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voltage divider bias combination

teachlit

New Member
I have a regenerative receiver working and receiving stations. It amplifies RF through a 2N3904 transistor. There is a potentiometer (500 ohm) on the emitter of the 2N3904 and a fixed resistor of 1.8 K ohm in series leading back to negative potential. When this potentiometer is closed (not sure is that is the term but it now provides no resistance to the AC signal), then the receiver will begin to oscillate. A buzz is then heard in the earpiece. If this potentiometer is then turned so its resistance increases a little, the buzz stops in the earpiece and any station tuned with the variable capacitors on the turning circuit will be audible.
I copied the circuit for this receiver from a website and have little idea of how it was designed. I can see it has a figure of eight network of resistors which voltage-divide the bias into the base. I think I just got lucky that this combination of resistor values delivered a Q-point that was in the right place on the load line.
Or did I get lucky?
I would appreciate any comments you might make about my circuit - does it deliver to the correct place on the load line or do I need to work some more?
I am also curious about how the fixed capacitor values around the transistor can be calculated. xtalrfamp.gifxtalrfamp.gif
 

Nigel Goodwin

Super Moderator
Most Helpful Member
Most capacitor values are VERY non-critical, and are simply used for decoupling or for coupling - the left hand most 0.01uF for example is a DC blocking coupling capacitor, and at 0.01uF is massively larger than the minimum it 'could' be (as it couples the RF), but a 0.01uF is physically small, so it makes sense to use a standard value rather than multiple different ones.

Presumably there's something in the layout that makes that a regen?, as there's seems no source of positive feedback around the transistor - the 500 ohm pot simply alters the gain by applying local negative feedback.
 

audioguru

Well-Known Member
Most Helpful Member
The buzz is low frequency audio, not high frequency RF. I think it is caused by having an almost dead battery.
 

teachlit

New Member
Thanks Nigel. That is also my understanding - that the capacitors are used to block DC voltage from any previous or subsequent stage in the circuit upsetting the bias arrangements around the transistor but at the same time allow AC signals through. But I am still interested in how to I could work out the minimum values for the capacitors in my circuit.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
Thanks Nigel. That is also my understanding - that the capacitors are used to block DC voltage from any previous or subsequent stage in the circuit upsetting the bias arrangements around the transistor but at the same time allow AC signals through. But I am still interested in how to I could work out the minimum values for the capacitors in my circuit.
Simply by working out the reactance of the capacitor at the required frequency - you need to know the load impedance (either calculate it, or measure it) - and the reactance of the capacitor along with the load impedance makes a simple attenuator at that frequency. You're usually looking for the reactance to be the same as the the load impedance at the required frequency (or lower, to better pass lower frequencies).

For the feed from transistor to diode (ignoring the loading of the diode) it's about 590Hz - which as it's RF is fine.
 

teachlit

New Member
“Simply by working out the reactance of the capacitor at the required frequency” reads to me like you are speaking a foreign language. I read over this statement twenty or thirty times and still cannot get the meaning.

Let me paint you a picture of the level of ignorance you are attempting to deal with here. First of all, I do know how to work out reactance. I have learned that much. Capacitor reactance is the reciprocal of 2 times pi times frequency times the value of the capacitor. However, in the circuit I gave you above, what frequency should I be applying? And why should I be applying that freqnency? You mention 590 Hz but say that frequency is okay as it is RF. Again, you have lost me. Why is 590 Hz allowable and other frequencies are not?

It could be that you only want people who have similar knowledge to yourself on this forum and that’s okay. I joined because I am very enthusiastic to learn and I do apologise if that enthusiasm irritates and you have to tell me things are ‘simple’ but please can you break it down a little for me? Perhaps show me an example of how the reactance formula is applied in this case.

This stuff is anything but simple for me and that's why I am on this forum so that I can ask "simple" questions and learn from the answers.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
A capacitor can be considered an 'AC resistor' for some purposes (as here), where it's resistance (reactance) varies inversely with frequency.

For the section in question there's a 0.01uF and a 27K - so you're looking for the frequency that the capacitor equals 27K - 590Hz in this case.
 

Ramussons

Active Member
I have a regenerative receiver working and receiving stations. It amplifies RF through a 2N3904 transistor. There is a potentiometer (500 ohm) on the emitter of the 2N3904 and a fixed resistor of 1.8 K ohm in series leading back to negative potential. When this potentiometer is closed (not sure is that is the term but it now provides no resistance to the AC signal), then the receiver will begin to oscillate. A buzz is then heard in the earpiece. If this potentiometer is then turned so its resistance increases a little, the buzz stops in the earpiece and any station tuned with the variable capacitors on the turning circuit will be audible.
I copied the circuit for this receiver from a website and have little idea of how it was designed. I can see it has a figure of eight network of resistors which voltage-divide the bias into the base. I think I just got lucky that this combination of resistor values delivered a Q-point that was in the right place on the load line.
Or did I get lucky?
I would appreciate any comments you might make about my circuit - does it deliver to the correct place on the load line or do I need to work some more?
I am also curious about how the fixed capacitor values around the transistor can be calculated. View attachment 126115View attachment 126115
Could someone tell me where the regeneration comes from? I see it only as a variable gain amplifier.
 

teachlit

New Member
A capacitor can be considered an 'AC resistor' for some purposes (as here), where it's resistance (reactance) varies inversely with frequency.

For the section in question there's a 0.01uF and a 27K - so you're looking for the frequency that the capacitor equals 27K - 590Hz in this case.
But if the formula for capacitor reactance is equal to the reciprocal of two times pi times frequency times capacitance, then the reactance of 0.01uF at 590Hz is 26 ohms. - not 27k.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
But if the formula for capacitor reactance is equal to the reciprocal of two times pi times frequency times capacitance, then the reactance of 0.01uF at 590Hz is 26 ohms. - not 27k.
I suspect you're not using the right numbers? - the capacitance value in the formula is in farads.

I've not done the maths personally for many years, FAR easier to just use an on-line calculator such as this one:


What you've done is worked it out using microfarads instead of farads.
 

teachlit

New Member
I suspect you're not using the right numbers? - the capacitance value in the formula is in farads.

I've not done the maths personally for many years, FAR easier to just use an on-line calculator such as this one:


What you've done is worked it out using microfarads instead of farads.
Thank you. I tried the on-line calculator and entered 10 nanofarad at 590 Hz and got the figure of 27 K ohms as you predicted. I wasn't entering the needed number of zeroes for nanofarads on my computer calculator but got it right when I entered the word nanofarad into the on-line calculator. Much appreciated. Now I have got that right, and see the reactance of the cap is 27 K-ohms at 590 Hz, what is the significant number here - the reactance or the frequency? And why would that be the important one?
 

audioguru

Well-Known Member
Most Helpful Member
10nF in series feeding 27k to ground causes 590Hz to be reduced -3dB and lower frequencies cut at -6dB per octave. Therefore no bass.
Music goes down to 20Hz and my voice averages 100Hz during a conversation.
EDIT: I got it wrong. This is a crystal AM radio with an transistor RF amplifier.
 
Last edited:

teachlit

New Member
Thanks for that breakdown. Is this combinaton, capacitor feeding through to the next stage and resistor connecting to ground; can this be described as a high pass filter? That is, as frequency rises, the reactance reduces and lets more of the AC signal through to the next stage? So this would be a good design feature if I wanted to pass RF through to a further stage.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
Thanks for that breakdown. Is this combinaton, capacitor feeding through to the next stage and resistor connecting to ground; can this be described as a high pass filter? That is, as frequency rises, the reactance reduces and lets more of the AC signal through to the next stage? So this would be a good design feature if I wanted to pass RF through to a further stage.
Yes, simple high pass filter - but as I mentioned, as it's passing RF it doesn't really apply - the capacitor is massively 'larger' than it needs to be as it's simply proving DC blocking.
 

teachlit

New Member
Yes, simple high pass filter - but as I mentioned, as it's passing RF it doesn't really apply - the capacitor is massively 'larger' than it needs to be as it's simply proving DC blocking.
The RF frequencies are passing easily then because at their frequency, let's say 1,000 KHz in the commercial AM radio band and with a 10 nF capacitor, the reactance is only 16 ohms. And to take this further, at 3,500 kHz, the reactance drops to 4.5 ohms, virtually no resistance.
For the sake of research, I entered the same frequencies into the calculator to see what would happen if the capacitor were to be a power smaller - say 1 nF - and found the reactance at 1000 kHz was 159 ohms, and at 3500 kHz was 45 ohms. I also computed these frequencies at 100 nF and got results of 1.5 ohms for 1000 kHz and .45 ohms for 3500 kHz.
As these are all small resistances, I think you have made your point very well in telling me the capacitors are much larger than they need to be and I could have used a range of values.
Would all these considerations be the same for either coupling cap used in a transistor stage? I refer to either the cap leading to the base of the transistor or the one leading from collector. There is a third cap on the emitter pin which I guess would do exactly the same job and could also vary within the same range of values.
I am begining to understand this a bit. Thanks.
 

JimB

Super Moderator
Most Helpful Member
and found the reactance at 1000 kHz
As these are all small resistances,
Sorry to jump in, but I must emphasise that resistance and reactance are not the same thing, and, cannot simply be added together mathematically.
The "sum" of resistance and reactance is impedance, which is calculated by squaring resistance and reactance, adding them together and then taking the square root.
I digress.

There is a third cap on the emitter pin which I guess would do exactly the same job and could also vary within the same range of values.
Not quite.
Ignoring the 500 Ohm variable resistor, the 1.8k emitter resistor contributes to setting the bias point of the transistor, but it also creates negative feedback (NFB).
NFB is usually a good thing to have, but when you are trying to squeeze the last bit of gain out of an amplifier, it is usual to do away with the NFB, because NFB reduces gain.
In the case of your amplifier, the NFB is removed by adding the 0.01uF decoupling capacitor across the emitter resistor.
The idea of a decoupling capacitor is to remove any AC signal, which it does by "shorting out" the emitter resistor to AC (RF in this case).
So to be an effective decoupling capacitor, the 0.01uF must present a low reactance at the frequencies we are interested in. In other words the reactance should be much less than 1.8k Ohm.

JimB
 

audioguru

Well-Known Member
Most Helpful Member
Most radios have Automatic Gain Control so that weak distant stations are amplified up to normal for good sensitivity but strong local stations have the gain reduced to avoid overload like this one does. Most radios also have many tuned LC circuits for good selectivity so that the radio does not play a few stations at the same time like this one does.
 

JimB

Super Moderator
Most Helpful Member
Yes AG, very insightful.

But the fact that this is a good example of a very poor radio is not the point of the current discussion.
The topic under discussion is the biasing arrangements of the transistor and the selection of values of coupling and de-coupling capacitors.

JimB
 

teachlit

New Member
Sorry to jump in, but I must emphasise that resistance and reactance are not the same thing, and, cannot simply be added together mathematically.
The "sum" of resistance and reactance is impedance, which is calculated by squaring resistance and reactance, adding them together and then taking the square root.
I digress.


Not quite.
Ignoring the 500 Ohm variable resistor, the 1.8k emitter resistor contributes to setting the bias point of the transistor, but it also creates negative feedback (NFB).
NFB is usually a good thing to have, but when you are trying to squeeze the last bit of gain out of an amplifier, it is usual to do away with the NFB, because NFB reduces gain.
In the case of your amplifier, the NFB is removed by adding the 0.01uF decoupling capacitor across the emitter resistor.
The idea of a decoupling capacitor is to remove any AC signal, which it does by "shorting out" the emitter resistor to AC (RF in this case).
So to be an effective decoupling capacitor, the 0.01uF must present a low reactance at the frequencies we are interested in. In other words the reactance should be much less than 1.8k Ohm.

JimB
I understand your digression, I think. Z is used as a symbol of impedance and is calculated when you add the squares of the reactance and impedance together and compute the square root of the result. This gives a figure in ohms but it referred to as impedance and not reactance or resistance because it is combination of the two. I have got this idea I think.

But to get to the main idea again of the selection of values in the de-coupling and coupling capacitors in a transistor bias circuit, let's deal with the cap joined to the emitter in many bias circuits. My view of this would be that the cap can be the same value as the bias and collector connected caps and create the same small amount of reactance or lack of reactance. Is the underlying reason for this that the cap connected to the emitter has the purpose of cutting out the resistor connected to the same emitter pin for AC signals. My guess is that there is another process going on across something called the "load line" - I am feeling my way here - and that the said resistor somehow shortens this load line. With the capacitor connected to the emitter pin, this load line is not shortened and the signal is able to swing across a larger portion of the line than if the capacitor were not there. I may have this wrong but that's what I am imagining is happening and would appreciate being put right if I have it it muddled.
 

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