How Does the H Parameter Model Calculate Current and Voltage Amplification?

[null void]

Homework Statement

calculate current and voltage amplification

The Attempt at a Solution

Conversion to h parameter model,

and this is my working step,
https://www.filesanywhere.com/fs/v.aspx?v=8b6b628b5e646faf6e9b

 

[null void]

there are something i can't understand, in my calculation, i use equation 3 to find Ai, i get -132... But if i use equation 6 instead of 3, i get 132..., same number but different sign. I suppose there is something went wrong in my equation 6...

 

[NascentOxygen]

Your eqn(2) doesn't correctly show how h_fei_b divides between the two resistances.

I suggest that, in future, after you create something as a .docx file that you take a screenshot of your effort and post it as a jpeg instead of the .docx file. This will allow those who rely on a basic tablet to still follow the thread, using software add-ons no more sophisticated than a jpeg viewer.

A further suggestion: I think it's a good idea to informatively label dimensionless parameters, e.g.,
h_fe = ... amps/amp
h_re = ... volts/volt

It helps you to stay correctly focussed, and makes checking easier (especially by those of us who may have forgotten some of these. [/size]

 

[The Electrician]

there are something i can't understand, in my calculation, i use equation 3 to find Ai, i get -132... But if i use equation 6 instead of 3, i get 132..., same number but different sign. I suppose there is something went wrong in my equation 6...

Your equation 2 is Iout = hfe Ib + hoe Vout

Equation 3 is Iout = -hfe Ib () What are these parentheses?

Why is there a minus sign in front of hfe in equation 3 but not in equation 2?

Shouldn't equation 2 be Iout = hfe Ib + hoe Vout + Vout/Rc?

Also, I don't see anywhere where you have taken Rs into account. Rs has a very large effect on Av.

I would think that Av would be defined as Vout/Vs. Is your problem defining it as Vout/Vin, which is the same as Vout/Vb?

 

[null void]

yeah sorry messed up a few thing up there, this is my second attempts

so is this the correct ways?

Also, I don't see anywhere where you have taken Rs into account. Rs has a very large effect on Av.

I would think that Av would be defined as Vout/Vs. Is your problem defining it as Vout/Vin, which is the same as Vout/Vb?

I think the arrow in the diagram is telling us the Vin is Vb

 

[null void]

This is for the voltage amplification, but i can't really figure out how to include the Rs in the calculation

A further suggestion: I think it's a good idea to informatively label dimensionless parameters, e.g.,
hfe = ... amps/amp
hre = ... volts/volt

Do you mean by putting amps/amp behind the term like a unit?

 

[NascentOxygen]

For your Iout/Iin expression, the algebra looks right, though there's a transcription error: you show RB hie where it should be RB + hie. I haven't checked your arithmetic, and haven't checked your voltage equation.

You may need a sign change. The arrow direction for i_out is indicating i_out = -[/size][/color] v_out / R_C

 

[NascentOxygen]

If VB is correctly denoted as Vin, then as you observed earlier, Rs does not enter into your calculations.

 

[The Electrician]

In post #5 and #6, you have used a value of .87k for hie, but in post #1 its value was given as .86k; which is correct?

...i can't really figure out how to include the Rs in the calculation

An easy way is to calculate the input resistance at the Vin node, and use that impedance plus the 1k Rs to form a voltage divider. The voltage division ratio from the Vs node to the Vin node can be multiplied times the Av you already calculated to get the overall voltage gain from Vs to Vout.

 

[null void]

You may need a sign change. The arrow direction for iout is indicating iout = - vout / RC

Oh yeah, I get what u mean.

hah sorry hie it is actually 0.86k ohms

An easy way is to calculate the input resistance at the Vin node, and use that impedance plus the 1k Rs to form a voltage divider. The voltage division ratio from the Vs node to the Vin node can be multiplied times the Av you already calculated to get the overall voltage gain from Vs to Vout.

I find it is a bit tricky to find the input impedance even tough Rs is not included because of the HreVout, it causes great pain for me to kill off Ib , Iin and Vb.

 

[The Electrician]

I find it is a bit tricky to find the input impedance even tough Rs is not included because of the HreVout, it causes great pain for me to kill off Ib , Iin and Vb.

In post #1, you said your problem was to "calculate current and voltage amplification", but you didn't say whether voltage amplification, Av, was defined as Vout/Vs or Vout/Vin.

Have you decided which it is?

Do you still need to calculate Vout/Vs?

 

[null void]

You may need a sign change. The arrow direction for i_out is indicating i_out = -[/size][/color] v_out / R_C

from the

-h_feI_b = V_out (h_oe + 1/R_c)

i should change it to

h_feI_b = V_out (h_oe + 1/R_c)

then at the end, my A_i,

A_i = R_B/ [ (R_ch_oe + 1)( R_B + h_ie )/h_fe + h_reR_c ]

With this change, i get positive 132.4439 amplification