RC phase shift oscillator, confusion, gain v VCE per DC load line

Thread Starter

soc7

Joined Dec 19, 2019
20
I'm designing an RC phase shift oscillator. (Although, I'm faced with the same issue when designing a Colpitts oscillator.)
I'm confused about what looks, to me, like a conflict between achieving the required amplifier gain and the Q point for VCE.

The amplifier uses a Common emitter Class A amplifier using:
NPN BJT, voltage divider bias, emitter bypass capacitor.
VCC of 14.1 V, 1 mA emitter (collector) current.
All the voltages and currents for the amplifier portion are as expected for a common emitter Class A amplifier.

Given what I think I understand about a DC load line and Q points, the Q point for VCE should be about 7 V. (Which my circuit achieves.)

Here's where I'm confused:

If I use the known, required amplifier gain of 29 and the collector current of 1 mA (keeping in mind that I'm using a emitter bypass cap), the calculated collector resistance should be:
r prime e = 26 mV divided by collector current = 26 mV divided by 1 mA = 26 ohms.
RC = Av times r prime e = 29 times 26 ohms = 754 ohms.
Using a resistor with a value anywhere near the 754 ohms will result in a voltage drop across RC of less than 1 volt.
This does not allow for much of a peak to peak signal voltage.
Plus, this resistance differs greatly from the resistance calculated via the DC load line approach.

So, what am I missing? What formula should I use to calculate collector resistance? (I'm not ready to use an RF choke, and the tutorials I've listened to say that a collector resistor can be used for both an RC and a Colpitts oscillator.)
Please do not use any images in your explanation. I am blind, so I can't access images.

Thhanks,
soc7
 

LvW

Joined Jun 13, 2013
1,160
Question: While finding and analyzing the load line and fixing the DC operational point - did you consider ther voltage drop at the emitter resistor correctly? This resistor drastically reduces the slope of the DC load line.
 

Thread Starter

soc7

Joined Dec 19, 2019
20
Yes, I considered the voltage drop across the emitter resistor when determining the VCE Q point.

But my confusion really has to do with the big difference between calculating the collector resistance via the required gain and r prime e versus calculating the collector resistance via the DC load line:

RC = 29 times r prime e (given a collector current of 1 mA.) = 29 times (26 mV divided by 1 mA) = 29 times 26 ohms = 754 ohms.

versus

VCC of 14 V,
Desired Emitter voltage of 1.5 V and emitter current of 1 mA resulting in a emitter resistor of 1.5 kOhm.
So, VCE = (14 V minus 1.5 V) divided by 2 = 6.25 V
Hence, RC = 6.25 V divided by 1 mA = 6.25 kOhm.
So, gain is: RC divided by r prime e = 6.25 kOhms divided by 26 ohms = 240.4

So in the first case, I achieve the required gain of 29, but because the collector voltage is so close to 14 volts, won't that cause clipping?
And in the second case, VCE will be where it should be (for a Common emitter Class A amplifier, but the gain is far too high for the RC oscillator.

So, I think that the real question is how do I determine the value of the Collector resistor?

Please straighten me out.

Since I'm blind, I'm limited to audio books, and they aren't very helpful. And the youTube-type videos are directed toward sighted folks, so there's a lot of, "this resistor goes here, and this is connected to this." Not much help to me.

And finally, I'm not a student. I've already had my 40 year career in software development. I don't have a teacher to ask, and I thought this forum was the right place to go.
 

Thread Starter

soc7

Joined Dec 19, 2019
20
The bjt amplifier is part of an RC phase shift oscillator. (About 800 Hz.)

I plan to use this rc oscillator in at least two ways. In one case, it will be used along with (and I haven't figured this out yet) with an already working 1 MHz Colpitts oscillator. My goal here is to create a simple, very simple, AM transmitter.

Separately, the rc oscillator might have an earplug as its output.
 

michael8

Joined Jan 11, 2015
107
So the collector load at AC is going to include the input to the phase shift network as well as the output (whatever the oscillator is driving). What are these resistances? Is the gain still 240?
 

Thread Starter

soc7

Joined Dec 19, 2019
20
So, since the AC signal is the issue, it sounds like the capacitive reactances of the 4 capacitors has to be taken into account?

If so, does that mean that the collector resistor is, not only in parallel with the output load resistance, but it is also in parallel with the total impedance of the feedback network?

If so, (the RC feedback network has 4 stages), do I calculate the capacitive reactance of 4 caps in series with one another, and the equivalent resistance of 4 resistors in parallel with one another?
 

Audioguru again

Joined Oct 21, 2019
2,374
You did not show a schematic for a phase shift oscillator so I found one.
Use 3 or more RC phase shifters but usually 3 are used.
Use another transistor with a high input impedance at the output of the oscillator output to drive an earbud or transmitter.
 

Attachments

michael8

Joined Jan 11, 2015
107
Autioguru attachment/schematic: typical phase shift oscillator, 3 RCs, series capacitor, resistor to ground. No values provided.

Emitter of transistor has resistor and capacitor to ground with note at emitter point: connect series resistor here to reduce gain to 29.
 

Thread Starter

soc7

Joined Dec 19, 2019
20
I should probably include in all my messages that I am blind. I can't create or, if someone sends one, view a schematic.

Please explain what you meant, Michael, when you said in your previous message to connect a series resistor at the emitter node. (I already have an emitter resistor and bypass capacitor, in parallel, connected to the emitter. Are you saying there is a second resistor connected in some way at the emitter?

AudioGuru, the calculations are probably easy for someone who is sighted and so can see the the images of schematics which so, in addition to the schematic, values for the components. Or those who are sighted are not hindered by videos in which the presenter is pointing at a drawing of the circuit and points sayingthings like "this resistor is connected here, and this goes here." Unfortunately, I have to rely on audio books which have a host of problems from schematics without component values to readers who make mistakes when describing what they see. Many of the audio books do not even describe figures. The book, and hence the reader, can only read a vague caption. So, if I'm going to learn, I ask here for guidance. It's slow and frustrating for me and those who take the time to answer questions, but it's all I've got.
 

michael8

Joined Jan 11, 2015
107
> i should probably include in all my messages that I am blind.

And others should read the entire thread before replying...

A transistor with an effective Re of 25 ohms is very non-linear
for any more then very small inputs. Adding some more Re
will lower the gain (negative feedback) as well as make it
more linear.

I've run a ltspice simulation of an approximation of your circuit. V+
is 14.1 volts, the NPN transistor is a 2N3904, the base biasing resistors
are 120K from V+ to base, base to ground is 22K. The emitter is connected
to a 1.5K resistor to ground and a 100uF capacitor to a 74 ohm resistor
to ground. So the Re is about 25+74 about 100 ohms.

The collector resistor is your 6.25K.

I've stuck in a feedback phase shift network, 3 stages, each a series
.01uF capacitor and a 10K resistor to ground. The collector drives
the first .01uF capacitor and the output is coupled via a DC blocking
capacitor of 1uF to the base bias network (and the base).

To start the oscillator (since simulations are perfect) I've
added a pulse to the base from a current source of a 1uSec
pulse of 1uA.

My RC phase values result in about 600 Hz.
 

Thread Starter

soc7

Joined Dec 19, 2019
20
Thank you for such a thorough and clear reply.

> A transistor with an effective Re of 25 ohms is very non-linear for any more then very small inputs. Adding some more Re will lower the gain (negative feedback) as well as make it more linear.

All of the examples in the textbooks use only an emitter resistor and a bypass capacitor in parallel with the resistor -- just those two, so all of the AC signal passes through the capacitor with no loss.

> I've run a ltspice simulation of an approximation of your circuit...

A final question, I hope, on this topic.

The organization that makes audio recordings of the textbooks I have access to require their narrators to read the text and describe figures, including schematics and formulas as the text and figures appear in the book; even if the text or figure is in error. The narrator of the particular audio book that contains the RC phase shift oscillator schematic seems to be familiar with the subject matter. (Unfortunately, an unusual occurrence.) And when he comes across something that is in error, sometimes, it's obvious to me that there's an error, he reads the text or describes the figure in a hesitant, halting manner. He does this when he describes the final RC stage in the feedback network.

I've typed out his description of the feedback circuit, below. Is the third stage correctly described, or is my follow-up version right?

The collector is connected to the right end of capacitor c1.
The left end of capacitor c1 is connected to the top end of resistor r3; the other end of which is connected to ground.
The left end of capacitor c1 is also connected to the right end of capacitor c2.

The left end of capacitor c2 is connected to the top end of resistor r4; the other end of which is connected to ground.
The left end of capacitor c2 is also connected to the right end of capacitor c3.

The left end of capacitor c3 is connected to the right end of resistor r5.
The left end of resistor r5 is connected to the base of the transistor.

Is the final RC stage connection, above, correct, or should it be similar to the connections of the first two stages:

The left end of capacitor c3 is connected to the top end of resistor r5; the other end of which is connected to ground.
The left end of capacitor c3 is also connected to the base of the transistor.

Again, I question his reading of the third stage because he reads it with hesitation. Plus, I thought each stage was an RC lag circuit.

(I want to understand what I'm doing, not just follow along blindly. Sorry.)

Thanks
 

michael8

Joined Jan 11, 2015
107
In the r5 connected between c3 and the base of the transistor case
there has to be something else connected to the base of the transistor
to provide a DC path for bias since c3 blocks DC. That will likely
provide some resistance to ground for the 3rd phase shift stage.

I simulated the r5 to base with my biasing components also connected to
the base. 120K to V+ and 22K to ground. The parallel combination of
120K and 22K results at AC of about 19K to AC ground.

I also took out the gain control resistor in series with the 100uF emitter
bypass capacitor. The oscillation amplitude appears to be limited by the
transistor collector current getting to zero during part of the cycle.
The waveforms are distorted too.

Changing the c3 and r5 connections so that c3 feeds the base and r5 is
connected from the c3 and base to ground as you described will result
in r5 being a DC path between the base and ground. The resulting DC
path will disrupt the transistor bias or else r5 needs to be part of
the bias network.
 

Thread Starter

soc7

Joined Dec 19, 2019
20
> In the r5 connected between c3 and the base of the transistor case...
> Changing the c3 and r5 connections so that c3 feeds the base and r5 is connected from the c3 and base to ground...

In my previous message, I left out mentioning two things:
1) the voltage divider DC bias network made up of R1 between V+ and base and R2 between base and ground.
2) the DC blocking capacitor between the RC network and the node where R1, R2 and the base meet.

In the following two scenarios, "node 1" refers to the node where R1, R2 and the base meet.

So, does that mean beginning with stage 3, either:

The left end of capacitor c3 is connected to the right end of resistor r5.
The left end of resistor r5 is connected to the right end of the DC blocking capacitor.
The right end of the DC blocking capacitor is connected to node 1.

OR:

The left end of capacitor c3 is connected to the top end of resistor r5; the other end of which is connected to ground.
The left end of capacitor c3 is also connected to the right end of the DC blocking capacitor.
The left end of the DC blocking capacitor is connected to node 1.

Your explanations help a lot. The light bulb over my head is starting to come on. Thanks for your patience.
 

Audioguru again

Joined Oct 21, 2019
2,374
I am sorry to hear that you are blind. I was blinded with cataracts in both eyes and saw only blurred light. I could not drive, read or watch TV. Then the doctor removed the faulty lenses in my eyes and replaced them with perfect synthetic lenses and my vision suddenly became better than when I was younger.
 

BobTPH

Joined Jun 5, 2013
2,729
I am sorry to hear that you are blind. I was blinded with cataracts in both eyes and saw only blurred light. I could not drive, read or watch TV. Then the doctor removed the faulty lenses in my eyes and replaced them with perfect synthetic lenses and my vision suddenly became better than when I was younger.
Just out of curiosity, why did you wait until they were that bad? My doctor recommended surgery when I could barely notice it.
Or did they develop unusually rapidly?

Bob
 

Audioguru again

Joined Oct 21, 2019
2,374
Hi Bob.
I got closeup "reading" glasses when I was about 40. Then I got 3-way (close, medium and far) parts on the glasses lenses when I was about 50. When I was about 66 I noticed speckles and thought my glasses got sandblasted but things looked the same without glasses. I became completely blind in a few weeks which was the waiting time for the surgery.

I had a heart attack when i was 64 and got fixed "better than new" with 2 stents which might have had something to do with cataracts. I got hearing aids when I was 70. Now I am young again but I am actually 75.
 

BobTPH

Joined Jun 5, 2013
2,729
I am glad that despite all your medical problems you are doing well. I have had similar experience. I have had a bad knee since I was 2 years old, caused by an odd non-malignant tumor. at 63, I finally got a knee replacement and, after a suitable recovery time, have a pain free knee after 60+ years of suffering with it.

Bob
 

michael8

Joined Jan 11, 2015
107
referring to soc #15 questions and his addition of the transistor bias network:

> So, does that mean beginning with stage 3, either:
> The left end of capacitor c3 is connected to the right end of resistor r5.
> The left end of resistor r5 is connected to the right end of the DC
> blocking capacitor.

That's a bit wierd as both ends of r5 connect only to capacitors.

So there is no DC path to the non-base end of the DC blocking capacitor
so there isn't any DC to block.

> The right end of the DC blocking capacitor is connected to node 1
> (the transistor base & bias resistors).
> OR:
> The left end of capacitor c3 is connected to the top end of resistor r5;
> the other end of which is connected to ground.
> The left end of capacitor c3 is also connected to the right end of the
> DC blocking capacitor.
> The left end of the DC blocking capacitor is connected to node 1.

Keep in mind I can't see the diagram in your "audio" book so I don't
know what the circuit being described really is.

Just considering the general phase shift oscillator circuit:

a. the transistor needs to be biased at DC.

b. the AC path has to feed the output of the phase shift network
into the transistor base without messing up the DC bias.

There are two obvious ways to do this.

1. Have a normal last phase shift stage (series capacitor to resistor
to ground, with the output at the capacitor and resistor junction).
AC couple this to the transistor base via a DC blocking capacitor
and include the usual two resistors to V+ and ground to bias
the transistor.

2. Merge the last phase shift stage with the transistor bias network.
The DC block is then unneeded and the last phase shift resistance
needs to be the AC resistance into the base & bias network.
This saves one resistor and one capacitor.

In case 2, assuming a target emitter current of 1mA and a emitter resistor
of 1.5K then the base bias voltage needed is 1e-3*1.5e3+.7 -> 2.2 volts.

With a supply of 14.1 volts this is a ratio of 2.2/14.1-> .156. Since the
bias network is also the resistor for the last phase shift stage it
needs to have a parallel resistance matching the phase shift stages.

I've used .01uF and 10K in the phase shift stages in my simulations.

As a first cut, try 10K for the base to ground resistor (will be
too low, but a starting place). Then the total bias resistance
(some of both resistors) needs to be 10K/.156 -> 64K.

So the V+ to base resistor would be 54K. The parallel resistance
part of the phase shift network with this combination would
be 10K in parallel with 54K -> 8.4K too low (looking for 10k).

Well, some algebra or more trys could get a closer solution...

Note the above assumes/ignores the transistor input resistance which
likely also counts if less than 50K or so (and depends on the transistor
emitter current and beta as well as any unbypassed emitter resistor.)
 
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