 17, 2007
22,183
1,728
Hello Alan,
I'll assume that you bought Radio Shack's 15 NPN transistor assortment.

You can't believe the specifications on the Radio Shack box; you have to look at the numbers on the transistors themselves, look up the datasheets, and then keep in mind that the transistors in the assortment most likely don't meet the manufacturer's minimum specifications.

The typical Radio Shack NPN assortment will contain some 2N4401's, 2N3904's, and perhaps even some 2N2222's (or their PNxxxx variants, which have similar characteristics) along with perhaps some house-marked transistors that won't cross-reference to anything, some Asian variants that you might find datasheets for, etc.

The 2N4401 and 2N3904 transistors have a useful maximum collector current of about 100mA. You'll find in datasheets that they're rated for up to 200mA or more, but that's really an exaggeration.

If you feed the base of a 2N2222/PN2222 transistor sufficient current, it can handle a collector current of 200mA+ comfortably. 15mA base current should do it for you. If you're driving it from 5v, that's a 330 Ohm base resistor.

If you're using a microcontroller to drive the base, keep in mind that many uC's are limited to 20mA current source/sink from any I/O pin, and also have a maximum source/sink current per device.

If you're planning on driving a lot of LEDs like that from a uC, consider using a driver IC such as a ULN2003 or ULN2803; seven and eight Darlington pairs, respectively. They'll save you space on your board, not to mention wiring up a bunch of discrete components. Drawback is a Vce of around 1.1v at your sink current.