transistor biasing

[monkey2410]

Please could someone explain to me what/how the calculations would be to bias the transistor in this distortion pedal circuit with a central quiescent Operating Point?

I know this should be straight forward to figure out but I have been racking my brains around it for too long reading from the internet and would understand it easier if someone could just show me how its done simply step by step.

This is the distortion pedal circuit that I would like to bias with a quiescent operating point:

[Plexi]

If you refer with "quiescent operating" to cleanest and "best" point: I'll try to biasing from R1 to get 4.5v at Collector.
I think any small (330R?) resistor from emiter to ground will improve that.

[monkey2410]

Thanks for the help, would you be able to expand on how you calculated R1 to be 330ohms though? It would be easier for me to understand if i could see the full calculations and formulas used.

[antonis]

Plexi proposes you to lift a little Emitter voltage (which now is zero volts) to leave a little space for Base-Emitter voltage variations and add a little "primitive" negative feedback...


Calculations are simple but you have to take in mind your specific transistor DC current gain to calculate Collector current and current through feedback resistor to estimate quiescent point and Emitter voltage with 330R resistor..

I say no more 'cause that kind of BJT biasing isn't included in my preferables..

[monkey2410]

I'm sorry I'm quite new to this so I'm struggling to understand what set formulas I need to be using so that I can calculate the quiescent operating point and set my circuit to operate at this.

The transistor being used is a 2n3904 so I know that the transistor gain is between 100-300 and thats literally it, I'm a newbie. If someone could just step by step explain the procedure of how to calculate the method for biasing this transistor circuit to the quiescent operating point it would be super helpful for me!

[robthequiet]

Hi, not a full answer but a starting point: Ohm's Law.

According to Ohm's Law, the voltage drop across a resistor is related to the current flowing through that resistor. The trick is to pick the resistance that gives you the desired voltage drop given the available voltage.

All resistances in a chain will drop voltages proportional to the ratio of the individual resistor to the total voltage available to the chain. From ground to V+, in a transistor circuit you must count the emitter and collector resistances as well as voltage drops across the transistor itself.

The 330(?) ohm resistor is an educated estimate, as Plexi has some experience with circuits and can probably eyeball the solution. With 9V available, the current through a 330 ohm resistor will be roughly 9V/330 ohms, giving 0.02727273 amps, or ~27.3 mA. Of course, in this case you only have one resistor dropping the whole load. (Feel free to check my math, forum)

Put another way, if you have two resistors in series from V+ to GND, and your V+ is 10V, and my resistors are 4K ohms and 6K ohms, they will drop, respectively, 4V and 6V. The current would be the same for all components.

The biasing in your circuit depends mostly on R2, 100K. If you were to breadboard this, you could measure the voltage at the base, emitter and collector of Q1. The voltage across the emitter to collector would help you to calculate the current. As Plexi notes, Q1 might need a resistor from the emitter to ground to limit the current going through the transistor. This will affect the voltages on each leg of the transistor to set the operating point, in conjunction with R2. Your question is how to do this with calculations before you build, so look at the chain from ground through the transistor and resistors in the chain, use maybe 1.2V drop for the transistor for planning purposes, and that gets you in the general range. Components vary, so ten 2N3904 transistors may not perform equally. Using resistors to eat some of the current in each leg makes the transistor less accountable for the total circuit behavior, thus more predictable.

Using the datasheet for the transistor in question, you would want to calculate the total current you need to get through the transistor to get it going in a happy range that turns it on but does not burn it, with the base voltage set where it accepts your signal and gives you the desired output. The desired output of a Fuzz Face may be different from the output of a clean boost. Calculating gain I will leave for the next poster. You can look at how other circuits are built in order to get a sense of what is generally done.

I'm sure other forum members would like to weigh in, but I feel safe recommending Ohm's Law as a start. Not intense math at the basic design level. Hope this helps.

[robthequiet]

deleted, minor arguable point -- PPS, take a look at Electra Distortion 

[PRR]

Any emitter resistor here makes no difference to DC bias point until it is so large it makes trouble.

The concept you want is that R2 = R1*hFE

The desired op-point is near 0.45mA. For 2N3904, hFE at 1mA is min 70 and likely not over 210. Modern parts skew to the high end. Pencil hFE=150. Then 10K*150 is 1.5Meg.

While hFE may vary widely from part to part, this plan is quite self-correcting. You can readily figure this by hand (and yes, Ohm's Law). I thought it would be lazy to let my Idiot Assistant figure it. Well, I had to smack it pretty hard to get three "same" parts with different hFE (Beta); but it figures a >2:1 change of hFE makes a 16% change of bias point-- pretty darn stable against device variation.

[Jeema]

monkey, the short answer is that it's probably easier to determine by experimentation.  In any case you wouldn't be able to come up with an exact answer using just math unless you know the exact gain of the transistor you're using, and those typically vary a decent amount as you are already aware.

I advise you to build the circuit on a breadboard.  When you're done, power it up - if it is properly biased, then the voltage at the collector should be half the supply voltage - in this case around 4.5v.  Measure the voltage at the collector.  If it is not around 4.5v, you will probably want to experiment with slightly higher/lower values for R2 until you get close to 4.5v.  Then it should be biased correctly.

[diydave]

I've stopped bothering looking up hfe's in datasheets and crunching numbers.
If you have a multi-meter, it can be easy-peasy for any transistor you want to use in any which configuration.

In your circuit, replace R1 / connect multi-meter between voltage source and collector, set it to read milli-amps, and read how much current runs through the multimeter.
Say you read 0.5mA. If you want to bias collector at 4.5 volts, you divide 4.5v by the current of 0.5mA, which gives you 9 kOhms (8k2 resistor will do).

Any time you alter the emiter- or feedback-resistor R2... repeat.
Any time you change the transistor... repeat.
If you want a higher or lower quiscent voltage... lower or increase the collector-resistor.

[antonis]

Obviously, Monkey is now more confused.... 

As it is, Q1 has a gain of 180 (20 times Vcc) with no load (just after C3) but it's severely "loaded"  by diode pair..
(they are set in parallel with Collector resistance..)

Keep saying it's a bad design circuit, I'll propose to delete R2, place an Emitter resistor R2 and form a voltage divider R3/R4 from +9V to GND for base bias..

Formulas are: V_Base = 9*R4/R3+R4,  V_Emitter  = V_Base - 0.7,  I_Emitter = V_Emitter/R2, I_Collector = I_Emitter,  V_Collector = 9 - I_Collector*R1

Rules of thumb are: V_Emitter = 5-10% of 9V,  10*R3//R4 < h_FE*[R2+(0.025/I_Collector)],  V_Collector = (9+V_Emitter)/2,  Gain = R_Collector/R2,  R_Collector = R1//Load

(r_e = 0.025/I_Collector is taken into account only in case of very low  Emitter resistor value or grounded Emitter..)

Additional formulas/rules: C1 = 1/(6.28*f_cut-off*[R3//R4//h_FE*[R2+(0.025/I_Collector)],   C3 = 1/(6.28*f_cut-off*100k//D1/D2) (hmmm...)

Due to 2 HPF interaction, it's advisable to oversize capacitors values 2-3 times..


But, as I've said, the above are just a matter of taste...

[monkey2410]

much more helpful Antonis, Thanks! I think I understand it better from that but I will let you (everyone) be the judge. Here is the update of the circuit, let me know if this is what you meant (antonis):

(excluding the capacitors)


This is how I have calculated the new resistor values:
Vbb= 9 x (5.6k/5.6k + 27k) = 1.546V
Ve = 1.546v - 0.7v = 0.846V
Rb = 5.6k x (27k/27k + 5.6k) = 4.64k ohms
Ieq = (Vbb - Vbe) / ((Rb/beta+1) + R2) = 1.546v - 0.7v/ ((4.64K/181) + 470) = 1.71mA
Vceq = Vcc - Ieq (R1+R2) = 9 - 1.71mA (1.8k+470) = 5.12V
Ib = (Vbb - Vbe) / (Rb + R2 (beta+1)) = (1.546v - 0.7v) / (4.64k + 470 (181)) = 0.00934mA

Is this correct working out? The only problem I see is that Vceq isn't the exact 'rule of thumb' value e.g. (9+0.846)/2 = 4.9V

I have tested the circuit by re-amping a guitar signal through it and it works, I can't really appear to hear any clipping the guitar sounds clean but I'm guessing this is probably expected if the transistor is operating at its Q point am i right?
When i ran this circuit with 1n4148 diodes I was receiving a a large amount of inconsistent hum but when I then replaced the 2 diodes with 1n4007 diodes the hum had disappeared. Is this to do with the face that 1n4148 are germanium diodes and 1n4007 are silicon therefore have a higher voltage drop??

Also antonis, I know that you predicted the beta of 180 because it is 20 times the Vcc by 'rule of thumb' but how would I know the beta precisely by working it out, if I did not know the circuit values to perform Ic/Ib?

Thanks for tolerating my lack of electronics knowledge btw everyone!

[PRR]

> is now more confused....

So am I.

[boppy100]

Thanks for tolerating my lack of electronics knowledge btw everyone!

Monkey, Thanks for asking the question.  Many of us (me) are learning right along with you. This is why I love this place, the freesharing of knowledge, expertise and friendship.

[diydave]

IMHO R3 and R4 (which form the voltage-divider for the base) are to low. This lowers the incoming signal.
But if you change them (resistorvalue x 10), you will have to adjust Re and Rc as well.... and crunch more numbers  .

Adding a capacitor to ground, parallel with R2, will boost the output more. Not sure if it will boost enough to get clipping.
The gain of the circuit isn't all that high (Rc / Re = 3,8 ).

Both 1n4148 and 1n4007 are silicone. 1n4148 is fast switching (and maybe the reason of the hum???).

[antonis]

> is now more confused....

So am I.

That makes many of us...





P.S.
monkey asked for some formulas so I tried to give them as far as VBulletin codes allow to..
(obviously, some schematics would be more helpfull but I'm also a lazy guy..)

[antonis]

I have tested the circuit by re-amping a guitar signal through it and it works, I can't really appear to hear any clipping the guitar sounds clean but I'm guessing this is probably expected if the transistor is operating at its Q point am i right?

My bad..

Formulas & Rules are valid for BJT amplifier working on its linear region..
(NOT distorting as long as its gain is less than equal to Power Supply / Signal V_pp..)

You have to first understand basics about Q-point to then intentionally bias transistor out of its linear region..
(to result in symmetrical or not clipping..)

If you want signal clipping only from diode pair, then you have to stay focused on the above rules..

diydave already told you about values of components which is the key for a "right" amp stage so you have to take into account input & output impedances...
(from what is driven and what it drives..)
Almost all of single stage amps are also used for impedance (mis)matching between previous and following stages so you have to first determine Previous & Next impedance, estimate IN (R3//R4//h_FE*R_Emitter) and OUT (Collector resistance) impedance, set a working current according to BJT specifications & Power Supply and the others will easily arise..

I know that you predicted the beta of 180 because it is 20 times the Vcc by 'rule of thumb' but how would I know the beta precisely by working it out, if I did not know the circuit values to perform Ic/Ib?

I didn't predict anything.. 
(at least, nothing about transistor's beta..)

For a grounded Emitter amplifier, biased at Vcc/2, maximun gain is 20 times Vcc (actually it's 40 times Vcc/2..)
(it's calculated from Collector resistor and Emitter intrinsic resistor, the second calculated from Collector current (0.025/I_Collector) - some people try to rise that arrangemet's gain by raising Collector resistor value not taking in mind that it results in lowering Collector current, which in turn results in raisig Emitter intrinsic resistance by the same amount..)

Thanks for tolerating my lack of electronics knowledge btw everyone!

It's just KVL & KCL implemenation.. 

P.S.
Transistor bias issues can't be covered in a few posts (maybe not even in a few hundred posts..) so take it easy..
(elementary steps, each one at a time, and only after in depth understanding the previous one - or else you'll find yourself suffering from terrible headaches..) 

Alternatively, you can always use some "Idiot Assistant" application and let it be..

[robthequiet]

Extra credit: Join two of these stages in series.

[antonis]

Extra credit: Join two of these stages in series.

Extra work: With some kind of NFB between stages.
(second Collector to first emitter, second emitter to first base or - why not..?? - BOTH of them..  )