# Amplifiers Calculations

Discussion in 'Homework Help' started by Kayne, Oct 21, 2010.

1. ### Kayne Thread Starter Active Member

Mar 19, 2009
105
0
HI,

I have come to the problem attached and have been given the answer from the text book but not sure how to calculate hie.

The question asked for a value of input resistance of the circuit.

The formula I am using is
$Rie= R1//R2//(hie+hfeRe)$
where
$hie =\frac{0.026}{Ibq}$
So
$= 18k//180k//(hie+(560\times200)$

Now Ibq is the bais current which I need but an unsure which numbers to use in the figure.
Do I have to find out the current between the 180k and 18k resistors?

• ###### Q1.doc
File size:
51.5 KB
Views:
12
Last edited: Oct 21, 2010
2. ### Georacer Moderator

Nov 25, 2009
5,161
1,273
You forgot to attach the image.

Be sure to hit the "upload" button on the Attachment Manager screen.

3. ### Kayne Thread Starter Active Member

Mar 19, 2009
105
0
Not sure what happened then on my screen the attachment was there. I have done it again.

Thanks

File size:
51.5 KB
Views:
15
4. ### Georacer Moderator

Nov 25, 2009
5,161
1,273
Your formulas are correct, they have only a mild level of estimation (they assume $\alpha \approx 1$ which is ok).

In order to find the bias collector current, as you correctly pointed, you need to do a DC analysis of the layout. Open circuit all the capacitors, and by doing this, isolate the circuit from the AC signal. We wanto to work only with DC quantities for now.

You need to convert the voltage divider to its Thevenin analog to apply the above, so look at the next section, "Voltage Divider Bias", too.

5. ### Kayne Thread Starter Active Member

Mar 19, 2009
105
0
Looking though that link the formula I need to use is below

$IE= \frac{VBB-VBE}{\frac{RB}{Hfe}}$

So if I subsistute in my values I get

$IE= \frac{12-0.7}{\frac{(\frac{180k\times18k}{180k+18k})}{200}}=138mA$

I think that I am still doing something incorrect as I work further into the poblem the value for hie is incorrect compared to the answer.

6. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,610
466
Let me describe a rough, first-cut, procedure that can also serve as a sanity check.

You want the collector voltage, Vc, to be about half the supply voltage, Vcc.

Calculate the current in the collector resistor, the 6.8k, which will drop about half the supply voltage: Ic = 6 volts/6800 ohms = .88 mA. This ignores the drop across the emitter resistor, but since the emitter resistor is less than 1/10 the collector resistor, it's ok for a start.

If Ie is .88 mA, then re = .026/.00088 = 29.5 Ω, and hie = (β+1)*29.5 = 5929.5 Ω.

If Ie is .88 mA, then Ib = Ie/(β+1) = 4.37 μA.

Calculate the current in the bias divider: Id = 12/(18k + 180k) = 60.6 μA.

This is more than 10 times the base current, so for a first approximation (and sanity check) we can ignore the effect of the base current on the bias divider output voltage.

So, we can see that your value of 138 mA for the emitter current is probably wrong!

There have been several long threads on this topic: