Hello,

I'm working through the transistor section of the site and I'm having some problems with simple circuits I've built to test my understanding. Hopefully someone can enlighten me as to what I'm doing wrong.

I've attached a schematic of two simple circuits I've built. The idea is that as I turn the pot clockwise the led should go from completely lit to dark. (I've connected the leftmost and centre pin, looking from the top.) In practice the first version of the circuit (A) works as expected, version B doesn't (the led stays off whatever). Why not?

I've looked up a datasheet for the bc548b transistor, and have also confirmed the pins using the meter check method described here. Unless I've made a stupid mistake I think I'm hooking it up right.

Am I wrong to think they should both work?

Any suggestions are greatly appreciated, as I don't think I can progress without some pointers! Thanks in advance.

Luke

 

Circuit B is a voltage follower, since you are not using the pot as a voltage divider you are only feeding one voltage into the base, which will faithfully come out the emitter minus 0.7 volts.

Circuit A would work, except the variable resistor is 10X the value of the collector resistor. In other words, the gain of the transistor is so high that it is saturating even when it is tweaked to maxumum value. Try a 1MΩ variable instead.

 

When you were experimenting with circuit A, you may have accidentally burned out either your pot or the transistor if you made the resistance of the pot too low.

The base-emitter junction "looks" electrically like a diode, or a PN junction. If the pot was turned down to near-zero resistance, the pot and the base-emitter junction would be nearly shorting out your 5v power supply.

In circuit B, if you turned the pot to near-zero resistance, it would also be carrying the brunt of the current through the LED. Pots aren't designed to carry much current; they're generally very low wattage.

I suggest trying a new pot and a new transistor, but also use at least a fixed 330 Ohm resistor from the transistor's base to the pot so that you won't accidentally turn your pot or the transistor into a crispy critter.

 

If the transistors or pots are not burned out and the pots are set to max resistance then the transistors and LEDs will not be turned off. Then the LEDs will always be turned on because there is no resistance from base to emitter to turn the transistors off.

 

I don't think he burned out the pot since circuit A was the first and he say IT worked. I admit that the pot can burn out in circuit A. Circuit B should also work when the pot is set to nearly zero. Both are bad design anyhow.

 

Thanks for all the replies.

It seems my understanding of transistors is more deficient than I thought!

In circuit A I can see now why setting the pot to zero would short the power supply. Since I did this initially I've replaced the transistor. I've rebuilt it substituting various large resistors values in place of the pot (330R - 1M). As the resistor goes up in value the brightness diminishes - this intuitively seems what I should expect.

I'm not entirely clear why circuit B is wrong. If I had a suitably large value resistance where the pot is shouldn't it restrict current flow to the base, therefore causing the transistor to restrict current flow between collector and emitter? I thought this circuit was functionally similar to the 4th diagram on http://www.allaboutcircuits.com/vol_3/chpt_4/2.html, except with a resistor in place of the switch. (This diagram: http://sub.allaboutcircuits.com/images/03077.png).

Thanks for the help - it's appreciated.

Luke

 

Circuit B is a Common Collector, often called a voltage follower. It has just under unity voltage gain (a gain of one). The base to ground resistance is the ß X Re (emitter resistor). Basically it will follow whatever the voltage at the base is, minus the BE drop.

Lets say your ß is 100, your emitter resistance (for purposes of appoximation) is 100Ω. Your Base to Ground (via Emitter) resistance is 10KΩ. Many times the transistor ß is much higher.

So you have to have a very high resistance for the transistor gain to be swamped, and create a voltage divider between R3, the transistor gain, and R4/LED1. This is why you would be able to do better with a 1MΩ resistor for R3.

 

Ok, still struggling to get a grip on everything that's been said. But I've just wired up circuit C (see attachment) and it seems to do as I expect. Is this a sound circuit?

Thanks again.

 

That is a Common Emitter with negitive feedback. It isn't quite as straightforward as something like this...

but very similar. If you'd like I'll look up the biasing equations. The ones shown above are pretty simple, Ic (Collector Current) = Ib (Base Current) X ß (beta).

Your schematic will resist something called thermal runaway, although there are better ways to do it. If the transistor starts conducting the collector voltage drops, which reduces the base current, which in turn reduces collector current and raises the collector voltage.

 

BTW, the local online book, All About Circuits, also has a chapter on BJT (bipolar junction transistors) that is worth reading...

http://www.allaboutcircuits.com/vol_3/chpt_4/index.html

I studied using a book called "Transistor Approximations" by Malvino. Basically I went through the math and derived the various bias equations, using this book as a check. There is no substitude for deriving the equations yourself to have them down.

It is an approximation, but it will get you started. All the equations start with the premise Ic=ßIb.

 

Bill,

Isn't circuit #2 in your last post but one the same as my original circuit A, except with a resistor in place of the pot (and therefore no possibility of a short circuit)?

Yes, I'm working through that chapter. I'm also working through the transistor section of a book called electronics for inventors. I've just worked through a bunch equations and worked examples, and it's beginning to make a little more sense. It's not as intuitive as I'd hoped! Hopefully it'll become intuitive if I stick at it.

Thanks again.

Luke