Quiescent point transistor

[Amoebius]

Hello,

I have difficulties to understand the calculation of the quiescent point of transistor in a simple class a amp.

I have read several tutorials and got 2 ways of doing it.

Some say: Icq = ((Vcc-Vre)/2)/Rc

others:      Icq = (Vcc/2)/Rc

Which method allows for maximum "swing"?

I attached a circuit i am experimenting with. R4 =Rc, R3 = Re

[Zero999]

Why not try both and simulate each one?

The base voltage, sets the voltage across the emitter resistor, causing the transistor to act like a constant current sink. The voltage across the emitter resistor is equal to the V_B-V_BE.

[MikeK]

Both of these are worth watching:

[Amoebius]

Ok, thanks for the answers.

I simulated both and couldn't see much difference. But my understanding of electronics is still very limited...

At this point I think, the voltage, that the amplified signal has to "swing" ist Vcc - Vre when the transistor is active. So one schould bias at half of Vcc - Vre for max swing.

[MrAl]

Hello,

I have difficulties to understand the calculation of the quiescent point of transistor in a simple class a amp.

I have read several tutorials and got 2 ways of doing it.

Some say: Icq = ((Vcc-Vre)/2)/Rc

others:      Icq = (Vcc/2)/Rc

Which method allows for maximum "swing"?

I attached a circuit i am experimenting with. R4 =Rc, R3 = Re

Hi there,

You use (Vcc/2)/Rc as long as you have it DC biased to 1/2 of Vcc as you would think.
The reason is because anything that flows through the collector goes though Rc and Rc drops voltage according to Ic.

If you happen to have it biased to some other voltage, then the collector current is:
Ic=(Vcc-Vc)/Rc

You never have to consider the emitter voltage unless of course it is not biased right.
You also never have to consider the collector to emitter voltage unless it is not biased right.

One thing i am working on now and then is if it is better to calculate 're' (lower case) along with the DC bias point rather than calculate the bias point without 're' and then add 're' later.  I was just about to post some formulas on another site for this very circuit, and should be able to do that soon.
If you like i can post at least one formula here for your actual circuit.  You probably will want to use some math software though to do the calculations because when it comes to the gain for example every component you have in the circuit comes into play so they all must appear in the final formula.  For the DC output bias point though the formula is a little simpler but you probably want to use math software anyway so you can calculate results faster and even graph some voltages or currents as you change something like the input AC voltage.

[Amoebius]

Thanks, Mr AI.

Please, post the formula, I am very interested. With Vre I meant V emitter resistor, not 're. I should have written it VRe i guess.

Thanks for the reply

I made the circuit on a breadboard and tested it with a smartphone for input and 250 ohm headphones and to my surprise it sounded almost descent.

[MrAl]

Thanks, Mr AI.

Please, post the formula, I am very interested. With Vre I meant V emitter resistor, not 're. I should have written it VRe i guess.

Thanks for the reply

I made the circuit on a breadboard and tested it with a smartphone for input and 250 ohm headphones and to my surprise it sounded almost descent.

Oh that's great. Nice when something works.

Here is the formula (B is Beta):
VoDC=(B*E2*R3*R6+E2*R3*R6+B*E2*R2*R6+E2*R2*R6+B*E3*R3*R4-B*E2*R3*R4+B*E3*R2*R4+E2*R2*R3)/(B*R3*R6+R3*R6+B*R2*R6+R2*R6+R2*R3)

Note you have to sweep Beta to verify your design.
The circuit components shown represent just one particular design.
R4 was chosen to give exactly 5 volts DC out at the collector but you can specify another R4 and adjust one or both of the input resistors to do the same.

and the circuit: