Biasing the base pin voltage of a boost pedal?

[carboncomp]

Looking at the voltage divider on the base pin of a bunch of single transistor bost circuits there seems to be a different array of voltages from 0.8 - 2v depending on the circuit in question.

Is there an optimum value that can be obtained by the datasheet of the transistor or is it just one factor and the Base-Emitter voltage is the more important value?

[PRR]

> optimum value that can be obtained by the datasheet of the transistor

No.

It's not about the transistor. It is about the circuit.

[carboncomp]

> optimum value that can be obtained by the datasheet of the transistor

No.

It's not about the transistor. It is about the circuit.

Can you expound on that a little, surely there has to be a right and a wrong way to bias a transistor, regardless of what comes after the transistors? In boost pedals, it's mainly just a decoupling output cap, and a resistor to ground for the volume, surely the values cant be arbitrary?



For example from input to output you just have a coupling capacitor and resistor for Highpass, then the voltage divider in question for biasing, resistors R3 & R5 to set the gain across the collector and emitter, and decoupling cap and a resistor for volume.

[antonis]

What Paul tells you is any circuit based on transistor h_FE (or any other item's parameter of wide spread) is a bad circuit..
See below variations of voltages due to different h_FE (beta)


First of all, we determine Collector current which is considered equal to Emitter one (not true for BJTs of very low h_FE)..
Emitter voltage is set by Base voltage (minus about 700mV) and Emitter resistor (if any) sets Emitter current, hence Collector one..

[antonis]

Can you expound on that a little, surely there has to be a right and a wrong way to bias a transistor, regardless of what comes after the transistors? In boost pedals, it's mainly just a decoupling output cap, and a resistor to ground for the volume, surely the values cant be arbitrary?

Just apply the calculation formulas posted above..

edit:
Base voltage is  9 X 100k/(1M+100k) = 818mV (actually a bit loer due to voltage drop across Thevenin equivalent resistance but let it be..)
Emitter voltage is 0.818 - 0.65 = 0.168..
Emitter current is 0.168/390 = 430μA..
Voltage drop across Collector resistor is 430μA X 10k = 4.3V..
Collector voltage is 9V - 4.3V = 4.7V..
Piece of cake, isn't it..??

Volume pot's value is a compromise between R3 ('cause is effectibelly set in parallel with Collector resistor) & next effect input impedance ('cause upper lug - wiper part is set in series with next input impedance)..

[Rob Strand]

First of all, we determine Collector current which is considered equal to Emitter one (not true for BJTs of very low hFE)..
Emitter voltage is set by Base voltage (minus about 700mV) and Emitter resistor (if any) sets Emitter current, hence Collector one..

At low current VBE can end up 600mV to 650mA.  IIRC at 1mA it's more like 650mV.

With small voltage drops across RE a more accurate VBE makes a difference on VC.

[carboncomp]

See below variations of voltages due to different h_FE (beta)

Thank you, that is very helpful.

[PRR]

...Can you expound on that a little...

There are oceans of books on this. If you want private lessons, it will co$t you.

Semiconductor Circuit Approximations: An Introduction; Albert Paul Malvino, ISBN: 0070398984
Transistor Circuits and Applications, Cowles, Laurence G., 1968
Source Book of Modern Transistor Circuits, Cowles, Lawrence G., Prentice Hall, 1976

Cowles is not a teacher but a technician deep inside Texas Instruments. Malvino is probably the better introduction. Get THIS exact title: Malvino's later books rapidly go-off into integrated circuits. Good for getting a job, not what you need here.

[antonis]

IIRC at 1mA it's more like 650mV.

I always consider 650mV for my claculations (which are nearby 1mA)

0.7V said to meet posted image value..
(not enough time for posted image editing..)

[Rob Strand]

I always consider 650mV for my claculations (which are nearby 1mA)

0.7V said to meet posted image value..
(not enough time for posted image editing..)

No problem.  I only mentioned it in passing.

[carboncomp]

If you want private lessons, it will co$t you.

What's your hourly rate?