Seeking answers about transistors in audio amps.

Thread Starter

Sillywan

Joined Jun 18, 2021
4
I am an old TUBE guy and I am having the hardest time understanding how to work with transistors as I did tubes. No matter how many times you say it a transistor is in no way a tube. I replace a tube and I have a set of simple parameters to meet. Beam tube, pentode, triode get the voltages where they belong and its pretty much swap and go. Biasing power tubes excluded. There is some knowledge involved there. I replace a transistor its a crap shoot. every single component has to be biased in order to function properly if at all. No wonder it takes triple the components to build a solid state amp.
Transistors are a limited edition type of part. There are only so many of that part and when they are gone you are possibly screwed if you are trying to fix your amp. First you have to pray that there is a replacement part that will fit the electronic requirements. If you find it you have to pray that even though it fits and matches requirements, is it going to work. That is where I am now. I know it has something to do with the biasing of the three transistors I replaced, I am just not sure what. They were all NTE replacements but well within the range of the parts they replaced.
I replaced every diode and every electrolytic cap on the board and measured every resistor 1 meg and above. The unit lit up and when the instrument was strummed the amp played the sound to a peak and went silent for a second or two and then made sound again as it faded. I am over driving the second transistor in the out put some how , right? Sorry for the dumb questions. I have built plenty of tube amps but it is luck if I get a SS amp repaired without original parts. Thanks for your input I would love to figure out the big secret to making transistors work.
 

MrChips

Joined Oct 2, 2009
30,701
Actually, no.

A properly circuit designed using transistors is largely independent of the transistor's parameters.
We design circuits to accommodate different transistor specs. It is very easy to repair a unit using transistors that are not the same as the original design.

Give us your situation and your question and we will try our best to assist you. Be certain to post the circuit schematics.

BTW, I am an old tube guy and I can work with tubes and transistors, no problem.
 

LowQCab

Joined Nov 6, 2012
4,022
The problem is that you are dealing with a really BAD Amplifier Design.
Musical-Instrument-Amplifiers,
whether they are Tube or Transistorized,
are ALL really bad designs ............... on purpose.

A certain "Character" of Distortion is the goal of the Musician,
Dirt-Cheap Production-Costs are the Goal of the Musical-Instrument-Amplifier manufacturers,
and they will cut-corners that nobody else could even imagine, just to save a few Pennies.

Get your "Tone" from Software on your Computer,
there are many top-notch Softwares to choose from.
Then go shopping at "Beringer", and get a "Commercial-Grade" Amp with
at least twice the Power that you think you'll ever need.

Then select a pair of reasonably High-Fidelity "PA" Speakers,
( they can be quite small and portable,
~10" X 12" X 8" will work fine,
the next size-up would be better though ),
because you'll be playing in Stereo now,
with HUGE-WIDE Reverb and other Stereo-Effects.

Or, if you're in a Band, you can simply send a Stereo-Feed to the Mixer,
directly from your Computer,
and have your Guitar mixed into the Stage-Monitors.

Now you won't be "stuck-with" the single "Tone" you have now,
and you can feel much more confident that your Amp is not going to take a dump on you.
.
.
.
 

Audioguru again

Joined Oct 21, 2019
6,671
I was glad to sell my tubes amplifier and tuner in 1964 because the tubes frequently failed and needed replacements.
I bought a solid state stereo receiver that lasted 26 years. Its switches wore out, Its Japanese replacement also had its switches wear out. None of the transistors and ICs failed.
 

Ian0

Joined Aug 7, 2020
9,667
You can make a valve amp with a triode or pentode on the input, a triode phase splitter, and two output pentodes and a big expensive output transformer. You can make a transistor amplifier with three small signal transistors and two power MOSFETs, and no transformer, and it will have <0.1% distortion.

Most silicon Transistors that have ever been used in power amplifiers ( eg the ubiquitous 2N3055) are still available.

I suspect that the amplifier that died was overbiassed. An overbiassed valve amplifier is just as likely to fail, but it would give more warning, such as the anodes glowing red!
 

Thread Starter

Sillywan

Joined Jun 18, 2021
4
Actually, no.

A properly circuit designed using transistors is largely independent of the transistor's parameters.
We design circuits to accommodate different transistor specs. It is very easy to repair a unit using transistors that are not the same as the original design.

Give us your situation and your question and we will try our best to assist you. Be certain to post the circuit schematics.

BTW, I am an old tube guy and I can work with tubes and transistors, no problem.
Thank you for your help. Yes sir I agree transistor circuits are are designed for the components that will be used. I understand that part of electronics and design theory as a whole. My problem is I had mentors when growing up thru the tube years and I learned from them the ins and outs of grid caps and voltages,bias supplies, TUBES, how forgiving they can be when designing a tone on the fly. Actually voicing tubes to get "something unique" or extra. The only mentor I ever had involving SS is a book. And I have read a bunch. What I have been able to learn is transistors are not forgiving at all. Instead of amps and volts I am looking at miliamps and milivolts. I am just not sure where I am looking at these values to adjust them. I know how to adjust the values with resistors I am just bench stupid in just where and how I should be looking for them. I will gather up all my notes and get pics of the circuit. This is a late 80's taiwon practice amp, I doubt there will be a schematic available. I will have the original part numbers and the parts i replaced them with. I am just trying to learn how to work with transistors. I have done the work and replaced them, and replaced all of the antique electrolytics and leaky diodes. I "THINK" I need to adjust the bias for the new components so the output transistor is not over driven which is causing the cutoff of the audio until the signal drops low enough for the component to work again as the sound fades away. Any way I will gather it all up and see if you can help me cross this transistor bridge I can't seem to cross. Thanks again.
 

Ian0

Joined Aug 7, 2020
9,667
You'll probably find a basic transistor power amplifier, and a preamp. The preamp will probably incorporate a circuit (maybe just two diodes) to create distortion that might sound a little bit like the distortion made by an overdriven valve power amplifier.
There will probably be a circuit diagram on the internet!
 

nsaspook

Joined Aug 27, 2009
13,079
What's up with the wooden rack? Some sort of vibration dampening system?
When I rebuilt the room for sound I need to hide the heating pipe, so I made a fake 'fire-place' to hide the amps. There is damping inside to reduce resonance. The structure was wood so spacing a few 2X4's for 19 inch racking was easy.

 

MrChips

Joined Oct 2, 2009
30,701
I will give you some tips and pointers on working with transistors.

We will examine two common applications:
1) a transistor as a switch
2) a Class A voltage amplifier.

Since you understand how vacuum tubes work (valves for U.K. readers) let us review a Class A audio amplifier.
AAC tube amp1.jpg
As you already can guess, we need to select a tube and the HV supply voltage. (I have left out the heater supply as this is taken for granted. For the uninitiated, tube filaments are commonly powered by 6.3VAC or 12.6VAC depending on the tube.)

We supply a load R2 between +HV and anode (plate). Input signal is coupled to the grid via an AC coupling capacitor C1.
The tube is already conducting. Keep this in mind when we get to discussing enhancement and depletion modes of FET (field effect transistor).

We need to apply a negative bias to the grid if we want to reduce the tube current. The voltage at the anode is 180° out of phase with the input. We couple the signal to the next stage through a DC blocking capacitor C2.

AAC tube amp2.jpg
Since it is a nuisance to have to generate a negative bias, we raise the potential of the cathode by inserting a resistor from cathode to ground. Now the grid is biased at 0V but is at a negative voltage with respect to the cathode.

R3 serves two purposes. It provides a negative bias on the gate. It also provides negative feedback which is always a good thing. This helps to make the amplifier less non-linear, i.e. a better linear amplifier.

With negative feedback comes a reduction in the gain of the amplifier. We can regain some output voltage amplitude by placing a bypass capacitor across R3. This combination of R3//C3 (means R3 in parallel with C3) has frequency dependency which we can use to our advantage.
AAC tube amp3.jpg
When we come to describing a transistor amplifier you will find the configuration to be very similar.

What do we mean by a Class A amplifier?
The bias point of the amplifier determines the classification.
For our discussion we have:
Class A - the active device is biased in the linear region
Class AB - between Class A and Class B
Class B - the active device is biased close to the cut-off region
Class C - the active device is biased in the cut-off region

In the graph shown below the active device is biased as Class A so that the operating point or Q-point (quiescent point) is midway on the operating range of the device. This Q-point describes the DC operating current and voltage when no input signal is applied to the circuit.

The red line is called the load line.

1624137671500.png


Don't be afraid to ask questions when you need clarification.
 

Thread Starter

Sillywan

Joined Jun 18, 2021
4
I will give you some tips and pointers on working with transistors.

We will examine two common applications:
1) a transistor as a switch
2) a Class A voltage amplifier.

Since you understand how vacuum tubes work (valves for U.K. readers) let us review a Class A audio amplifier.
View attachment 241694
As you already can guess, we need to select a tube and the HV supply voltage. (I have left out the heater supply as this is taken for granted. For the uninitiated, tube filaments are commonly powered by 6.3VAC or 12.6VAC depending on the tube.)

We supply a load R2 between +HV and anode (plate). Input signal is coupled to the grid via an AC coupling capacitor C1.
The tube is already conducting. Keep this in mind when we get to discussing enhancement and depletion modes of FET (field effect transistor).

We need to apply a negative bias to the grid if we want to reduce the tube current. The voltage at the anode is 180° out of phase with the input. We couple the signal to the next stage through a DC blocking capacitor C2.

View attachment 241695
Since it is a nuisance to have to generate a negative bias, we raise the potential of the cathode by inserting a resistor from cathode to ground. Now the grid is biased at 0V but is at a negative voltage with respect to the cathode.

R3 serves two purposes. It provides a negative bias on the gate. It also provides negative feedback which is always a good thing. This helps to make the amplifier less non-linear, i.e. a better linear amplifier.

With negative feedback comes a reduction in the gain of the amplifier. We can regain some output voltage amplitude by placing a bypass capacitor across R3. This combination of R3//C3 (means R3 in parallel with C3) has frequency dependency which we can use to our advantage.
View attachment 241699
When we come to describing a transistor amplifier you will find the configuration to be very similar.

What do we mean by a Class A amplifier?
The bias point of the amplifier determines the classification.
For our discussion we have:
Class A - the active device is biased in the linear region
Class AB - between Class A and Class B
Class B - the active device is biased close to the cut-off region
Class C - the active device is biased in the cut-off region

In the graph shown below the active device is biased as Class A so that the operating point or Q-point (quiescent point) is midway on the operating range of the device. This Q-point describes the DC operating current and voltage when no input signal is applied to the circuit.

The red line is called the load line.

View attachment 241703


Don't be afraid to ask questions when you need clarification.
So this is a basic triode amplifier in transistor form? Different component values to accomodate the low voltage transistor instead of high voltage tube. I would measure the bias point for the value I would be setting with R1 and R2?
 

MrChips

Joined Oct 2, 2009
30,701
So this is a basic triode amplifier in transistor form? Different component values to accomodate the low voltage transistor instead of high voltage tube. I would measure the bias point for the value I would be setting with R1 and R2?
Let’s hold off on the answer to that until we get to a transistor amplifier and I will let you judge for yourself.
 

Audioguru again

Joined Oct 21, 2019
6,671
The value of Re also sets the bias point.
Many audio amplifiers today do not use transistors. Instead the active parts of many amplifiers is on an integrated circuit.
Here is an LM3886 stereo amplifier that produces 68W into 4 ohms or 38W into 8 ohms with low distortion when powered with +28V/0V/-28V.
 

Attachments

MrChips

Joined Oct 2, 2009
30,701
Before moving on to a Class A transistor amplifier let us briefly discuss a BJT (bipolar junction transistor) application being used as a switch.

Suppose we have a relay that we wish to control (turn on and off) from a digital 0-5V signal.
Let us assume that the specifications of the relay call for 50mA @ 5V in order to activate the relay.
AAC transistor switch.jpg

Here we use an NPN 2N3904 BJT to act as a power switch to activate the relay.

In this example, the relay is placed on the positive 5V power supply side of the BJT. In this case the BJT is called a low-side switch.

With no active control signal we want the transistor to be in the cut-off mode.
With an active control voltage of +5VDC we want to drive the transistor ON in the saturation mode.
Thus this is very different from a Class A linear amplifier that operates in the active region.

1624146290888.png

How much input signal (current and voltage) is required to operate this circuit?
For this, we need to examine the transistor's data sheet.
2N3904 datasheet

The base-emitter junction of a BJT resembles that of a P-N diode. It takes only 0.7V to forward bias the diode and turn on the transistor. A typical current gain (beta or hFE) is 100 to 300.

This means that:
collector current = beta x base current
Thus, if beta = 100 and we need collector current Ic = 50mA
we need IB (base current) = 50mA / 100 = 0.5mA

Beta changes with operating current and voltage.

As mentioned earlier, we don't design circuits to rely too much on device specifications. We don't leave it up to chance that we will get a champion part or the best of the manufacturer's production lot. We take a worst case approach.

The rule of thumb when designing a transistor to be used as a switch is to assume a current gain (beta) = 10. This will guarantee a working switch with just about any suitable transistor.

Thus, IB = Ic / beta = 50mA / 10 = 5mA.
Input resistor R1 sets the base current.

The correct calculation is R1 = (Vin - VBE) / IB

For a close enough approximation R1 = 5V / 5mA = 1kΩ

(Diode D1 is required to suppress the back e.m.f. when an inductive load is switched off.)

The take-away here is we learn:
1) Device load line
2) Three operating regions on the load line, cut-off, active, saturation
3) Use device parameters as a guide, assume worst case scenario.
 

Thread Starter

Sillywan

Joined Jun 18, 2021
4
This is cool. I've done this. I used this same rule of thumb theory to make a small preamp for a transducer for a mandola. It looked exactly like the LPB circuit. probably because it was, basicly. I used a 3904 for that. How strange is that I have that drawing here somewhere. But in the circuit above the diode stops the backdraft from the coil when the field collapses. and the signal is voltage/amperage, not audio. Am I in the ballpark, or am I heading for the parking lot?
 

MrChips

Joined Oct 2, 2009
30,701
Ok. Now that we are aware of what a load line looks like and that the transistor can be biased somewhere along the load line, cut-off region, saturation region, active region, let us examine some simple BJT linear amplifiers.

As with a tube circuit, we select a device, supply voltage, and load.
AAC transisto amp1.jpg

The graph below is the I-V characteristics of a typical device.
While the numbers on this graph might not match that of the 2N3904 BJT we can still use this as an example.

1624150703308.png

R2 is our load resistor.

When the transistor is in the saturation region, VCE = 0V, Ic = Vcc / R2 = 12V / 150Ω = 80mA
This gives us the upper left point A on the y-axis.

When the transistor is in the cut off region, Ic = 0mA, VCE = supply voltage Vcc = 12V.
This gives us the lower right point B on the x-axis.

We draw a straight line between these two points to give us the Load Line shown in red.

When the circuit is operating as a linear amplifier, the transistor current and voltage must lie on this load line.

We select our Q-point halfway along the load line at Ic = 40mA @ VCE = 6V.
In order to arrive at this DC bias point, we need to supply a base current of IB = 60μA. (This number comes from the device parameters).

Now we are able to determine the bias resistor R1.

R1 = (Vcc - VBE) / IB = (12V - 0.65V) / 60μA = 189kΩ

As an approximation
R1 = Vcc / IB = 12V / 60μA = 200kΩ

200kΩ is good enough.

This is a simple example and not recommended in a real design.
As the temperature increases, so does the base current and hence the collector current.
We have a positive feedback situation or run-away situation.

In the next step, we will incorporate negative feedback to stabilize the DC operating conditions of the circuit.

Points to note
1) The bias voltage on the base-emitter junction is almost constant, from 0.6-0.8V.
2) The input impedance of the circuit is low compared to that of a tube amp circuit.
3) The output impedance is low compared to the tube circuit.
4) We bias the tube amp with high input impedance with voltage. The low impedance input BJT gets biased with current into the base.
 
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