Transistor Biasing Rule of Thumb?

[phaeton]

First off, a huge thanks and applause to R.G. for this nugget of joy that I managed to dig up.  That clears up a lot of things, even if it creates more questions.  Mind you, that relates to an 'emitter follower' type, and my question(s) mainly deal with something I'm probably going to mistakenly call a collector follower.

I was looking at things like Joe's Easy Drive, EH's  LBP2, Screaming Bird, , Hog's Foot, and stuff.  For awhile there I assumed that transistors (at least in the first 'stage') were biased so that their quiescent base voltage is always half the supply voltage-  Gives the biggest voltage swing top and bottom, right?

Well, now that (i think) I've figured out how voltage dividers work, It appears that the base input in all these are biased off center.  I'm starting to see a pattern here (at least in some EH designs), of a 10:1 ratio in resistances which would suggest a bias point consistently at or near +.818V (with a 9V supply).  Other places have the collector/emitter resistances at either a 10:1 ratio or 100:1 ratio.

Am I misinterpreting what I'm seeing here?  I.e. in the Screaming Bird/Hog's Foot, are those 430K/43K resistors not biasing the base input?  Is the base biased instead from the emitter/collector resistors, or is it a combination of all four?

Thanks for any input, directions or smacks upside the head.

[johngreene]

It maybe easiest to understand this way:

The base->emitter 'connection' is a diode (for NPN). The 'arrow' indicates the diode. So the voltage drop from base to emitter is fixed at .6V because of the diode.

The gain of the transistor is determined by the ratio of the collector and emitter resistor. If the emitter is grounded or bypassed with a capacitor, you get the maximum gain that transistor can provide but the input impedance will be at a minimum.

The voltage at the base determines the voltage that will appear at the emitter. From this you can calculate the current flow through the emitter resistor. The current through the collector is going to be very close to the same as the current through the emitter. Close enough to call it the same for this exercise. Therefore you know how much voltage will be dropped across the collector resistor.

So, if we use the Hog's Foot as an example. The base is biased at 9*(43K/473K) = .818V.
The emitter should be around .818V-.6V = .218V
The current through the emitter would then be .218/390 = 559 uA
The voltage drop across the collector resistor would then be 10K * 559uA = 5.59V
So the voltage at the collector of the transistor would be 9 - 5.59 = 3.4V
The gain of this configuration would be around 10K/390 = 25.6

Now, if you biased the base at 1/2 the supply the emitter resistor would have to be increased to get the same amount of current through the transistor (to retain the same voltage drop across the 10K). That would be 4.5-.6/559uA = 6.98K. Now you would only have a gain of 10K/6.98K = 1.4

So, in order to get any gain you would have to bypass the emitter resistor but this would dramatically reduce the input impedance of the transistor. But you could use a 390 in series with a 6.6K resistor and bypass just the 6.6K with a capacitor and basically have the equivalent of the original design but with 2 (3 actually to get 6.6K) additional parts.

One of the drawbacks of biasing the base low like this is what happens as the battery voltage gets lower. You only have .218V at the emitter to work with. The transistor will be biased off if the battery drops below 7 volts or so.

--john

[GFR]

Transistor biasing by rule of thumbs:

1- Choose a collector current. Let's say 1mA.

2- It's nice to have ~1V on the emitter for better stability (temperature, hfe variations, etc). So make RE=1V/1ma=1K. Bypass this for maximum gain with a big cap (or bypass a portion of it like John said, for less gain). You can use a lower Ve, for more headroom, but the bias will be less stable.

3- Set the collector voltage at 1/2 the Vcc minus the emitter voltage. So if Vcc=9V, the emitter is at 1V, you've got 8V left. Half of 8V is 4V, so the collector should be at 4V above the emitter, that is 5V. Dropping 4V at Ic=1ma means a 4K collector resistor. RC=3.9k is OK.

4- Since the emitter is at 1V, the base will be o.6V (one diode drop) above or 1.6V. Make a voltage divider that will give you 1.6V at the base. That gives you (R1/(R1+R2)) = 1.6/9 or R1=0.18*(R1+R2). Make the current through the voltage divider at least 10x the base current. If hfe is at least 100 then Ib<(Ic/100)=0.01mA, and you need 10x0.01mA=0.1mA in the voltage divider. For Vcc=9V, (R1+R2)=9V/0.1mA=90K. So, R1=0.18*90K=16.2K and R2=90k-16.2k=73.8K. Using R1=15K and R2=75K should be OK. If you want higher impedance here you need less Ic and/or transistors with higher hfe.

[davebungo]

Transistor biasing by rule of thumbs:

1- Choose a collector current. Let's say 1mA.

2- It's nice to have ~1V on the emitter for better stability (temperature, hfe variations, etc). So make RE=1V/1ma=1K. Bypass this for maximum gain with a big cap (or bypass a portion of it like John said, for less gain). You can use a lower Ve, for more headroom, but the bias will be less stable.

3- Set the collector voltage at 1/2 the Vcc minus the emitter voltage. So if Vcc=9V, the emitter is at 1V, you've got 8V left. Half of 8V is 4V, so the collector should be at 4V above the emitter, that is 5V. Dropping 4V at Ic=1ma means a 4K collector resistor. RC=3.9k is OK.

4- Since the emitter is at 1V, the base will be o.6V (one diode drop) above or 1.6V. Make a voltage divider that will give you 1.6V at the base. That gives you (R1/(R1+R2)) = 1.6/9 or R1=0.18*(R1+R2). Make the current through the voltage divider at least 10x the base current. If hfe is at least 100 then Ib<(Ic/100)=0.01mA, and you need 10x0.01mA=0.1mA in the voltage divider. For Vcc=9V, (R1+R2)=9V/0.1mA=90K. So, R1=0.18*90K=16.2K and R2=90k-16.2k=73.8K. Using R1=15K and R2=75K should be OK. If you want higher impedance here you need less Ic and/or transistors with higher hfe.

This contribution should be pinned up in the FAQ area somewhere if it isn't already.

[Joe]

The resistance of the base-emitter junction is a function of net base resistance and beta (hFE):
Rbe = Rb / (Beta + 1)

Then to find the emitter current:
Ie = (Vb - 0.7) / (Rbe + Re)

Collector current is:
Ic = Ie * (Beta / (Beta + 1))
(usually skipped, since the values are so close.)

From that, the collector voltage or resistance is calculated:
Rc = (Vcc - Vc) / Ic
Vc = Vcc - (Rc * Ic)

Once biased, Vb and Ve drop because of the lowered Rbe.

[GFR]

This contribution should be pinned up in the FAQ area somewhere if it isn't already.

Hmmm... I can scan an ETI article which is basically the above but in a more step-by-step way, if someone wants to host it. I've also got a similar article for designing 78xx regulated power supplies by rules of thumb. Good reading especially for begginers.

[phaeton]

Ok...

Thanks for the schooling, johngreene and GFR.....  Of course, this raises plenty mo' questions, but for right now, I'm going to only ask a few:

1)

So, if we use the Hog's Foot as an example. The base is biased at 9*(43K/473K) = .818V.
The emitter should be around .818V-.6V = .218V

and/or

Choose a collector current. Let's say 1mA.

Are either of these going to be "the ideal"?  Since this appears to be the starting point in both strategies, how do you determine the "ideal"?





2)  In R.G.'s GeoFX article on guitar effects debugging he states (for NPN transistors, 9V supply): 

For linear amplifying, ...... the base must be more positive than the emitter by about 0.4 to 0.7V for silicon, and 0.0 to 0.3 for germanium

Am I misunderstanding something, or would that suggest that the Hog's Foot is misbiased?
Also, is it even possible for the base and emitter voltage to differ anything other than a diode drop?






3) Finally, (and this might be to involved to answer simply)- 
a) Assuming I built the LBP2 (very similar to Hog's Foot) and,
b) it had a gated sound where the attack was fine, but the decay would get blatty then suddenly drop off

my first thought should be that the transistor is biased with the base voltage too low, and that I can fix this by increasing the 43K resistor or decreasing the 430K resistor?




Otherwise, I'm reading and re-reading the above stuff and trying to let it sink in.

[johngreene]

Ok...

Thanks for the schooling, johngreene and GFR.....  Of course, this raises plenty mo' questions, but for right now, I'm going to only ask a few:

1)

So, if we use the Hog's Foot as an example. The base is biased at 9*(43K/473K) = .818V.
The emitter should be around .818V-.6V = .218V

and/or

Choose a collector current. Let's say 1mA.

Are either of these going to be "the ideal"?  Since this appears to be the starting point in both strategies, how do you determine the "ideal"?

The "ideal" will depend on your design 'constraints'. Are you designing for:
a. low power?
b. high gain?
c. max linearity?
d. max signal swing?

are just a few parameters and each will require different things to be optimized and almost always at the expense of others. So, how you design, or what you choose to start with, is a function of what you are trying to accomplish.

2)  In R.G.'s GeoFX article on guitar effects debugging he states (for NPN transistors, 9V supply): 

For linear amplifying, ...... the base must be more positive than the emitter by about 0.4 to 0.7V for silicon, and 0.0 to 0.3 for germanium

Am I misunderstanding something, or would that suggest that the Hog's Foot is misbiased?
Also, is it even possible for the base and emitter voltage to differ anything other than a diode drop?

It is not possible for the base to be any more than a diode drop higher than the emitter. It can be lower, which would turn the transistor off.

3) Finally, (and this might be to involved to answer simply)- 
a) Assuming I built the LBP2 (very similar to Hog's Foot) and,
b) it had a gated sound where the attack was fine, but the decay would get blatty then suddenly drop off

my first thought should be that the transistor is biased with the base voltage too low, and that I can fix this by increasing the 43K resistor or decreasing the 430K resistor?

My guess would be that you are correct in your thinking.

Otherwise, I'm reading and re-reading the above stuff and trying to let it sink in.

Sounds like you are 'sponging' it pretty well!

--john

[phaeton]

Sounds like you are 'sponging' it pretty well!

Well, i'm trying.  I'm giving it an honest to goodness go-round, really I am,  but there's a lot of abstract and intangible stuff to comprehend all at once.

Speaking of the LPB2, it's a true story. I actually used 47K/470K because that's what I had on hand.  I measured a base voltage of 1.2V, and as I went to measure other things it died.  Dunno if I accidentally shorted some leads with the probe and blew up the transistor or what.  Ran out of time before I could fiddle with it.  I'll have to mess with it s'more later.

Otherwise, with the 10uf bypassing the emitter resistor it actually got some nice distortion for a single-transistor unit.  Whether the distortion was a result of overdriving the transistor or a biasing problem (i.e. the blatty/gated decay) i dunno, but I seriously want to work with this.

[GFR]

Ok...

Thanks for the schooling, johngreene and GFR.....  Of course, this raises plenty mo' questions, but for right now, I'm going to only ask a few:

1)

So, if we use the Hog's Foot as an example. The base is biased at 9*(43K/473K) = .818V.
The emitter should be around .818V-.6V = .218V

and/or

Choose a collector current. Let's say 1mA.

Are either of these going to be "the ideal"?  Since this appears to be the starting point in both strategies, how do you determine the "ideal"?

The "ideal" will depend on your design 'constraints'. Are you designing for:
a. low power?
b. high gain?
c. max linearity?
d. max signal swing?

are just a few parameters and each will require different things to be optimized and almost always at the expense of others. So, how you design, or what you choose to start with, is a function of what you are trying to accomplish.

Let me add a few more:

e. max battery life
f. lower noise
g. widest frequency response

And as John said, you can't have all at the same time.

[phaeton]

> You can't have it all at the same time.

Right.  I realize it's a trade-off, always.

In this case, I'm after high gain.

Didn't get to mess with anything last night like I planned.  Dog was sick and anhillated my room