So I want to make an audio amp like this one.

http://www.allaboutcircuits.com/vol_3/chpt_4/5.html

Not those same values of course. In fact here is the circuit I did.

Ok the microphone does not put out 1.5 volts like in the picture in AAC. I used 0.2 Volts as a made up example. (took the mic out of a telephone, checked that it works) The output of a microphone without an amplifier is millivolts or less isn't it?


I notice that when I close the switch, the speaker does emit a crackling sound at the moment of connection and the diaphragm physically moves, i.e. there is a current in the collector. I double checked that by using an LED. My circuit is correct.

But alas I do not hear my voice, or heck not even noise, nothing at all from the speaker but silence. I tried yelling into the mic and even gently tapping the diaphragm. Nothing. Silence....

As you see I biased the transistor like this, the way its connected its the sound signal plus the 561.92mV which comes from my connection to the collector node. So If I assume that the output of the microphone is 0.2 volts (hypothetically) then the transistor never leaves active mode as my simulator says.

Now of course it might not be biased properly because I have no clue what the peak voltage output of the microphone really is, and my meter does not pick up something that small.

So can anyone help?

This is a great example of a real life use of a BJT amp. I want to make an intercom system with this and have it go to my basement. (just a joke, I am just learning the basics of transistors here, if I heard ANYTHING I would be happy)

 

Hello,

The circuit in the AAC ebook is only for theoretic use.
Here is a practical schematic for an intercom:
http://www.mitedu.freeserve.co.uk/Circuits/Audio/doorint.htm

There are more audioschematics in the links of this page:
http://www.educypedia.be/electronics/circuitsbyaudio.htm

Greetings,
Bertus

 

What? Why does it have to be so complicated. Op amps, capacitors and transistors. You know that its things like this that might annoy the heck out of an electrical engineering student and make him mad.

Okay class and now after we learned all this stuff for the past 5 months I want to boldly assert that everything you learned is only theoretical and if you want to build it in real life, you need 6 semesters of quantum mechanics.

Man I refuse to believe it. It has to be possible to make a real life amplifier as simple as the one in my schematic.

I don't need an actual intercom, I just want to hear my voice out of a speaker when I speak into a microphone. Its simplicity itself Watson. People made similar circuits in 1895 with a simple vacuum tube. Why over complicate something so simple?

 

Hello,

What are you using as a microphone?
Can the microphone handle the current?
I see you have a 10 Ohm in the base line, wich is VERY low.
What transistor are you using?
What is the voltage at the collector of the transistor when the switch is closed?

As I told you the circuit is for theoretic use only.

Greetings,
Bertus

 

you said that your collector is at 561mv on the collector? that means that your BJT is saturated - ie it is fully on, and has lost its ability to amplify.

so you can increase your 10ohm resistor to a value where the collector is roughly at 3v. I would try a 1k resistor there first.

 

Is it a good idea to put DC on a speaker like that? Wouldn't AC coupling with a capacitor be better?

 

If you are not using a carbon mic you will probably need two transistors. Couple to the speaker using an audio output transformer or any small power transformer you have.

 

Actually that is similar to one of the circuits I learned when I was in high school, except the mic was a carbon type, such as old telephones use. I agree it is not practical, but there is no reason it can not be made to work, as long as the limitations are explained.

First of all, this particular design is a classic bias design, and every weakness the transistor has will show its ugly head. For example, there is something called thermal run away. As the transistor warms up (it is a resistor in this layout) it conducts more, as it conducts more it heats up more, until the loop causes it to burn up.

No two transistors are quite alike, which means designing with specific resistors is out. Normally designs are used that compensate for this variation, but in a raw design like this all the differences are accented, not supressed.

Speakers are very low ohmage, which makes it only worse. They were never meant to be designed to connected directly across the power supply, which this design basically does. The 15VDC power supply guarantees this will be a major issue.

While the current is much lower, the same can be said for the microphone, which is extremely similar to the speaker in this design. Microphones were not meant to carry current.

Making it work

You can make this design work, but it will never be any good. So lets go through the exercise.

First of all, get that power supply down, way down. Try 3V instead. This will reduce the amount of heat generated by the transistor. It will still get warm, maybe hot.

Since the exact gain the transistor has is unpredictable, use a 1 KΩ pot, maybe a 10KΩ for the base bias resistor, with a 470Ω resistor.

Start with the resistor high, then slowly adjust it down. You will hear something, but the gain of a single transistor is also pretty low overall, so it won't be overwhelming.

If you have a DVM measure the voltage on the collector. It will be somewhere around 2.5V, more than that the currents through the speaker and collector of the transistor start getting too high. As is they will be pretty large, if you're using a battery it won't last.

As has been mentioned, a carbon mic works best. This is because it is basically a variable resistor, which is exactly what this circuit needs. It is also much more sensitive to audio, which means it will give the transistor a larger signal to amplify, but it has a very limited frequency response. Telephones didn't need fidelity, so it was perfect for the application.

The reason I showed a speaker for the mic is basically that is what a dynamic mic is. You can also use a conventional speaker for a mic with good results.

Most circuits are more complex for audio amps. They incorporate negative feedback to suppress as many of the undesirable characteristics as they can, plus adding the gain of several transistors means you will have enough gain, while keeping the current to the circuit and the heat it generates strictly under control.

Pesky editor dropped a lot of typing during the edit, so you will note changes as I worked on it.

 

there is a form of negative feedback in this particular design: you will notice that the base resistor is tied to the junction of the speaker and the collector. so when the collector current goes up, the voltage at that junction will go down, which lowers the current going through the base, which lowers the collector current.

you CAN make this circuit work but it will be device dependent.

 

Actually, this design has no negitive feedback, which is the problem. Don't confuse the invertion with feedback. You start getting feedback when you add an emitter resistor, not shown. Basic electronics 101. In the OPs original design the power sources for the base and collector are even separated, with is not generally practical.

You might get another form of feedback, which will prove it is working. If the mic and speaker are too close it will squeal.

 

Actually, this design has no negitive feedback, which is the problem. Don't confuse the invertion with feedback. You start getting feedback when you add an emitter resistor, not shown. Basic electronics 101. In the OPs original design the power sources for the base and collector are even separated, with is not generally practical.

a Re resistor will bring you feedback, so does tying the base to the collector.

the feedback is stronger with a high resistance load (for DC). nonetheless, it is there.

a more refined form of this topology makes a great preamp or headphone amp, and is the basis for some amps by Nelson Pass.

 

Like I said, basic electronics. We are going to have to agree to disagree on this one, as there is nothing on the collector that affects the base bias or signal level on the output (the definition of feedback).

Add an emitter resistor and everything changes, this is the first of several changes that introduce negitive feedback, which also brings thermal run away under control and limits the gain of the transistor (another definition of negitive feedback).

The second classic negitive feedback mechanism on this kind of transistor design is a voltage divider on the base. This interacts with the emitter resistor to lock the transistor down tightly.

With the original design, the transistor gain sets the total gain, there is no control on it.

 

Like I said, basic electronics. We are going to have to agree to disagree on this one, as there is nothing on the collector that affects the base bias or signal level on the output (the definition of feedback).

OK, let's go through that thought process.

let's say that your collector current goes up (maybe the transistor is warmed up). what will happen to the collector voltage? it will go down, right.

Once the collector voltage goes down, what will happen to the base current (in the 2nd schematic where the base is tied to collector via a resistor)? the base current is essentially the collector voltage - Vbe divided by the base resistor, right?

so your base current will go down. what will happen to your collector current? it will go down too.

Oops, that's opposite of what initiated the whole process.

a classic definition of "negative feedback".

this circuit is essentially the basis for "miller cap". a miller cap is nothing but a form of negative feedback (between the collector and the base) that varies with frequency.

Add an emitter resistor and everything changes, this is the first of several changes that introduce negitive feedback, which also brings thermal run away under control and limits the gain of the transistor (another definition of negitive feedback).

all true.

But there is negative feedback even without a Re resistor: when Ic goes up, Vbe necessarily goes up. As Vbe goes up, Ib goes down, which brings down Ic as well.

that effect always exists. the only question is how strong it is. For some devices (lateral mosfets), the effect is so strong that you don't need Re to stabilize Id. for others (vertical mosfets), the effect is only so strong at very high Id (>10amp for example) that it is too late to rely on this effect to save a device.

With the original design, the transistor gain sets the total gain, there is no control on it.

there is, through the negative feedback introduced by the resistor between the collector and the base.

 

Check your books again, this is not how it works. You're trying to treat a transistor as a voltage controlled device. Works in some models, but not this case.

Thermal run away is exactly opposite the way your stating. This design would likely burn itself up unattended.

And the base current is in no way tied to the collector current. The base BE drop will vary a little according to temperature, between .6 and .7V, but the resistors set the bulk of the current, and this doesn't vary. Temperature variation is not considered negitive feedback either.

Ideally, the equation Ic=β Ib best describes this circuit. It does not describe thermal effects, but then we generally try to eliminate those.

 

Check your books again, this is not how it works. You're trying to treat a transistor as a voltage controlled device. Works in some models, but not this case.

I am not sure which part of it is "voltage controlled" but if you think there is a problem with the thought process, please tell me which element of it you think wouldn't work.

 

I just did, Vbe does not change that much. Less than .1V in most cases.

The voltage controlled comment deals with some other threads that got quite hot on the subjuct. There is a voltage controlled model out there, but it is generally not used for basic design.

 

Thermal run away is exactly opposite the way your stating. This design would likely burn itself up unattended.

as I said earlier, the negative feedback will always exist. Whether it is strong enough to overcome 2ndary breakdown, or other impact, depends on the devices.

And the base current is in no way tied to the collector current. This is false.

1st of all, the base current is tied to the collector current via hFE, by definition.

2ndly, the base current is tied to the collector voltage, Ib=(Vc-Vbe)/Rb; and the collector voltage is tied to, among other things, the collector current: Vc=Vcc-Ic*Rload.

so base current IS tied to collector current.

if it were not, it would have been false.

 

As I said, we will have to agree to disagree. The textbooks teach otherwise.

Your formula is off too, like I said Ic = β Ib

Basic transistor theory. Want some references?

http://www.allaboutcircuits.com/vol_3/chpt_4/4.html

Each curve on the graph reflects the collector current of the transistor, plotted over a range of collector-to-emitter voltages, for a given amount of base current. Since a transistor tends to act as a current regulator, limiting collector current to a proportion set by the base current, it is useful to express this proportion as a standard transistor performance measure. Specifically, the ratio of collector current to base current is known as the Beta ratio (symbolized by the Greek letter β):

 

And the base current is in no way tied to the collector current. The base BE drop will vary a little according to temperature, between .6 and .7V, but the resistors set the bulk of the current, and this doesn't vary. Temperature variation is not considered negitive feedback either.

let's work it out.

Vc=Vcc-Rload*Ic.

and Vc=Ib*Rb+Vbe

and Ic=Ib*hFE.

so Vcc-Rload*hFE*Ib=Ib*Rb + Vbe.

Ib*(Rb+Rload*hFE)=Vcc-Vbe

or Ib=(Vcc-Vbe)/(Rb+Rload*hFE).

and Ic=(Vcc-Vbe)/(Rload+Rb/hFE).

Ib's tempco is
dIb/dt=-dVbe/dt/(Rb+Rload*hFE).

since dVbe/dt = -2mv/c,

dIb/dt=2mv/c/(Rb+Rload*hFE).

this is for the schematic that the original poster tried.

for the first schematic, where the base is tied to Vcc via a resistor, the equations are simpler:

Vcc=Ib*Rb+Vbe

Ib=(Vcc-Vbe)/Rb.

dIb/dt=-dVbe/dt / Rb = 2mv/c / Rb.

so the tempco would be BIGGER (more positive) than the one in the 2nd circuit, because this design lacks the negative feedback mechanism we talked about.

obviously, the difference is bigger, or the negative feedback is stronger, in cases where hFE is bigger (high hFE darlingtons), or Rload is bigger.

 

Your formula is off too, like I said Ic = β Ib

which part of it do you think is off?