# Common emitter amplifier with collector feedback gives unexpected behavior.

#### kc5tja

Joined Mar 22, 2014
1
Hello! This is my first post here, so hopefully I'm complete and clear in stating my question.

My goal is to build a noise bridge to facilitate some experimentation with antennas. I've come across a plurality of designs on the Internet, categorically all of which rely on a common-emitter amplifier design relying on collector feedback instead of emitter degeneration. Here's just one example: http://newenglandqrp.org/files/noise-bridge-schematic.gif

I can build the circuit verbatim, but I figured I'd take this opportunity to explore this amplifier configuration. I've done CE amplifiers with emitter degeneration before, and I feel I have a firm grasp of those configurations. Collector feedback, however, has me utterly mystified.

I went out onto the web and sourced a number of articles which all say the same basic thing: to design your amplifier, you'll want to put Vc at Vcc/2 (or as close as feasible), so pick an Ic and Rc that gives the closest approximation. Then, solve this equation to find Rb, and you'll magically have your amplifier:

Rb = ((Vcc-Vbe)/Ib)-(beta*Rc)

So, I design an amplifier like so:

1. I chose Ic = 9mA, as with a 9V battery, I can use reasonable resistor values to get close to 4.5V collector voltage.
2. Rc = 463 (measured)
3. Thus, if I pick beta=70 (per the datasheet at Ic=10mA, closest to 9mA desired), Rb = ((9V-0.65V)/128.5uA)-(70*463) = 32.5K.

The closest resistor I had was 47K (measured to be 46.3K), so already I knew my amplifier parameters would be somewhat off. This schematic shows the circuit configuration I came up with: https://gist.github.com/sam-falvo/133b730d7353b77172e4

"Slightly off" and "completely out of the ballpark" are two different things, however. What I found was that instead of 9mA Ic, I only have 2.6mA, which means instead of 4.5V across Rc, I have 1.2V, and the transistor drops the rest of the voltage. Additionally, instead of Ib=128.5uA, I actually have closer to 160uA. This gives a forward gain close to 16.25. That gain is significantly lower than expected, and I'd love to know why. Vbe was measured to be 0.65V.

Searching on the Internet-at-large turns up nothing useful, and for days, I thought every website was lying through their teeth. Until only recently, I found this link on AAC: http://forum.allaboutcircuits.com/s...t=collector+feedback+amplifier+common+emitter

The behavior of my amplifier is, in some respects, consistent with the discussion found in that thread, despite having a different circuit:

1) Low measured hFE implies the transistor is in saturation somehow,
2) hFE near 20 suggests a European-made transistor (not that this is relevant, but...),

However, I'm not sure why the transistor is in saturation: the base current is in the microamp range, and the collector current is nowhere near its documented limits on any of the transistor data sheets that I can find, regardless of vendor. Moreover, Vce is close to 8V, not 0.2V like what the datasheet says it ought to be when saturated.

The take-away of the aforelinked AAC article is, flatly, never trust hFE, and avoid using it in any kind of equations. However, clearly, people are making working amplifiers without emitter degeneration, and I'd love to know the process of how it's done.

Can anyone offer some insights on this? Where am I going wrong, and what wisdom is missing from what seems like every single website available on the topic?

Thanks!

#### crutschow

Joined Mar 14, 2008
32,064
The operating point for that simple circuit with collector feedback and no emitter resistor is dependent upon the hFE of the particular transistor you are using and that value often has a 3:1 variation or more between different units of the same part designation. So you either have to live with that variation in collector bias point or add an emitter degeneration resistor. You can bypass the resistor with a capacitor if you don't want it to reduce the AC gain of the stage.

#### alfacliff

Joined Dec 13, 2013
2,458
I have used that type amp for years from audio to rf. I have never had the thing saturate on me, if the current gets too high, the bias is reduced due to drop across the collector resistor.

#### LvW

Joined Jun 13, 2013
1,646
Rb = ((Vcc-Vbe)/Ib)-(beta*Rc)

So, I design an amplifier like so:

1. I chose Ic = 9mA, as with a 9V battery, I can use reasonable resistor values to get close to 4.5V collector voltage.
2. Rc = 463 (measured)
3. Thus, if I pick beta=70 (per the datasheet at Ic=10mA, closest to 9mA desired), Rb = ((9V-0.65V)/128.5uA)-(70*463) = 32.5K.
Here is my calculation: For Vce=Vcc/204.5V and hfe=70 and Ic=9 mA
Rb=(Vce-Vbe)/Ib=(4.5-0.7)/0.13mA=29.2kohms.

Please note that with an input coupling C you will have dc stabilisation (operating point), but for a certain amount of signal feedback you need a signal source with an internal resistance.

By the way: Dont you think that a current gain of only hfe=70 is rather conservative?

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#### Alec_t

Joined Sep 17, 2013
13,443
I'm not sure why the transistor is in saturation
It isn't; you have Vce = 9V-1.2V = 7.8V.

#### crutschow

Joined Mar 14, 2008
32,064
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By the way: Dont you think that a current gain of only hfe=70 is rather conservative?
Not really. The minimum stated in the data sheet is 40. That would be a conservative value.