miller effect circuits

[mstechca]

I am getting somewhat confused.

A number of you have seen my story on my superregen receiver posted here in different threads over the past few months, and a number of you have made good suggestions.

So after I realized that my design had distortion at loud volume, I thought I should take a different approach similar to what a number of you recommend.

In every amplifier stage, except for the power amplifier stage, I have added a 4.7K resistor and a capacitor in series to form the "miller" circuit.

It seems that the sound quality is much better, but the volume is a little bit reduced.

I have chosen resistor values based on the voltages at the base of each NPN transistor.

I think that the emitter current and collector current must be the same for optimal transistor operation.

I am going to try to choose different value resistors to obtain more gain.

Right now, my detector has an 18K pull-up resistor, an 82K 1% resistor from base to the 18K pull-up resistor, and a 4.7K resistor between emitter and the tank circuit that is connected to ground.

The circuit won't work if the voltage is under 6. It will work at 3V if the pull-up is changed to about 10K.

I plan to decrease the pull-up resistor, but I want to know my limits before creating component "bombs" (you know, circuits that blow up when they start). Also, I want to make the amplitude larger, and I'm afraid if the 82K resistor goes too low, then I think the gain will go too low.

The transistor I am dealing with is a Pn3563.

I'll see what I can personally come up with
I intend to add 10 or more uF bypass capacitors to the 4.7K resistors.

 

[akg]

in this case how can anyone help w/o a ckt dia ?? :?

 

[mstechca]

the circuit is shown below. The parts in cyan (turquoise) are what I intend to deal with much later. The parts in yellow are what I need help with.
Right now all my resistors highlighted in yellow are set to 4.7K and the capacitors next to them are set to 0.33uF.

I also added a 270pF capacitor between the left-most 18K pull-up and ground.

I am able to pick up CH-TV (air channel 11), but the volume is too low. (lower than if I shorted each miller effect circuit). The yellow circuits are the "miller effect" circuits.

If I remember, electronics-tutorials.com claims that the emitter resistor is calculated as follows:

<base voltage> - 0.65 / <current of collector>

Now my base voltage happens to be substantially high, because the lowest value used for a pull-down resistor is 470K.

The collector resistor is 18K.

If the collector current and the emitter current are supposed to be equal, and my thinking is correct, then the emitter resistor must be close to the collector resistor, or I strain the base by adding a pull-down resistor that is lower than 82K, and if I do that, wouldn't my gain be cut greatly?

so what I was wondering is, how can I achieve the highest gain (or audio volume) with the least distortion?

The 22uF connected to the cyan part is for now connected to the speaker. Instead of 22uF, I am testing with 1000uF.

I still think I should lower the resistors in yellow, because making the capacitance value too high will just waste board space.

 

[Nigel Goodwin]

The yellow circuits are the "miller effect" circuits.

Perhaps you read about 'miller effect' somewhere, and liked the sound of it?.

Those components aren't anything to do with 'miller effect', they are simply bypass capacitors for the emitter resistors - bypassing them in that way increases the gain, by reducing the negative feedback around the transistor. All altering the value of the capacitor will do is alter the amount of bass, reducing it as the capacitor gets smaller.

 

[Roff]

The yellow circuits are the "miller effect" circuits.

Perhaps you read about 'miller effect' somewhere, and liked the sound of it?.

Those components aren't anything to do with 'miller effect', they are simply bypass capacitors for the emitter resistors - bypassing them in that way increases the gain, by reducing the negative feedback around the transistor. All altering the value of the capacitor will do is alter the amount of bass, reducing it as the capacitor gets smaller.

He mentioned a resistor and capacitor in series, then proceeded to highlight some parallel RC networks. I'll have some of whatever you're having, Mstechca.

 

[mstechca]

The yellow increases the gain, by reducing the negative feedback around the transistor. All altering the value of the capacitor will do is alter the amount of bass, reducing it as the capacitor gets smaller.

That's weird, because when I tested it, I didn't notice an increase in volume. The circuit I explained above produced a clear audio signal, but at a ridiculously low volume.

He mentioned a resistor and capacitor in series, then proceeded to highlight some parallel RC networks. I'll have some of whatever you're having, Mstechca.

Ron, I'm asking a decent question.

The point is that I want to reduce distortion and obtain maximum volume

I will try (without anyones assistance) to lower the resistance values and if that makes great improvements, then another equation has entered the garbage can!

 

[Russlk]

That is a wierd circuit! Is there an input somewhere? What are those diode-connected 2n3904 doing? I assume this is a superregen receiver, where is the antenna?

 

[Nigel Goodwin]

I assume this is a superregen receiver, where is the antenna?

It started off as one, but it ceased to be so a LONG time back, random modifications have made it worse and worse :lol:

 

[mstechca]

That is a wierd circuit!

It is not a final circuit.

Is there an input somewhere?

I can connect an antenna to the Pn3563's emitter. or I can connect the antenna to +ve, which provides somewhat better reception.

What are those diode-connected 2n3904 doing?

The 2n3904, and the capacitor connected to it make a poor-man's varactor I'm using 2 of them and connecting their inputs to the same source through resistors, so that I can gang-tune my circuit without the need of two separate trim capacitors.

I assume this is a superregen receiver, where is the antenna?

I forgot to include it in the circuit, but I can still pick up the TV station without an antenna!

 

[audioguru]

The point is that I want to reduce distortion and obtain maximum volume

A super-regen is an AM radio. TV sound is FM. FM is very distorted on an AM radio, even when you tune to the side of the FM carrier and "slope-detect" the FM. The slopes of your tuned circuit are nowhere near linear so of course the sound is distorted!

Your audio amp leaves a lot to be desired if you want low distortion:
1) If it has an 8 ohm speaker then with only a 6V supply, its max output power at clipping is only 1/4W.
2) With only 22uF feeding the speaker, frequencies below 909Hz are cutoff.
3) With 18k as the base bias resistor for the NPN output transistor, it can rise only 0.25V from an idling voltage of 3V, because an 8 ohm speaker needs much more current than the current gain of the transistor can supply with an 18k base resistor. It would be much better with a 2.2k resistor but then the voltage gain of the driving transistor would be reduced. It should have a bootstrapped resistor for high gain and low distortion.
4) Only a single diode is used so the output has crossover distortion from a single transistor.
5) The amp doesn't have overall negative feeddback.

 

[audioguru]

Now my base voltage happens to be substantially high, because the lowest value used for a pull-down resistor is 470K.

Of course. 470k is much too high for an emitter resistor. The transistor would be practically cutoff, and its voltage gain would be vey low.

The collector resistor is 18K.

If the collector current and the emitter current are supposed to be equal, and my thinking is correct, then the emitter resistor must be close to the collector resistor, or I strain the base by adding a pull-down resistor that is lower than 82K, and if I do that, wouldn't my gain be cut greatly?

How is the base going to be strained???
What is "a pull-down resistor"???
The voltage gain of a common-emitter transistor stage that has a high resistance load is roughly the collector resistor's value divided by the emitter resistor's value. The emitter resistor can be removed or bypassed with a capacitor for much more gain which is determined by the collector resistor's value divided by the value of the built-in emitter resistance of a transistor.

 

[mstechca]

Of course. 470k is much too high for an emitter resistor. The transistor would be practically cutoff, and its voltage gain would be very low.

I have experimented with the resistor. (from base to ground), and it seems that a value too low will considerably reduce the range of the receiver, which means I can't pick up signals from distant transmitters.

The collector resistor is 18K.

If the collector current and the emitter current are supposed to be equal, and my thinking is correct, then the emitter resistor must be close to the collector resistor, or I strain the base by adding a pull-down resistor that is lower than 82K, and if I do that, wouldn't my gain be cut greatly?

How is the base going to be strained???
What is "a pull-down resistor"???

The "pull-down" resistor I am talking about is the resistor connected between base and ground. If the resistor is too low, the sensitivity will be too low.

The voltage gain of a common-emitter transistor stage that has a high resistance load is roughly the collector resistor's value divided by the emitter resistor's value. The emitter resistor can be removed or bypassed with a capacitor for much more gain...

I figured that one out, but it seems better to add a resistor and a by-pass cap at the emitter instead of a short. I find the quality is better, but I seem to have a problem with volume.

 

[audioguru]

it seems better to add a resistor and a by-pass cap at the emitter instead of a short. I find the quality is better, but I seem to have a problem with volume.

Your emitter resistors are 4.7k but your bypass capacitors across them have a value of only 0.33uF.
At low audio frequencies the gain of each stage is only 18k/4.7k= 3.8.
At 103Hz the gain of each stage is 3.8 x 1.414= 5.4.
At 1030Hz the gain of each stage will be 54.
At about 2060Hz and higher the gain of each stage will be about 100.
No bass whatsoever. Increase the value of the bypass capacitors to about 22uF for flat response down to 47Hz.

When you short an emitter resistor then the transistor will have more base current and will conduct more collector current. Therefore the value of the base bias resistor must be increased to compensate in order for the transistor to be biased correctly.

I don't think you will ever understand how a transistor amplifier stage works without seeing the results on an oscilloscope.

 

[mstechca]

At low audio frequencies the gain of each stage is only 18k/4.7k= 3.8.

This I understand.

At 103Hz the gain of each stage is 3.8 x 1.414= 5.4.

Where did you get 1.414 from?

At about 2060Hz and higher the gain of each stage will be about 100.

I don't get what equation you are using here.

All altering the value of the capacitor will do is alter the amount of bass, reducing it as the capacitor gets smaller.

No bass whatsoever.

Whether I do or do not get bass, that isn't the concern.

Increase the value of the bypass capacitors to about 22uF for flat response down to 47Hz.

I plan to do that.

I don't think you will ever understand how a transistor amplifier stage works without seeing the results on an oscilloscope.

Right now, I can't buy an oscilloscope.

 

[audioguru]

"At 103Hz the gain of each stage is 3.8 x 1.414= 5.4."
Where did you get 1.414 from?

1.414 is the root of 2. At 103Hz, the capacitive reactance equals the emitter resistor's value so the gain is increased 3dB which is times 1.414.

"At about 2060Hz and higher the gain of each stage will be about 100."
I don't get what equation you are using here.

A single RC network at the emitter boosts the gain two times for each doubling of frequency or boosts the gain 10 times for each 10 times the frequency. Therefore at 1030Hz (10 times the frequency that has a gain increase of 1.414) the gain would be 54. Doubling this frequency to 2060Hz then the gain would be 108 but it is max which is about 100 for a single simple transistor stage.

"No bass whatsoever."
Whether I do or do not get bass, that isn't the concern.

You said the volume was too low. Most of the volume of sound is in the bass frequencies. A tweeter speaker that is playing by itself isn't very loud.

 

[mstechca]

A single RC network at the emitter boosts the gain two times for each doubling of frequency or boosts the gain 10 times for each 10 times the frequency. Therefore at 1030Hz (10 times the frequency that has a gain increase of 1.414) the gain would be 54. Doubling this frequency to 2060Hz then the gain would be 108 but it is max which is about 100 for a single simple transistor stage.

I'm real lost :?

Let's say that my RC network has a 1K resistor, and a 1 uF capacitor, and the frequency being received is 200Mhz.

How do I calculate the gain according to your math?

 

[Nigel Goodwin]

A single RC network at the emitter boosts the gain two times for each doubling of frequency or boosts the gain 10 times for each 10 times the frequency. Therefore at 1030Hz (10 times the frequency that has a gain increase of 1.414) the gain would be 54. Doubling this frequency to 2060Hz then the gain would be 108 but it is max which is about 100 for a single simple transistor stage.

I'm real lost :?

Let's say that my RC network has a 1K resistor, and a 1 uF capacitor, and the frequency being received is 200Mhz.

How do I calculate the gain according to your math?

You don't it was an audio amplifier!.

 

[audioguru]

A parallel RC network by itself doesn't have gain. If it is a transistor's bypassed emitter resistor then the transistor has more gain at higher frequencies due to the reduced capacitive reactance of the capacitor at higher frequencies.
Review that the voltage gain of a common emitter transistor amplifier stage is the ratio of the collector resistor to the emitter resistor. Then if the emitter resistor is bypassed with a capacitor, the gain becomes the collector resistor divided by the capacitor's reactance.

A 1uF electrolytic or polyester film capacitor is probably inductive at 200MHz. A 1uF ceramic disc cap with very short leads would have a very low capacitive reactance at 200MHz, 0.0008 ohms. Transistors have a built-in emitter resistance much higher than that so the gain would be max at about 100.

You say that the emitter resistor is 1k but don't say the collector resistor's value. Assume it is 10k. Therefore the voltage gain of the transistor stage is 10 at low frequencies. The 1uF bypass capacitor's reactance equals the value of the 1k emitter resistor at 160Hz where the gain is 10 x 1.414= 14.14 times. The gain rises 10 times for each ten times the frequency so at 1600Hz the gain is 100. The transistor's built-in emitter resistance limits the gain to about 100 so at frequencies above 1600Hz the gain is 100.

 

[Roff]

This is really going to get interesting when you get into calculating the input impedance of the following stage, and the effect that impedance has on the gain.
Oops. Maybe I shouldn't even open that can of worms.
Ah, what the heck. :twisted:

 

[audioguru]

It gets tricky to calculate the gain when the following stage has negative feedback from its collector to its input. It is tricky to calculate the gain of that kind of stage too.

It also gets tricky to measure the gain of a high gain transistor stage that doesn't have negative feedback, so its output is extremely distorted and is nearly rectified.