1
\$\begingroup\$

I'm learning about transistor configurations (particularly in relation to audio circuits) and am having difficulty analysing this fairly straightforward common emitter stage from the Instrument input of a Roland RE-201 Space Echo. Specifically, I am confused how to find the DC bias voltages for this circuit. If R3 220k was connected to Vcc this would be a standard 4 resistor setup but with it being tied to the collector (which is connected to Vcc through R8 27k) I am having trouble calculating the voltages. Every explanation I have looked at so far brings in the transistor beta at some point in the calculations but am wondering if it is possible without using this (or even without estimating beta) with the known resistors and Vcc of 17v.

I tried looking at it in terms of a voltage divider but don't end up with the 5v Vc that the schematic shows:

i.e.

Vb = 17 * ( 47k / [47k + 220k + 27k] ) = 2.71v

Ve = Vb - 0.65v = 2.06v

Ie = 2.06 / 560R = 3.6mA

Ic = Ie

Vc = Vcc - (27k * 3.6mA) = wrong!

I think I'm perhaps incorrect with my first line of calculation as there may possibly also be a resistance to ground from the collector through the transistor (?) which I imagine should be considered, but I don't know how to factor this in.

Can anyone offer some advice?

Many thanks in advance

EDIT: schematic re-uploaded to show that C6 / R10 are connected to other audio inputs which go through a separate amp stage and so bypass Q11.

enter image description here

\$\endgroup\$
5
  • 1
    \$\begingroup\$ Did you forget the schematic? \$\endgroup\$ – Transistor May 17 '20 at 22:32
  • \$\begingroup\$ Oops yes thanks for pointing that out, I deleted to edit and forgot to re-add. Have uploaded now. \$\endgroup\$ – SteveUK May 17 '20 at 22:55
  • \$\begingroup\$ @SteveUK Where does R10 and C6 go to, exactly? \$\endgroup\$ – jonk May 18 '20 at 3:31
  • \$\begingroup\$ @SteveUK Never mind. I found the site for the manual. Nice site. \$\endgroup\$ – jonk May 18 '20 at 4:11
  • \$\begingroup\$ I've annotated and re-uploaded the schematic for clarity \$\endgroup\$ – SteveUK May 18 '20 at 9:53
0
\$\begingroup\$

R3,R6 and Q11 forms a kind of VBE (Vth for MOS) multiplier and almost all the supply current will flow through the collector and emitter, therefore we can write $$ (V_{cc}-I_c \cdot R_8) \cdot R_6/(R_3+R_6)=V_{BE}+I_c \cdot R_7 \, .$$ From that you get Ic, since VBE is approximately constant due to the exponential BJT transfer characteristic. Once you got Ic, you get the emitter and collector voltage, and thus the base voltage as well.

In first estimates you can assume that beta is infinite. At least for discrete implementation. Solve the equations to get an insight into the circuit, and introduce beta in the second round only if it is needed. Usually beta is not needed. For calculation done only to gain insight and understand the circuit behaviour you never take beta into account. So I am little surprised that you have seen it in every explanation so far.

PS: your first equation would be only valid, if no transistor would be there. The current of the transistor, which is way higher than the current of the resistive divider will develop a current drop on R8, which is not reflected in your first equation.

\$\endgroup\$
0
\$\begingroup\$

This circuit does have feedback, to "stabilize" the operating point.

That the ratio of Rcollector and R_collector_base is 10:1 is an indication that BETA >> 10 will mostly be ignored.

Notice I wrote "mostly".

\$\endgroup\$
0
\$\begingroup\$

The driving voltage for the voltage divider R5-R6 is the DC collector voltage. In the first line of your calculation you have completely ignored the DC collector current. Start with Ie=Ic (unknown value).

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.