Understanding differential amplifier design example from AoE

Thread Starter

cake4all

Joined Oct 19, 2018
6
Hello, currently I try to teach myself the basics of analog circuits. Actually I already had to attend a circuitry course at university, but it's already some time ago and I didn't learn as much as I anticipated. It was mostly about circuit analysis and small signal models, no circuit synthesis.

For this purpose, I've bought the book "Art of Electronics, 3rd Edition". However I am having some trouble to comprehend the design example for the differential amplifier on page 103. I've attached the schematic , if you don't have access to this book(please scroll to the second page, somehow latex wants a blank page first).

The following formulas are derived:
G_diff = v_out / (v_1 - v_2) = R_C / (2 * (R_E + r_e))
G_CM = - R_C / (2 * R_1 + R_E + r_e)
CMRR is approximately = R_1 / (R_E + r_e)

The author writes, he wants the following things:
1) V_out = 0.5 * V_cc
2) Quiescent current of 100µA through R_C.
==> R_C = 2.5 V / (100µA) = 25 kOhm ==> I can totally follow that.

3) R_1 is chosen to give total emitter current of 200µA, split equally between both sides.

My Questions:
A) How is R_E calcualted?
B) How is R_1 calculated?

My thoughts to R_1:
I_{R_1} = 200µA
U_{R_1} = V_out - V_{CE,2} - V_{R_1} + V_EE = 2.5V - V_{CE,2} - V_{R_1} + 5V = ???
R_1 = U_{R_1} / I_{R_1} = ???

My thoughts to R_E:
I_{R_E} = 100µA
U_{R_E} = V_out - V_{CE,2} - V_{R_1} + V_EE = 2.5V - V_{CE,2} - V_{R_1} + 5V = ???
R_E = U_{R_E} / I_{R_E}


In the exercise 2.18 on the next page I am supposed to design a differential amplifier on my own, but I can't do that if I do not even understand the example.
Kind regards
 

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Jony130

Joined Feb 17, 2009
5,145
The authors clearly stated that:
Exercise 2.18. Then design a differential amplifier to run from ±5V supply rails, with
Gdiff = 25 and Rout = 10k. As usual, put the collector’s quiescent point at half of VCC.
So, from this, because we need Rout = 10kΩ hence Rc = 10kΩ and Ic1 = 2.5V/10kΩ = 250μA and the total emitter current will be Iee = 500μA.

re = 25mV/250μA = 100Ω

And the voltage gain we need is equal to Gdiff = Rc /(2 * (RE + re) ) = 25 V/V therefore:

10kΩ/25 = 400Ω ---> RE = 400Ω/2 - 100Ω = 100Ω

And using the KVL we can find R1:

Vee = Iee*R1 + Ic*Re + Vbe (without input signal the voltage at base is 0V)

R1 = (Vee - Vbe - Ic*RE)/Iee = (5V - 0.6V - 25mV)/500μA = 8.75kΩ we can use 9.1kΩ from E24.

Any additional questions?
 
Last edited:

Zeeus

Joined Apr 17, 2019
508
Maybe not really important but just to know : The formulas derived are correct ONLY if the quiescent current on both transistors are equal

If not equal then formula not correct. For small signal :

\[ Vi = V1 - V2 \]

Quiescent I1, I2
\[ gf = (I1*I2)/(IQ * VT) \]

signal current
\[ I1 = (gf) * Vi \]
\[ I2 = -I1 \]
 

Thread Starter

cake4all

Joined Oct 19, 2018
6
Thank you very much!
Vee = Iee*R1 + Ic*Re + Vbe (without input signal the voltage at base is 0V)
I completely forgot, that the input voltages are 0 for the quiescent point. This makes actually sense, because this way \( V_{CE} \) can be omitted.
Cheers!
 

MrAl

Joined Jun 17, 2014
7,139
Thank you very much!

I completely forgot, that the input voltages are 0 for the quiescent point. This makes actually sense, because this way \( V_{CE} \) can be omitted.
Cheers!
Hi,

I hate to ask, but are you sure he said Vout=Vcc/2 rather than Vout=(Vcc-Vee)/2 which would make Vout=0?
That's a more usual way to bias this, but whatever you know you should use.
 

Thread Starter

cake4all

Joined Oct 19, 2018
6
Hi,

I hate to ask, but are you sure he said Vout=Vcc/2 rather than Vout=(Vcc-Vee)/2 which would make Vout=0?
That's a more usual way to bias this, but whatever you know you should use.
Yes, he definitely wrote \[ V_{out} = \dfrac{V_{cc}}{2} \] in the exercise and in the design example.
 
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