Hello
I am trying to figure out the input capacitor to an emitter grounded amplifier that is biased by a voltage divider,with a Rc and Re resistors.The Re is not bypassed.What I trying to do is use 60 hz as a testing freuency for the amplifier.I am using a transformer that has a output of 12acv.I had to place two resistors in series to get a small voltage across the second resistor which is about 150 ohms that had a drop of about .009 volts. Iam assuming that the series resistors that is the output of the ac transfomer would be the generator resistance.The input to the grounded emitter which would be according to forum information that Rin (tot) =R1|| R2|| re'.How would I calculate for the input capacitor to the transistor? Also what about the D.C. input resistance,how is that used for the calculation.

 

You should draw a schematic of what you actually have showing all the values.

At 60Hz the capacitive reactance will be [2653/C(in uF)] Ω. So 10uF will give a reactance of 265 Ω which is probably significantly less than the amplifier ac input resistance.

BTW the ac input resistance contribution from the emitter seen at the base would be approximately β(re'+Re) - not re'. Presumably you are denoting the dynamic emitter resistance as your term re'. Of course you would have to add the contribution from the base bias resistances - R1 & R2 in your case...? That's why a schematic would help ....

Also this is probably not a specific radio and communications question.

 

Hello

Thid image is a sample of what I have in experiment.The only difference is that the a.c. voltage is 12acv and the D.C. is 9 volts.
R1 is on the a.c. side is 220 k and R2 on the a.c. side is 150 ohms.
The D.C. side with 9 volts has R1 as 220 k and the R2 is 47k. I did
put the transitor in there but Rc is 1k and Re is 220 ohms. I was wondering if the two series resistors in the a.c. circuit affected the coupling capacitor,including the input resistance of the transistor,in this case the R1 and R2 in the D.C. circuit of the 9 volts? Some where in the forum I understood that the coupling capacitor can be reversed in polarizartion if the power rating is to large(3volt using a 16 volt rating).It seem as if the R2 on the a.c. side is in parallel with the capacitor and the D.C. side of the capacitor is in parallel as far as the diagram circuit.But with the transistor following the coupling capacitor R1,R2,and Rin(tot) would be in paralle.Do the voltage on each side of the coupling capacitor have a important affect on the chargeing and discharging (time constant).

 

That drawing is much too small to make out.

 

That drawing is much too small to make out.

That's why we have the Preview Post button.

 

Hi
What is Preview button ? I am not using an high IQ cam drawing system,but a microsoft paint software that came with the system. I copy it to my E- mail then to allabout circuits up load. How it get so small that's ???? to me.but let me explain my confusion in text again until I find how to to enlarge the config.
The Z of the transformer and the Z of the series resistors that is use to attenuate the 12 a.c. volts to about 9 milivolts equal the R-generator,I assume.(This transformer is plug into the wall socket of 120 a.c.v. with output of 12 a.c.v.)The divider for D.C. bias of the emitter ground of the amplifer and the input resistance of the transitor plus the Rg,does this contribute to time constant of the coupling capacitor between the Rg and the input of the transistor.
Looking at this from another angle ,There is resistance on one side of the coupling capacitor and resistance on the other side of the coupling capacitor, are they considered as controlling factor toward the Time constant ? If so how should they be calculated ? Not considering the coil of the Primary and secondary winding of the transformer but only the two series resistors and input to the transitor including the two bias resistors for the transistor.
Also I not sure what part the D.C. voltage plays inthe time constant of the coupling capacitor. It is always stated about the a.c. formula regarding the input resistance but not the effect of the D.C, if there are any.

 

It's right next to the Submit Reply button.

 

Wow I did't see that.I must be losing fucus or something.But what about resistance on each side of the coupling capacitor ?How should that be calculated? I wonder If they are in series on one side and Pallellel on the bias side of the transitor.If the equavalent is two resistor in series with the coupling capacitor.I don't know.

 

Wow I did't see that.I must be losing fucus or something.

Losing fucus can be very frustrating.

 

Losing fucus can be very frustrating.

Yeh I know. that's one of my biggest problem in this situation ,I got three minutes on the library computer I must continue this at a later date.Don;t forget the resistance on both sides of the coupling capacitor I need to understand that to advace .

 

He didn't get it.

 

Hello

I am trying to get a drawing of the circuit scanned to be posted. But I only was trying to find out if the resistance on the a.c. side of the coupling capacitor and the input resistance to the transistor including the voltage divider bias have a affect on the time constant of the coupling capacitor.Since the E-book only shows a capacitor in series or parallel to a resistor with both connected to the same power source,either D.C. or a.c. and sometimes the ac and D.C.source would be in series.But in this case,The resistance of the a.c.source which is made of the transformer coil and two resistors in series to produce a millivolt of output upon one side of the coupling capacitor and the other side of the coupling capacitor which is connected to the bias voltage across the base resistor (which is R2 in the voltage divider of the bias circuit).
One side of the capacitor has a very small a.c. voltage (millivolts) and the other side has a large voltage of one voltage which is D.C.The problem is what amount of a.c.voltage would be transfered through the coupling capacitor that would cause a perceived fluctuation that would not be attenuated by too much of bias voltage and resistance from source resistance and and bias resistance of the transistor.
Would the coupling capacitor be in series with the R-generator and Transistor input resistance to produce the time constant for the coupling capacitor?

 

Hi

Here is the circuit in regard to the previous threads about the output resistance of the transformer and the input resistance of the transistor. Do both resistance on each side of the capacitor influences the time constant and voltage across the C1? If so,how would it be calculated.

 

If the transistor is a 2N3904 then its HFE is 230 and its emitter and collector currents are about 3mA. The datasheet for the 2N3904 has a graph that shows its tyoical input impedance is 1.3k ohm plus the emitter resistor increase which is the hfe of about 150 x 220= 33k in series with the 1.3k of the transistor.

The 34.3k is parallel to the 220k of the base bias resistor which is 29.7k which is parallel to the 47k base bias resistor which makes 18.2k.

The signal source resistance is about 148 ohms so the coupling capacitor's cutoff frequency should be calulated with a resistance of 18,348 ohms.

 

The signal source resistance is about 148 ohms so the coupling capacitor's cutoff frequency should be calulated with a resistance of 18,348 ohms.

How did you get 148 ohms for the signal source resistance ? When using the 148 ohms I calculated (if the formula is correct) C = 1/(2 pi)( f)( R) =
1/(6.28) (60 hz)(18,348) =0.0000001 or 1 nano. The reactance,Xc=(2)(pi)(f)(c) or1/(6.28)(60)=25,595.744 ohms.Would you agree ? And one more item,what about when resistance equal reactance ? Does this influences the cut-off frequency of 60 hz.

 

The signal source is a high impedance (I guess) transformer winding. It has 220k in series with 150 ohms which is 149.9 ohms, not 148 ohms. So 149 ohms is the source impedance.

The total resistance seen by the coupling capacitor is the 149 ohms plus the 18.2k ohms that I calculated before.

What do you want for a low frequency response? Do you want 60Hz to be the cutoff frequency which is a level of -3dB (0.707 times)? If so, then the capacitor value is 1 over 2 pi x 18.349k x 60Hz= 0.145uF.
I have never seen a 150nF capacitor but they are available somewhere. I would use a common 220nf capacitor (0.22uF).

 

How did you get 148 ohms for the signal source resistance ? When using the 148 ohms I calculated (if the formula is correct) C = 1/(2 pi)( f)( R) =
1/(6.28) (60 hz)(18,348) =0.0000001 or 1 nano.

How did you get 1nF from that?

 

I think his calculator ran out of numbers.

 

I think his calculator ran out of numbers.

I think so too. If he didn't use exponential mode (which his calculator may not have), 145nF rounds to .0000001F. Of course, that's still not 1nF.

 

I think so too. If he didn't use exponential mode (which his calculator may not have), 145nF rounds to .0000001F. Of course, that's still not 1nF.

Actually, I did not bring the decimal over three places to the left to simplify the measurement to 100 nanos. Also, It was not specified that the the two resistors was taken as parallel connection that produced the 149 ohms.
I would like it to cut -off at 60 hz with a very low voltage drop across the C1. Maybe a 90% to 100% input transfer voltage from Rg2 to the total input of the transistor.I guess I can use another 220 nano capacitor in series with one the prods of my analogy meter to pick off the gain of the input signal at Vc ?