Well, I'm still in high school. I'll be entering college in two years.

However, my dad holds a degree in electrical engineering. But he's gone until 6 P.M., which is an awfully long wait for me. Besides, helping me out with electronics all-evening-long is not something he really craves... .

And even when I get to college, I'm not sure I will be able to afford to study electronics. I'm not sure what my major will be, but it's likely that half of my schedule will be filled with the required courses to qualify for that major. On top of that, if the college has any sort of a liberal arts tendency, there will be more required courses. And on top of all that, there are several other subject areas I have a high interest in. I'm not sure if there will be any room left... .

But I agree, some interactive help from a teacher would be nice. For now, I'll have to subsist on my dad. He's actually great at explaining some difficult concepts to me, but the amount of the time he can devote to me is quite limited.

 

It looks like it will work... but tying the bases together like that defeats the reason for the AND gate. Might as well just use one transistor. And carry that one step further, you're back to the switch, resistor and LED idea.

Edit: Whoops, I goofed. I looked at the drawn schematic wrong.

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All those who believe in psycho kinesis, raise my hand.

 

the reason it looks like ohms law doesn't work here is because the LED has a voltage drop (Vf from the datasheet) which "consumes" voltage. Also, the transistors have a voltage drop (Vcesat from datasheet). The resistor takes care of the rest. Vf for the diode is, say 2V, and Vcesat for a small signal transistor at low current is going to be around .3 V. So the resistor "sees" 9 - 2 - .3 - .3 or 6.4V. Use that in place of 9V to determine current.

The Vf for the LED and Vcesat for the NPNs are just guesses on my part. read the datasheets but the numbers won't be dissasterously off.

 

No not like that. Usually you have your positive power supply rail at the top and the ground or negative at the bottom. Your transistors would then be drawn the other way up so that the emitters pointed downwards. That's not to say the way you've done it is wrong, just that my way fits more with convention and allows everyone to be on the same page when we talk about these kinds of circuits.


You haven't allowed for the voltage drop across the LED. The LED will have about 2V dropped across it when forward biased, and this leaves 7V which will be dropped across the resistor. Do your maths again with this in mind, and you'll come up with my answer.

You want to provide enough voltage bias on the base of the transistor to bring it out of it's linear region and into it's saturation region. In other words, you want to turn it hard on. This typically happens when 0.7V is applied across the base emitter junction. You need to work out the resistor value which will cause this to happen, and you do so as follows:
Firstly, you need to know what current will be flowing from collector to emitter. If we assume your LED is taking 8.5mA (which we worked out before) then we can use this to determine roughly the current flow into the base provided that we know the current gain (HFE) of the transistor. I don't know what transistor you're using so I can only guess, but a typical current gain would be about 150. I'll use that for now and you can adjust it to suit your transistor.
Ok so 8.5mA flows from collector to emitter so 8.5mA/150 = 56.6uA flowing into the base. This current will flow through your base resistor, and hang on to this because you'll need it to calculate your base resistor value.
Now, we already know that we want 0.7V across the base-emitter junction in order to bring the transistor into it's saturation threshold. If 0.7V will be across the base-emitter junction, then that means 9-0.7V = 8.3V will be dropped across the resistor. Use your current calculation from earlier and ohms's law will tell you that the resistor value you need it 8.3V/56.6uA= 146K.

Of course, it's more typical for an LED to take between 10ma and 20mA so you may want to go back and adjust all your resistor values and calculate everything again.

So was I, once. So were we all!

The Art of Electronics is THE Electronics book - you'll enjoy it.
It's no problem. It's nice to have someone asking tidy questions and providing evidence of work they've already done for once, believe me!

Brian

 

Thank you immensely for the explanation. This is actually beginning to make some sense to me!

As for picking the correct resistor for the transistor base... it's giving my head a little spin, but I'll just approach it mathematically for now. Perhaps one day it will make more conceptual rather than mechanical sense, but I'm quite happy for now.

I would like to verify a couple things though:

Within a single series circuit, the voltage can vary throughout, differing from one component to the next. But current is the same throughout? So for example, if I have two LEDs, and I want to run 10mA through one of them, but 20mA through the other -- there is no way to connect them in series?

Also, the amount of current that is drawn from the battery is determined "by request" of the circuit? The more components I have, the more current will be drawn?

Finally, I'm not sure I understand what "transistor HFE" is. From searching, it looks like it's called "current gain" and "transistor beta", but I'm not finding anything that resembles a more detailed and beginner-friendly explanation...

Anyways, thanks for the help again. I'm feeling much better about myself already...

 

yes, a single series of components all see the same current. however, LEDs will draw as much current as you can give them (and burn out if enough current). that's why you use a resistor.

yes, the circuit determines the current, not the power source. however, note that all power sources have resistance. In many cases it can be ignored, like this one.

Hfe is the transistor's DC current gain. a Hfe of 100 and a base current of 10mA will allow up to 1A of current flowing from collector to emitter. I say "up to" because load components (LED + resistor) will determine the current draw.

 

Hfe = Ic / Ib

Ic = Collector Current
Ib = Base Current


Also, let me reaffirm
please please please draw your transistors per convention emitters pointing down and +ve at the top

Dotnet

 

I know I'm late but here are some basics:
Kirchhoff's voltage law:
The sum of the voltage drops around a DC series circuit equals the applied voltage. In this circuit, the VD (voltage drop) across the two transistors, the 820 ohm resistor and the LED equals 9 volts.
Kirchhoff's current law:
The current flowing to a point in a circuit must equal the current flowing away from that point. In other words, if two LEDs are in series, the current flowing through one must flow through the other, unless an alternate path is provided to draw the current around one of the LEDs.
Now, about those NPN transistors:
The base thru emitter current path is a forward biased diode. While the equation for current vs. voltage drop in a diode is complicated ( i=Is(e^(qv/nkT-1)) ) we usually 'assume' the VD to be about .6 or .7 v.
Now on the collector to emitter current path we take to be a current source. Here is where hfe comes in. For every electron going thru the B-E causes a cascade of hfe electrons thru the C-E pathway. So, if I have a hfe of 150, and a base current of 1 milliamp, I'll have 150 milliamp thru the collector and 151 milliamps thru the emitter (the current thru the base plus the current thru the collector). As I increase the base current I will continue to get this gain until 'saturation' which is when other limitations in the circuit prevent the C-E from being able to keep up.
Because the VD thru the base-collector is an exponential relationship, the gain is not linear with respect to the base voltage.
For the designer, the current thru the transistor is the first consideration. If you plan for 20 ma thru the transistor and the data sheet give a minimum hfe of 100, then you need a base current of 200 micro amps (200 micro amps X 100 = 20,000 micro amps or 20 milliamps). I usually double this to ensure saturation. The base resistor is critical. The easiest way to destroy a transistor is to allow too much current thru the base. With a 9v battery, a desired current of 100 micro amps and .6 v drop thru the base we have:
E= I/R
(9 - .6) = (400 micro X R)
8.4 / 400 micro = R
R = 21 k ohms
This represents the base resistance.
So, now that you have the transistors turned on, you look at the LED circuit. Specifically, the VD of the LED:
From Wikipedia:
Typical voltage drops
Although there are small variations between individual diodes, in general the voltage drop depends on the color of the LED and the brightness. The following table summarized some of these voltage drops for various LEDs.
LED Voltage Drops
Diode type Typical Voltage Drop (Volts)
non-high-brightness red 1.7 volts
high-brightness, high-efficiency and low-current red 1.9 volts
orange and yellow 2 volts
green 2.1 volts
bright white, bright non-yellowish green, and most blue 3.4 volts
bright blue and some specialized LEDs 4.6 volts
Most manufactures recommend 10 milliamps for the 430 nm blue diodes, 12 mA for the 3.4 volt types, and 20 mA for lower-voltage LEDs.
The LED resistor is:
E=IR
(9 – 1.7) = 20 milli x R
7.3 / 20 milli = R
365 = R
Try these values and see what you get. Usually I check the pins and hfe on a transistor just prior to installing it. It is easy to get different types mixed up. Lots of Luck!

 

Thanks again for the help guys. You gave me a lot to digest...

 

I want to applaud you for being genuine in your endeavor to experiment and try to understand your task at hand. It's refreshing to see someone actually try something on their own merits as well as inquire for assistance in an earnest attempt to learn something in the process. If you stay on track, you show promise of success.
Oh, and even though the circuit in question was a basic one, it wasn't one of wanting to screw up tv reception, jam cell phone communications, or knock out the power grid of the entire eastern US seaboard!

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Don't make me reach through this monitor to slap you a good one!

 

Get a student Loan. You could always stay in college and get many degrees instead of one. I have to take courses (like you said) before I can get into electronics, but its worth the wait.

Do you buy any chance have an electronics (or tech) division at your school? You could try going to them.

 

Please note that the load ( Led and it's resistor) must be placed before all NPN type transitors.

4 wrongs with the below pic.

No wonder you are confused.

Attached Images

4wrongsA.gif (20.1 KB, 16 views)

 

I thought it didnt matter where the resistor on the collector/emitter goes, since Ic=Ie

 

Hmm nice circuit image! Might I ask , how the programm is called you used to create them ?

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i² = j² = k² = ijk = -1

 

With PNP type, the loads (Leds and it's resistor) are place after the transitor.

When you learn how to use your parts correctly, then you can start to use math formulas.



I edited them with Microsoft Paint / Mspaint.