# Current source improves linearity on collector BJT

I'm learning about analog electronics. I'm looking into amplification stages at the moment.

Question 1: Single BJT C-E Amplifier. Usually you put a resistor between collector and V++, so as $I_b$ changes, the collector current changes and causes a voltage drop across this resistor, which then is the amplified signal.

I found circuits that use a constant current source instead of a resistor to improve linearity. How does this work? I don't get it, doesn't the amplification arise from the change in current in the collector/emitter path? How does a constant current do any good for this? (Actually the universe should explode when the current through the transistor is fixed, while $I_b$ tries to control it... Thank god we have supply rail limits:-))

Then I have a second question which might be related: How do current mirrors in differential pairs work? Isn't the point of the two current paths ($I_{c1}$, $I_{c2}$), that the difference in current causes the difference in the output voltage at the usually used resistors? How can they change $V_{out\;diff}$ if they are cemented to be equal? Again, the universe should explode...

• Welcome to EE.SE, jjstcool! I'm an editor/reviewer on this site and I would like to suggest that you post your questions separately. You can probably keep the first two questions here and ask the 3rd separately. That will fit better in the site's Q&A format. – Ricardo Jun 30 '14 at 1:41

I found circuits that use a constant current source instead of a resistor to improve linearity. How does this work?

All physical current sources are approximations of ideal constant current sources so there is no danger of the destroying the universe.

We model physical current sources as ideal current sources with some finite (though large) parallel resistance.

In the AC small-signal model of a constant current source loaded CE amplifier, the small-signal collector current is through this parallel resistance. This means that the small-signal gain can be quite large though at the expense of a very high output impedance.

I don't think it's true that a current source collector load will improve linearity per se. The fact that the circuit has high gain doesn't imply greater linearity for large signals.

However, the high open-loop voltage gain can be traded for linearity via negative feedback.

• Thanks, that helps. Do I get this right: The gain of a BJT-stage is proportional to the collector current. The voltage gain increases with the collector-resistor. So I can improve gain by increasing both using the constant current source. After your explanation it is essential to know the output resistance of the source and I cannot just idealize it? Is there a paper about it or sth? – jjstcool Jun 30 '14 at 2:22
• The small-signal transconductance $g_m$ is proportional to the DC collector current $I_C$ and the small-signal voltage gain depends on $g_m$. For the true CE amp (emitter at AC ground), the open-circuit voltage gain is $-g_m (R_C||r_o)$. If you idealize the constant current source load, the gain is limited by $r_o = \frac{V_A + V_{CE}}{I_C}$ where $V_A$ is the Early voltage for the BJT. In other words, $-g_mr_o$ is the maximum voltage gain available and equals $-\frac{V_A + V_{CE}}{V_T}$ – Alfred Centauri Jun 30 '14 at 2:28
• Ah, thanks. One last thing: When I suppose a very high Early voltage, the slope of $I_{ce}$ vs $V_{ce}$ gets very "un-steep". How can one possibly set a quiescent point so that the collector voltage does not "run" into one of supply limits but ideally stays at the middle (with varying temperature etc.)? – jjstcool Jun 30 '14 at 2:38
• @jjstcool, typically, the high gain amplifier is enclosed within a global negative feedback loop where the DC feedback is 100%. This stabilizes the operating point. – Alfred Centauri Jun 30 '14 at 17:20
• yes I came up with this thought as well then. I will calculate it through and I guess I will get this as a result. A harder question will be the calculation of the amount of distortion, but from here I can do this myself. Thanks for the help. – jjstcool Jul 1 '14 at 6:02