BJT transistor amplifiers

[animez4me]

Hi there
My partner and I are currently having trouble with a common emitter amplifier. Just for testing purpose in the labs, to see if our calculations were right. We set up Vcc = 6v, Ra = 1k, Rb = 100k, Re = 100, Rc = 470 and load is a 8ohm speaker. Our Vin was 5v (pk-pk). Two 10microF capacitors at input and output of BJT amp. Our calculation shows that we'll get a voltage gain of 16 however our output was even lower than 4v. We're not sure what is going on there?
Thanks

 

[colin55]

The resistance of the speaker is too low.

 

[crutschow]

You calculations apparently did no include the speaker impedance as part of the collector load, which swamps the 470Ω collector resistor.

 

[audioguru]

You didn't attach your schematic so we are guessing about what Ra and Rb connect to.
If the transistor is biased properly and has no external load then its voltage gain is about 4.5, not 16.
The voltage gain is 470/(100 plus 4.7) where 4.7 ohms is the internal emitter resistance of a silicon transistor when its collector current is 7mA.

With the 8 ohm speaker shorting the output then the voltage loss is huge and the output swing is almost nothing.

 

[animez4me]

Yeah sorry about the vague description, I cant produce a schematic at home, I'll try put one up in the labs. I think I'll explain my project a bit more since we're really stuck right now. We were asked to use a 555 timer and any type of BJT amplifier to produce a sound out of the speaker. So right now we're working on a common emitter design.

The Ra and Rb are voltage dividers before the BJT. Here is my calculations, from the theory we were taught.
VE = Vcc*(Rb/(Ra+Rb)) = 5.24
IE = VE/RE = 0.0524
re' = 25mv/IE = 0.477
rc = Rc//RL = 7.87
Av = rc/re' = 16.5

If you guys could point out where i went wrong in the calculations that would be great. Didnt I already include the speaker impedance when calculating rc?
Also how can I bias the amplifier the right way so i can get a better sound from the speaker after the amplifier compare to straight from the timer?

 

[audioguru]

Everything is wrong in your calculations:
1) The collector resistor is 470 ohms and the emitter resistor is 100 ohms so when the transistor is turned on as hard as it can, the current is 6V/(470+100 ohms= 10.53mA which will be the peak current with no load. You want the collector current to be half which is 5.26mA. Then the collector voltage will be 3.53V will be able to swing up 2.47V and swing down 2.47V.

2) Without a signal, the emitter has a current of 5.3mA in the 100 ohm emitter resistor so the emitter voltage is 0.53V. You had the transistor saturated all the time with the emitter voltage at 5.24V which is impossible.

3) The base voltage is about 0.7V higher which is 1.23V. Make a voltage divider for the base with Ra and Rb that provides 1.23V with a current that is 10 times the base current of the transistor that is calculated from its hFE (beta).

4) The internal emitter resistance is 26mV/5.3mA= 4.9 ohms. The voltage gain without a load is Rc/Re which is 470/(100+4.9)= 4.48 and is lower when there is a load.

 

[The Electrician]

Your 555 is already putting out 5 v P-P. That is more than enough voltage to get a "sound" out of a speaker (assuming the frequency is mid band audio). You don't need any voltage gain; you only need current gain. Use an emitter follower (common collector) amplifier.

 

[audioguru]

Your 555 is already putting out 5 v P-P. That is more than enough voltage to get a "sound" out of a speaker (assuming the frequency is mid band audio).

No.
With a 6V supply the output of a 555 with no load is +4.7V. An emitter-follower will reduce it even more to about 4.0V or less.
I don't know if the speaker is fed DC and the current into an 8 ohm speaker is +500mA or if a coupling capacitor is feeding plus and minus 250mA.

 

[The Electrician]

No.
With a 6V supply the output of a 555 with no load is +4.7V. An emitter-follower will reduce it even more to about 4.0V or less.
I don't know if the speaker is fed DC and the current into an 8 ohm speaker is +500mA or if a coupling capacitor is feeding plus and minus 250mA.

No?

What do you mean, no? No, what?

You say that the output of a 555 with 6 v supply is 4.7V. The OP says he was getting 5V P-P. Don't argue with me about that; take it up with the OP. Maybe he was using the CMOS version of the 555.

And, as I said, 5v P-P is more than enough to get a "sound" out of a speaker. If 5 v P-P is more than enough, 4 V P-P is still enough. After all, the requirement is "We were asked to use a 555 timer and any type of BJT amplifier to produce a sound out of the speaker."

Your own calculation indicates that "...the current is 6V/(470+100 ohms= 10.53mA which will be the peak current with no load." with the common emitter configuration the OP was trying out. That will be the maximum current in one direction with a load, and with RE of 100 ohms, he will only get ~60 mA in the other direction (with no speaker coupling cap).

Even if he only gets 250 mA with capacitor coupling and an emitter follower, that sure beats 10.53 mA one polarity and 60 mA the other polarity doesn't it?

He could bias the common emitter amp for much higher quiescent current, but an emitter follower (with no output cap; a few hundred mA of DC in the speaker won't hurt it unless it is very small) is the obvious way to match the relatively high impedance of his 555 output to the low impedance of the speaker, when no additional voltage gain is needed, as is the case here. Maybe the lesson the instructor wants him to understand is that no voltage gain is needed here; it's an impedance matching problem.

And, with only a 6V supply, he can't possibly get more than 6V P-P with any configuration of BJT amplifier operating on that supply, and if he can get 4V P-P with the emitter follower, why bother trying to get more output voltage? There won't be much difference in SPL with 4V P-P compared with 6V P-P.

Also, the problem didn't say that the BJT amplifier has to be linear. The simplest solution would be to connect the speaker as the collector load, allow DC in the speaker, have no emitter resistor and use the BJT as a switch. Insert a suitable resistor between the 555 and the BJT base and possibly another resistor in series with the speaker if there is danger of overheating the voice coil.

 

[audioguru]

The transistor is not needed. Simply connect the 8 ohm speaker to a 1000uF coupling capacitor connected to the output of the 555. The supply voltage is only 6V so the output swing of the 555 at 200mA peak is about +1.5V to +4.5V. The output of the coupling capacitor will be plus and minus 1.5V at 188mA peak.

 

[The Electrician]

The transistor is not needed.

The transistor is needed because the OP was instructed to "...use a 555 timer and any type of BJT amplifier to produce a sound out of the speaker."

 

[audioguru]

The transistor is needed because the OP was instructed to "...use a 555 timer and any type of BJT amplifier to produce a sound out of the speaker."

The teacher doesn't even know that a transistor is not needed.
If a transistor and base resistor are used as a switch then the 3V loss of the output of the 555 can be reduced to about 0.3V so the sound will be slightly louder than without the transistor.
You need to ask the teacher if "slightly more power" is useful amplification.

 

[The Electrician]

The teacher doesn't even know that a transistor is not needed.
If a transistor and base resistor are used as a switch then the 3V loss of the output of the 555 can be reduced to about 0.3V so the sound will be slightly louder than without the transistor.
You need to ask the teacher if "slightly more power" is useful amplification.

You don't know what the teacher knows or doesn't know. It seems apparent to me that the main purpose of the exercise is not just the production of a sound, but rather to teach something about the use of discrete transistors as amplifiers or controllers of electricity.

I'm sure the teacher knows full well that a bipolar 555 can make a sound when directly driving a speaker.

animez4me, would you ask your teacher whether he knows that the 555 could produce a sound if directly driving the speaker, and if he does in fact know that, then why did he ask you to use a BJT between the 555 and the speaker, and report back to us his answer?

 

[animez4me]

Hey guys, thanks for the enthusiastic replies. Yes after testing it ourselves plus the teacher is fully aware that the voltage that comes out of the 555 is more than enough to drive the speaker. The Electrician, you're pretty much dead on for the purpose of this project, working with BJTs Its pretty dumb since we're trying to amplify something that doesnt need to be amplified lol... but hey what can we do XD
One thing to note though, he doesnt want us to make a switch, he wants to see some amplification. Yes and the easiest way is just to use emitter follower since no voltage gain is needed. However the teacher said he designed a common emitter followed by emitter follower so we're giving that a go. Any tips on how to still get a good sound in the end?
btw we've made some notes with switches and resistors ranging from 261-523 Hz so we can play a tune out of the speaker

 

[The Electrician]

From what you've said in this last post, you're allowed to use more than one BJT, which I was wondering about.

Since your supply voltage is only 6V, you can't possibly get more than 6V P-P (without a transformer, for example).

However, since the output of the 555 is something like a square wave, if you have a lot of gain in your amplifier, the waveform will be clipped and it won't look any different than what comes from the 555 directly.

What I would do to make it interesting, and to impress your instructor, is to follow the 555 output with an approximately 10:1 divider. Connect a 10k resistor to the output of the 555 and the other side of that resistor to a 1k resistor to ground. Now you have approximately 1/10 the output from the 555. Now, design an amplifier to bring that back up to several volts P-P. You might even parallel the 1k resistor with a capacitor to make the signal less square wave like; smooth off the corners, so to speak. Smoothing the signal to be amplified will also let you see if your amplifier is clipping the signal--do you know what clipping is?

 

[audioguru]

Hey guys, thanks for the enthusiastic replies. Yes after testing it ourselves plus the teacher is fully aware that the voltage that comes out of the 555 is more than enough to drive the speaker. The Electrician, you're pretty much dead on for the purpose of this project, working with BJTs Its pretty dumb since we're trying to amplify something that doesnt need to be amplified lol... but hey what can we do XD
One thing to note though, he doesnt want us to make a switch, he wants to see some amplification. Yes and the easiest way is just to use emitter follower since no voltage gain is needed. However the teacher said he designed a common emitter followed by emitter follower so we're giving that a go. Any tips on how to still get a good sound in the end?
btw we've made some notes with switches and resistors ranging from 261-523 Hz so we can play a tune out of the speaker

The output of a 555 is a square-wave and the same from a transistor driven from it. A square-wave sounds like a buzzer so it will not produce "good sound".

 

[audioguru]

The teacher might be impressed if the output voltage is almost doubled in a bridged amplifier. Then the output power is nearly 4 times as much.

 

[animez4me]

From what you've said in this last post, you're allowed to use more than one BJT, which I was wondering about.

Since your supply voltage is only 6V, you can't possibly get more than 6V P-P (without a transformer, for example).

However, since the output of the 555 is something like a square wave, if you have a lot of gain in your amplifier, the waveform will be clipped and it won't look any different than what comes from the 555 directly.

What I would do to make it interesting, and to impress your instructor, is to follow the 555 output with an approximately 10:1 divider. Connect a 10k resistor to the output of the 555 and the other side of that resistor to a 1k resistor to ground. Now you have approximately 1/10 the output from the 555. Now, design an amplifier to bring that back up to several volts P-P. You might even parallel the 1k resistor with a capacitor to make the signal less square wave like; smooth off the corners, so to speak. Smoothing the signal to be amplified will also let you see if your amplifier is clipping the signal--do you know what clipping is?

Clipping is when the voltage exceeds some sort of limit and the waveform is like cut off right?
For the 1k and 10k is that the voltage divider infront of the BJT transistor?

 

[The Electrician]

Clipping is when the voltage exceeds some sort of limit and the waveform is like cut off right?
For the 1k and 10k is that the voltage divider infront of the BJT transistor?

Right. The idea is to decrease the amplitude of the 555 signal by a factor of 10, and then use your BJT amplifier to increase the amplitude back to a reasonable level to drive the speaker, several volts P-P perhaps.

 

[colin55]

You can't connect anything more than 10u to an 8R speaker on a 555.