Transistor vs Resistor

koala

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Ok, someone tell me if this circuit with transistor is better at efficiency OR a resistor in series with LED. The circuit doesn't belong to me, I got it off a website which unfortunately I lost.

circuit.gif


Vince.
 

evan9162

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There's no difference. If the transistor is dropping a certian voltage at a certian current, and that voltage and current is the same as if the resistor were in its place, then they are dissipating the same amount of power, and it's all the same.

In fact, the transistor will be slightly less efficient, since there will also be current flowing into the base (from the resistor in the above schematic).

There's really no reason to build the above circuit. A resistor of appropriate value and power dissipation will make a simpler circuit, and will behave identically.

The BC639 transistor can dissipate up to 0.8W. They may be using this circuit because they just happen to have the transistor available, and not an appropriate power resistor. However, a 1W power resistor will be smaller than the transistor and resistor from above.
 

OddOne

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Using a pass transistor to drop the current into a LED is only really useful if you're driving the transistor with a circuit that can react to the current flow through the LED, and need to react to changes in supply voltage and/or LED current. In other words, using a transistor is only useful if it's a component of a regulator of some sort. Otherwise you're basically wasting a part best used in another manner.

If you DO want to use a transistor instead of a resistor, you'd be way better off using an N-channel power MOSFET, as the gate current for one is way lower than the gate current for a typical NPN transistor. Not much of an efficiency gain but every little bit counts...

oO
 

koala

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Urm.. I don't get it, aren't mosfet used as a switch? On and Off? My idea of using a transistor here is as current a regulation. Can a mosfet act as a transistor providing current regulation? If so, how can I wire it?

Forgot to mention, it does not need to be BC639.

Vince.
 

asdalton

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That circuit diagram illustrates how not to bias a bipolar junction transistor.

The problem is that a transistor's gain is not as predictable as, say, the resistance of a high quality resistor. So transistors are usually biased using a voltage divider, which contains either two resistors or a resistor plus a zener diode.
 

whiskypapa3

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Something is missing from the diagram. The transistor may either be nearly off, saturated or spit-sizzling hot depending on the transistor's Beta
 

Steelwolf

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According to another schematic I have, you could put a zener diode with 1 lead hooked up between the resistor and the transistor, and the other lead hooked up to a point after the transistor (or basically, hooked to the ground). This is supposed to provide a fair bit of current regulation. It was used in a 9V battery candle and according to the SPICE model, seems to hold regulation within a few mA despite a change from 9V to 5V. Cheap, simple and as efficient as a power resistor.

As to the question of a MOSFET vs Transistor, they are actually quite similar. The "FET" part of the acronym actually stands for "Field Effect Transistor". Now, a person with a degree in electronics might be able to explain all the differences between a MOSFET and a Transistor, all I can say is that a MOSFET gate is activated by level of current entering the base while a transistor gate is activated by the voltage difference between the base and the emitter (in the case of a N-type).

(I think I got that right. Someone correct me if I'm wrong. /ubbthreads/images/graemlins/tongue.gif )


My Bad: Sorry. Thanks Doug O for the correction.
[ QUOTE ]
FETs have *gates* that respond to *voltage*, bipolars have *bases* that respond to currents.

[/ QUOTE ]
 

Doug Owen

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Actually, this is a current regulator, not a very good one as many might judge, but one none the less.

Consider the base circuit for a minute. Three NiCd cells will make a battery of say 3.75 Volts. If the Vbe of our (silicon) transistor is .7 Volts, there is is 3.05 Volts 'across the 680 ohms'. 3.05 divided by 680 is 5.25 mA for base drive. The Beta (DC current gain) of typical small signal bipolar transistors is probably some thing like 30 to 50 at these sorts of levels, let's assume 40? Forty times 5.25 mA is 206 mA, this is the current in the collector circuit, provided Vf of the LED and Vbat don't force it into saturation (basically zero volts from Collector to Emitter). In practice .25 Volts is probably about as close to saturation as we can expect unless we pick a low Vsat part, meaning that we need the Vf of our LED to be under 3.5 Volts at 200 mA (which it should be).

Due to the relatively large voltage across the resistor, small changes in Vbat will cause only very small changes in Ibase, therefore very small changes in LED current, far smaller than a simple resistor with a fraction of a volt would cause. Changes in Vf of the LED will not change the current at all, unlike the simple resistor circuit (unless it gets much bigger of course), 'thermal runaway' (at least from the LED) is not possible here, but a very real problem with the simple resistor. Beta does go up with transistor temperature, but it should stabilize quite quickly (at least it did when I *tried* this suggested circuit......).

Once the value of the single circuit is 'selected' against the transistor and battery used, it should be far more stable than a simple resistor circuit.

There's a bit of well intentioned but incorrect advice in this thread. Those who would say 'this is no way to bias a transistor' must not have any formal training. It not only is a very common scheme, but being the simplest is the first method taught. Most every class A stage in cheap transistor radios (say the IF strip) is biased this way. While far from optimum, it's the simplest and therefore has a place. FETs have *gates* that respond to *voltage*, bipolars have *bases* that respond to currents. The one type of JFET that will work here, N channel enhancement mode devices, are very marginal at this low a gate voltage, most needing a bit more than 4 Volts to guarantee 'turn on', which is just what they'll do (go from no current to 'full blast'). No current regulation, a simple switch in a 'direct drive' LED circuit. Changing the resistor value will have no practical effect. As 'stiff' base supply (as suggested using a zener diode) will work *if* we also add a low value resistor to the emitter circuit so that Vbe plus the voltage drop on the resistor at design current equals Vz, but we'll need more battery voltage and an extra resistor.

The circuit is simple to the point of being primitive, but it does regulate. And it's simple enough to actually build so you don't have to guess about what goes on. Like I said, it seems I might have an unfair advantage here, not only did it teach this good stuff for a while, but I actually tried it........

Doug Owen
 

INRETECH

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A much better current circuit would be a LM317 in Constant-Current mode, R=1.25/desired current - and as long as input voltage is above the operating point of the circuit, it is not a factor in the operation - the LED will maintain itself at ALWAYS the desred current

For 20ma, the resistor would be approx 62 ohms

Circuit requires just two parts, a LM317 and the resistor
 

evan9162

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Wow, good info Doug.

I ran some numbers, and here is what I found. I assumed you started with a Vbat of 4V, with the above circuit, and a simple resistor circuit designed to deliver identical current to the LED at 4V. I assumed that the LED had a Vf of 3.3V.

Here is the data:
<font class="small">Code:</font><hr /><pre>
Transistor
Vbat Ib Ic
4 0.0048 0.194
3.75 0.0044 0.179
3.55 0.0041 0.167

Resistor
Vbat r I(LED)
4 3.6 0.194
3.75 3.6 0.124
3.55 3.6 0.069

</pre><hr />

So the transistor circuit does "regulate" better than the resistor. This is assuming, as Doug said, that the transistor remains unsaturated.

In reality, because of the way that Vf of a luxeon varies, the resistor circuit will be a little more stable.
 

asdalton

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[ QUOTE ]
Doug Owen said:
There's a bit of well intentioned but incorrect advice in this thread. Those who would say 'this is no way to bias a transistor' must not have any formal training. It not only is a very common scheme, but being the simplest is the first method taught.

[/ QUOTE ]

When I took a class in analog circuit design, we were told not to do this due to variations between the nominal and actual transistor beta. The Art of Electronics contains the same advice. I suppose that it's less of a problem if you measure the actual beta of the transistor that you are using, or if your acceptable current range isn't too strict.
 

Doug Owen

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[ QUOTE ]
asdalton said:
[ QUOTE ]
Doug Owen said:
There's a bit of well intentioned but incorrect advice in this thread. Those who would say 'this is no way to bias a transistor' must not have any formal training. It not only is a very common scheme, but being the simplest is the first method taught.

[/ QUOTE ]

When I took a class in analog circuit design, we were told not to do this due to variations between the nominal and actual transistor beta. The Art of Electronics contains the same advice. I suppose that it's less of a problem if you measure the actual beta of the transistor that you are using, or if your acceptable current range isn't too strict.

[/ QUOTE ]

I think you've correctly summed up the situation. Biasing amplifiers is always a compromise. As we try for tighter and tighter control things get more and more complex. In the end we resort to 'highly swamped' situations, but still an additional factor "r'e" ("r prime E", the small but problematic change in base impedance of the device itself) dominates.

But, here, where we literally select the final value of the resistor used (therefore the base drive), device to device differences aren't a factor.

This is not an elegant circuit, but it does win points for low parts count.

Doug Owen
 

INRETECH

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The gain of the transistor can also vary with heat, in either case - a linear circuit must dissipate the excess voltage into heat
 
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