Darn BJTs!!

[clipman3]

I don't post here too often cuz I can normally answer any questions I have with the Search button (and I don't have much knowledge to contribute), but today I'm stumped. Sorry for such a noob topic, but I'm having an absurdly difficult time biasing a '3904 in the active region, and would greatly appreciate some clarification and guidance. I've read countless articles, and searched multiple times but I still can't seem to grasp the idea. No matter what I do, the tranny still ends up in really farty saturation, with but a few hundred millivolts Vce. My most recent attempt (which proved to be pretty decently acceptable in calculation and sim) ended up in cutoff  . WTF.

Here's basically what I know: the base supplies the current, which is pulled in in varying amounts by the emitter, through Re, thus forward bias. Somehow the collector pulls up the current from going to ground with a positive bias? That makes no sense though, which is probably why I'm so confused. To make matters worse, this explanation says that Rc doesn't really matter all that much. I feel like I'm sorely confused through all the information I've read.

The most helpful things I've read so far on the topic have been the previously mentioned article and this PDF.

I'm just working on the typical NPN amplifier (ala LPB). Right now, I have a 330K/2.2K divider on the base, Resistors in parallel to get ~390R on the emitter, and resistors in parallel to get ~590R on the Collector. Nothing. Ended up with 9v on the collector, and a few hundred millivolts on the emitter/base.

Apologies in advance, and thank you for your time.

[bluebunny]

This is a pretty good explanation of what's going on in a simple booster and may help.  If you have the same 9V at the collector as you have coming out of your power source, then you have a short across your collector resistor (which sounds a bit low to me, btw, but don't pay too much attention to that amateur opinion!).  Gotta be a voltage drop across a resistor if there's current going through it (which there is).  Herr Ohm says so.

[clipman3]

This is a pretty good explanation of what's going on in a simple booster and may help.  If you have the same 9V at the collector as you have coming out of your power source, then you have a short across your collector resistor (which sounds a bit low to me, btw, but don't pay too much attention to that amateur opinion!).  Gotta be a voltage drop across a resistor if there's current going through it (which there is).  Herr Ohm says so.

Hey, thanks for the reply!
That is absolutely what I thought too. Couldn't find a short... But I don't think it's right anyway. I just tried to follow the tutorial in one of those articles 
I've looked at that image a few times... So is it right to guess that 10K is an acceptable Rc value for a 9v source, and that grounding the emitter would end up in saturation, while putting a larger value resistor would lower the amplification? Does the ratio of Rc to Re determine the gain, while the base's voltage divider just makes sure the signal sits at a certain point in the biasing range?

[GibsonGM]

This is a pretty good explanation of what's going on in a simple booster and may help.  If you have the same 9V at the collector as you have coming out of your power source, then you have a short across your collector resistor (which sounds a bit low to me, btw, but don't pay too much attention to that amateur opinion!).  Gotta be a voltage drop across a resistor if there's current going through it (which there is).  Herr Ohm says so.

Does the ratio of Rc to Re determine the gain, while the base's voltage divider just makes sure the signal sits at a certain point in the biasing range?

YES! 

You should get LT Spice (free download, use search engine) and take the 1, 2 hours req'd to learn how to run simulations...then you can mess around with these values and SEE what is happening!  Good questions, clipman.

[R.G.]

Here's basically what I know: the base supplies the current, which is pulled in in varying amounts by the emitter, through Re, thus forward bias. Somehow the collector pulls up the current from going to ground with a positive bias? That makes no sense though, which is probably why I'm so confused. To make matters worse, this explanation says that Rc doesn't really matter all that much. I feel like I'm sorely confused through all the information I've read.

The collector-base junction is reverse biased in normal operation. This is a kind of "trap door" that blocks the power supply from flowing, like a reverse biased diode. When you let current into the base by forward biasing the base-emitter junction, the charge carriers injected into the base have the side effect of letting the trap door open a little. This lets current in through the collector and it flows out the emitter.

And this tells you what you need to bias a BJT: collector-base reverse biased, base-emitter forward biased. For an NPN, this means the emitter is the lowest voltage, the base is higher than that by about one silicon diode drop (0.45-0.7V) and the collector is highest voltage of all three. If that is not true,

I'm just working on the typical NPN amplifier (ala LPB). Right now, I have a 330K/2.2K divider on the base, Resistors in parallel to get ~390R on the emitter, and resistors in parallel to get ~590R on the Collector. Nothing. Ended up with 9v on the collector, and a few hundred millivolts on the emitter/base.

It also tells you why what you're doing isn't working. Your base emitter voltage has to be more than 450mV for it to amplify at all. You say "a few hundred millivolts". How much iis it?

If that's too low, the base-emitter is not forward biased enough, and no current flows in the collector, so with zero current through it, the collector resistor has zero volts across it.  A quick calculation on 330K/2.2K as a voltage divider says that you have something like 59mV on the base, and that won't let the base conduct. Try raising your 2.2K to between 47K and 75K. This should bring the emitter up and the collector down - if the soldering, pinout, and wiring are OK.

[clipman3]

The collector-base junction is reverse biased in normal operation. This is a kind of "trap door" that blocks the power supply from flowing, like a reverse biased diode. When you let current into the base by forward biasing the base-emitter junction, the charge carriers injected into the base have the side effect of letting the trap door open a little. This lets current in through the collector and it flows out the emitter.

And this tells you what you need to bias a BJT: collector-base reverse biased, base-emitter forward biased. For an NPN, this means the emitter is the lowest voltage, the base is higher than that by about one silicon diode drop (0.45-0.7V) and the collector is highest voltage of all three. If that is not true,

If that's too low, the base-emitter is not forward biased enough, and no current flows in the collector, so with zero current through it, the collector resistor has zero volts across it.  A quick calculation on 330K/2.2K as a voltage divider says that you have something like 59mV on the base, and that won't let the base conduct. Try raising your 2.2K to between 47K and 75K. This should bring the emitter up and the collector down - if the soldering, pinout, and wiring are OK.

THANK YOU
It finally makes sense  thanks for taking the time to explain the concept R.G, it really helps me to understand what's going on in the process.

You should get LT Spice (free download, use search engine) and take the 1, 2 hours req'd to learn how to run simulations...then you can mess around with these values and SEE what is happening!  Good questions, clipman.

Definitely going to check it out now! I've been looking for a decent sim for a while!

[Keppy]

Good thread here: http://www.diystompboxes.com/smfforum/index.php?topic=93668.msg806813#msg806813

[PRR]

> You should get LT Spice

In my humble old-man opinion:

You should be able to rough-bias a transistor on a matchbook.

SPICE is good for 8-digit answers, but does NOT know what you want to do, and will never give you advice. You can put "ANY" values in it.... where do you start? Which way do you change? In complex circuits SPICE may be faster than a breadboard; in a 4-R 1-Q breadboard you can clip-lead a voltmeter to the circuit and live-swap resistors faster than you can point-click a mouse.

And doing it on a matchbook should (if you have any feel for numbers) soon lead to the insight that SPICE lacks.

> is it right to guess that 10K is an acceptable Rc value for a 9v source,

What is "acceptable" to you?

To the transistor: there is <1uA leakage, and a MAX limit maybe 50mA.

We often aim for collector near half, or a bit higher. Say 5V up from zero, or 4V across the collector resistor.

4V/1uA is 4Meg.
4V/50mA is 80 Ohms.
These are rough upper/lower limits for the transistor to act like a transistor.

Next guidance: Rc should be 2X to 5X lower than the load it drives. Since guitar-chain devices tend around 50K, 10K is a very fine value.

That 590r seems way-low for most audio purposes. Yes, 590r will drive a 50K load "gooder" than 10K will, but the difference is slight. Meanwhile the Supply Current is 16 times higher. In batteries, this sucks. In wall-power, we need more filter-cap to smooth the more power. Yes, supply current is often not a critical point, but never negligible, and it seems poor form to squander it without good reason.

We picked 4V across 10K as our aim-point. 4V/10K is 0.4mA.

The base-bias divider can be penciled as passing 1/10th of this current. This is because it should not sag from Base Current, and because most modern devices can have hFE>100.

So 0.4mA in the collector-emitter path suggests 0.04mA in the base bias divider. This is across the full 9V, so 9V/0.04mA= 225K total resistance.
(This is a fine point where the device matters and the designer's preferences matter. High-hFE devices don't need so much base-bias divider current. If performance is allowed to vary a lot from device to device {Hendrix going through a case of fuzz to find the best ones} then less current is needed.)

We would like now to know the Base Voltage. We know the Base-Emitter voltage is "about 0.6V", but what is "about"? In fact it could be 0.5V or 0.7V. So we can't just set "about 0.6V" and hope to work. Instead we put in an emitter resistor which will drop a voltage, which we pick to be "larger" than our uncertainty in Base-Emitter voltage.

A very conservative choice is 1.0V. This is much larger than our 0.2V uncertainty.

Above we aimed collector current at 0.4mA. The emitter current is "the same" (actually 1% high due to base current, but that's negligible). 1V at 0.4mA is 2.5K.

Now the base voltage has to be forced to 0.6V + 1.0V or 1.6V, with a total 225K resistance. 1.6V is 0.177 of 9V, so the bottom resistor will be 0.177 of 225K. 40K. The top resistor is the rest, 185K.

Now back-track. If 9V through 185K+40K divider puts 1.6V on Base, and Vbe is 0.5V to 0.7V, then we may really have 1.0V, 0.9, or 1.1V across Re (2.5K). These give 0.36mA to 0.44mA. And if Rc is 10K, these give 3.6V to 4.4V drop across Rc, Collector sitting at 5.4V to 4.6V. Which are perfectly respectable collector operating points.

[clipman3]

Thanks a ton PRR! That really helps me get through the calculations a bit better. My problem is I always wanna know the theory behind a circuit, and you guys really helped me a lot with this one.
And, I would use a breadboard for testing, but I don't have one  I need to get one. Replacing components on perfboard constantly gets really old really fast.

[GibsonGM]

You can find breadboards on Ebay very cheaply, Clipman.  $10 ea or less, cheap!   

Paul is correct about simulation software (and essentially all the other stuff he writes, LOL - guru); I suggest getting it so you can "see" what is going on, part of learning that theory you're interested in.   You can't get much from software if you DON'T learn the other things, it's just another tool.   It can be pretty hard to understand transistor action, and the many little pieces/phenomena associated with them; just keep working at it, and asking questions.  I too am an old man (42), and learn all kinds of new stuff on forums like this!     

Another good way to do the same thing, with YOU making the circuit on breadboard, is to invest in an oscilloscope.   Eventually, I did, and find it very useful.  Maybe after you've gotten in enough to know you want to keep doing this as a hobby.   In the end, it's Ohm's Law, over and over, in a few different configurations, that will get you thru! 

[nocentelli]

In my humble old-man opinion:

You should be able to rough-bias a transistor on a matchbook......

The rest of that post is a superb walkthrough, and I know rg, prr, gus, mh, et al have probably collectively explained the same dozens of times over in the past. It chimes perfectly with the gain stages I have memorized from existing circuits (e.g. lpb1, range master, hot silicon etc). I wonder if there is an online calculator that would give suggested values if provided with the other parameters. I'm not looking for something to do all the work for me, but in the same way one can understand rc filter calcs perfectly and still find Duncan's tonestack calculator useful.

However, if anyone could elucidate how the biasing arrangement in the BMP (with the base-collector resistor in place of base to +9v) affects things, that would also be great.

[greaser_au]

You should be able to rough-bias a transistor on a matchbook.

this is pretty much how I was taught to do it 

david

[jatalahd]

I also agree with PRR about the matchbook thing because doing it that way shows true understanding on how the circuit behaves. However, if one does like maths, one can derive design equations without knowing much about electronics. For a common-emitter amplifier with an emitter resistor I would do the design this way (with some help taken from electronics related textbooks):

1. Choose the value of the collector resistor Rc according to the output impedance and collector current you want to have. Typical values for Rc are 10k - 47k.
2. Normally for symmetrical output swing, the collector bias voltage Vc is desired at Vcc/2, hence the collector current can now be directly calculated as Ic = (Vcc - Vc)/Rc
3. Select the emitter resistor Re according to the gain you want to obtain. Gain from the common-emitter amplifier is approximately Rc/Re
4. A fist rule for the common-emitter design (bias stability requirement ) is that the parallel base-resistance Rb1||Rb2 <= 0.1 (Beta+1) Re. This is the basis for the following equations I have calculated:
4.1 Evaluate the base resistance Rb1 (from Vcc to transistor base) using the equation:

4.2 Once you know Rb1, then evaluate the base resistance Rb2 (from transistor base to ground) using the equation:

5. Ready 

The parameters you know beforehand are the operating voltage Vcc, the base-emitter voltage Vbe (about 0.6V) and the current gain factor Beta. Other unknowns one has chosen according to the need at hand.

Note that this step-by-step procedure only applies to the BJT common-emitter configuration with emitter resistor.

[clipman3]

Thanks Jatalahd for the step by step  This thread has been very enlightening, and the calculation examples provided along with help from LTSpice, I have been able to get the (mostly) clean gain I was looking for in my build 

However, if anyone could elucidate how the biasing arrangement in the BMP (with the base-collector resistor in place of base to +9v) affects things, that would also be great.

It's just another way to bias the base. It let's a certain amount of DC flow down to the base to make sure the Emitter-Collector junction stays open, and keeps the base floating a little above ground. As far as AC is concerned, it's essentially negative feedback at that point, because the signal gets inversed when it's amplified. It's not quite as dependable as biasing with a voltage divider, but that's the point in some circuits.

This page has a lot of info on the different ways to bias a BJT amplifier.

[PRR]

> base-collector resistor in place of base to +9v

If you _only_ have the collector-base resistor for bias...

If you want the collector "about half" of the supply....

Rb is hFe greater than Rc.

Re is not needed for bias, but usually needed to keep gain under control (or not, if you want over-load).

hFE varies over a huge range. However we do not need an exact value. If the transistor is specced hFE=100-400, then call it "200". If Rc is about 5K, then Rb is 1,000K (1Meg).

I won't ask you to trust my matchbook. Here is Idiot Assistant's neat computation:



hFE   Vc
125   6.3V
255   5.4V
510   4.7V

A 4:1 change of hFE has resulted in a 1.3:1 range of Vc. All the Vc are "about half" of the supply voltage (maybe 0.6V high because of base-emitter drop). Any transistor in the bucket will bias-up to a similar point.

A drawback of this method: the input impedance is shunted by Rb divided by stage voltage gain. If stage voltage gain is high, input impedance is low. Voltage gain for this configuration may be near 80. 1Meg/80 is 12K. There is also hIE multiplied by hFe, which for this stage will range from 7K to 30K. So input impedance could be as low as 4K.

When you come down to it... single BJT stages often will NOT do what you want, except as a puzzle in cleverness or frugality. If you want to make music, the direct-coupled Pairs are far more powerful.

[jatalahd]

However, if anyone could elucidate how the biasing arrangement in the BMP (with the base-collector resistor in place of base to +9v) affects things, that would also be great.

I am not sure what configuration is referred to here, but this is one typical (but rarely used) way to bias the BJT:


The step-by-step analytical design procedure to achieve a certain collector bias voltage Vc is:

1. Select a value for Rc to limit the collector current (gain and output impedance are affected by feedback via Rb, but still somewhat selectable with Rc). Suitable values for Rc are between 5k - 20k (for Vcc = 9V )
2. Choose the value for desired collector voltage Vc, typically for symmetrical output swing Vc is chosen to be Vcc/2
3. Calculate the value of Rb from the equation:

4. Round the obtained value to the nearest standard resistance value and you are done 

So the preconditions are again to know the operating voltage Vcc, base-to-emitter voltage Vbe (~ 0.6V) and current gain factor Beta beforehand and then select Rc according to the requirements. Then the value for Rb can be evaluated directly. The design equation above for Rb is obtained from the circuit configuration, there is no design rules involved. Because the circuit is so simple, it does not offer the same kind of stability and flexibility as the CE-stage with emitter resistor discussed earlier. Adding an emitter resistor to this circuit would be "interesting", because then there would be two feedback paths from Rb and Re. I have never tried that kind of setup.

[Kipper4]

I just found this it might not help but my understanding has increased and the guy has a few other useful tubes too

http://youtu.be/c6cmkm3UPUI

Sorry if it's no help