Why is my amplifier not working with my speaker?

[MacDaddio]

Hey guys so I'm going to use this template because it said so in the stickies but my question isn't a word problem kind of thing. I am building an amplifier using this tutorial and am having problems. I have attached my schematic for it, however I'm fairly new to circuits and this project has got me stumped.

Homework Statement

My amplifier doesn't work. The schematic with all the values are attached(it was the easiest way to show what I did). Attachment of the amp to the speaker could be the problem. When I hook the amp, I only get a faint static.

Homework Equations

V=IR

The Attempt at a Solution

I made an audio jack cable go from aux to two wires. One of the wires being positive and the other negative. So when I put these wires into a speaker, the speaker works with my iPod playing. However when I make the wires go through the amp, it doesn't work. I wired the + wire from the audio cable to the amp's input, the amp's output to the + side of the speaker, and the - wire to the - side of the speaker, but nothing... I apologize if I'm doing something completely ridiculous... this is my first homemade circuit! XD All constructive criticism is appreciated!

 

[berkeman]

Hey guys so I'm going to use this template because it said so in the stickies but my question isn't a word problem kind of thing. I am building an amplifier using this tutorial and am having problems. I have attached my schematic for it, however I'm fairly new to circuits and this project has got me stumped.

Homework Statement

My amplifier doesn't work. The schematic with all the values are attached(it was the easiest way to show what I did). Attachment of the amp to the speaker could be the problem. When I hook the amp, I only get a faint static.

Homework Equations

V=IR

The Attempt at a Solution

I made an audio jack cable go from aux to two wires. One of the wires being positive and the other negative. So when I put these wires into a speaker, the speaker works with my iPod playing. However when I make the wires go through the amp, it doesn't work. I wired the + wire from the audio cable to the amp's input, the amp's output to the + side of the speaker, and the - wire to the - side of the speaker, but nothing... I apologize if I'm doing something completely ridiculous... this is my first homemade circuit! XD All constructive criticism is appreciated!

Welcome to the PF.

Can you label each node in your circuit with the voltage that you get by measuring with a voltmeter (DVM)? That will help us to be sure that your biasing for the two transistors is working correctly. Measure each node's voltage with respect to the - side of the battery.

 

[rude man]

Just as a general statement, your base bias voltages are too low and your emitter resistors (10 ohms) are way too low.

 

[CWatters]

I would also check the calculations for the base bias resistors. eg Assume the base current is zero and work out what voltage the potential dividers produce. Are they reasonable?

Edit: Don't bother. Rude Man has already told you they aren't.

 

[MacDaddio]

Thanks for the welcome Berkeman! I will definitely label this with the voltages. I'm going to work soon so I will update my schematic after work. One thing, what do you mean by measure the voltage with respect to the - side of the battery? Like have the - wire of the multimeter on the - side of the battery as i measure all the voltages?

Also thanks rude man and CWatters. I thought the 10 ohms was weird but I was following that guy's example that I linked in the first post. He has the resistors labeled as 10R and when i looked it up online it said that 10 R = 10 ohms... So I will do my own math for when I get back. This is probably the problem I'm thinking.

 

[berkeman]

Thanks for the welcome Berkeman! I will definitely label this with the voltages. I'm going to work soon so I will update my schematic after work. One thing, what do you mean by measure the voltage with respect to the - side of the battery? Like have the - wire of the multimeter on the - side of the battery as i measure all the voltages?

I just meant for you to treat the - side of the battery as the "ground" for the circuit, and label all the other node voltages with respect to that "ground".

 

[MacDaddio]

Okay guys so I have been reading a lot on this stuff and am starting to understand it a lot better, however it is still really confusing. So I decided to simplify my amp for now(See picture) but I'm unsure of what values I should choose for Ic and Rc... I also am not sure how much voltage should be at the point right before the base of the transistor. The voltage divider sets it to a certain voltage right? Well at all the examples I looked at this value is 2-3 volts, but how can I calculate the best value for this?

The voltage drop from Vcc to ground is 9 volts let's say(using a 9v batter).
This means that:
Vcc(battery) = Vrc(from Vcc to collector) + Vce(from collector to emitter) +Vre(from emitter to ground) right?

So I need to pick a value for Ic that will keep the amp on but not fully on. I looked up the max value for Ic and it says its 600 mA. The gain is 100. In the tutorial that I was following, the guy used .5 mA for the Ic, is this normal? How do you know what value to use??

 

[Basic_Physics]

http://www.allaboutcircuits.com/vol_6/chpt_6/10.html

http://www.allaboutcircuits.com/vol_3/chpt_4/9.html

 

[rude man]

Okay guys so I have been reading a lot on this stuff and am starting to understand it a lot better, however it is still really confusing. So I decided to simplify my amp for now(See picture) but I'm unsure of what values I should choose for Ic and Rc... I also am not sure how much voltage should be at the point right before the base of the transistor. The voltage divider sets it to a certain voltage right? Well at all the examples I looked at this value is 2-3 volts, but how can I calculate the best value for this?

The voltage drop from Vcc to ground is 9 volts let's say(using a 9v batter).
This means that:
Vcc(battery) = Vrc(from Vcc to collector) + Vce(from collector to emitter) +Vre(from emitter to ground) right?

So I need to pick a value for Ic that will keep the amp on but not fully on. I looked up the max value for Ic and it says its 600 mA. The gain is 100. In the tutorial that I was following, the guy used .5 mA for the Ic, is this normal? How do you know what value to use??

I said before your emitter resistors were too low (10 ohms). So your move is to reduce them to zero?

ic = 0.5 mA is a good operating point. You might go up to 1-2 mA.

I suggest: Re = 1K, Rc = 3.9K, R1 = 2.4K, R2 = 10K giving

quiescent: ic ~ 1 mA, Vc = +5V giving you ~ 8V peak-to-peak swing at the collector.
V gain ~ 4

Then you can play around with changes.

 

[MacDaddio]

Hey rude man, so at quiescent isn't Ic = (Vcc)/(R1+R2)? Then Ic = .726mA for Vcc =9v or are you saying I should use a 12v value for Vcc to get that Ic value of about 1mA?

I was reading what Basic_Physics posted and it was very helpful, however I'm having difficulty finding the midpoint between saturation and cutoff for my transistor. I was looking at this:

http://www.csus.edu/indiv/t/tatror/projects/met%20highway%20safety%20project%202010/npn%20transistor.pdf

but I'm not seeing one distinct value for the saturation voltage... Sorry if the answer is obvious, I'm trying really hard to understand. :D

 

[CWatters]

Hey rude man, so at quiescent isn't Ic = (Vcc)/(R1+R2)?

No that's not Ic. Ic flows through the collector.

If the base voltage has been set to say X then the emitter will be at Ve = X - Vbe. That voltage appears across the emitter resistor and sets Ic. Actually it sets Ie but Ic≈Ie.

In the very original circuit the base voltage X was set too low (0.7 ish) making Ic very sensitive to changes in Vbe. This was probably done (over done) to try and maximise output swing available from the 3V supply. If Ve is large that reduces how low the collector voltage can go even with the transistor fully on.

 

[MacDaddio]

Well that's the thing, my biggest problem is finding the base voltage right now. I mean I'm looking at load lines, and I'm understanding that Vce = .5Vcc, but I can't solve these equations because I have to many variables. how do I solve for the base voltage? Or do I just pick something? I've noticed that most base voltages were 2-3 v...

 

[CWatters]

Ok this should help...

I hope he doesn't mind but Rude Mans design process probably went like this ...

Choose VCC = 9V
Decide that 1V of that will be "used up" across the emitter resistor to set Ic. So choose Ve = 1V.

Then decide Ic = 1mA (a very common value for this sort of transistor circuit)
That mean Re calculates as Re = Ic/Ve = 1mA/1V = 1K

To get Ve = 1V the base voltage needs to be biasde to Ve+Vbe = 1.7V

We choose R1 and R2 to give:

a) Vb = 1.7V
d) The current flowing in R1 & R2 >> Ib to allow Ib to be ignored.

If Ic = 1mA and hfe were say 100 then Ib would be 0.01mA

I believe Rude Man meant R2 = 2.4K and R1 = 10K which gives

Vb = 9V * 2.4/12.4 = 1.7V as required.

Then current through R1 & R2 is about 9/12.4 = 0.7mA which is >> than Ib so the effect of Ib on Vb can be ignored.

Then what about Rc.. Well the available voltage swing is from Vcc down to about Vcc - (Ve+Vsat) = 9 - (1 + 0.3) = 7.7V Call it 8V.

Half of 8V is 4V

So we aim for Vc = Vcc - 4V
If Ic is 1mA then Rc = 4V/1mA = 4K. Nearest preferred value is 3.9K

EDIT: You are right. Some values are indeed just "choosen". For example you could choose a different value for Ic within reason.

 

[MacDaddio]

Thank you so much CWatters! That actually helped a TON! Okay I get all of it until the Rc part. What is Vsat? Why is it .3v?

Also is it always assumed that Vc is about half of the voltage drop from Vcc to Ve?

 

[CWatters]

Vsat is the transistor saturation voltage. Basically transistors are not ideal devices, they are not ideal lossless switches. Take a look at the plot here..

It shows Ic vs Vce for various values of Ib. Note how on the left hand side the ability of the transistor to sink current (Ic) reduces at low Vce. In a circuit this has the effect of preventing Vce going below about 0.3V (it varies depending on the transistor and other parameters).

So in your circuit Vc can't go below about Ve + Vce no matter how hard you drive the base of the transistor.

Vsat sometmes has implications for the power dissipation. Not in this circuit but in power switches where Ic might be very large. In that case when fully ON the power dissipation in the transistor would be Vsat * Ic.

 

[CWatters]

Also is it always assumed that Vc is about half of the voltage drop from Vcc to Ve?

Not allways but if you don't center the quiescent Vc at roughly the mid point you limit the amplitude of the output. In the case of an audio amplifier you increase the risk that the output will be clipped by the Vcc rail or Ve+vsat.

If you picked quiescent Vc to be say 7V then the output could only swing +2V before it reached Vcc (9V). if you picked quiescent Vc to be say 3V then it could only swing -1.7V before Vc went below Ve+Vsay. To maximise the swing in both directions choose the mid point.

Obviously if you only expect the output swing to be a few mV this is less critical than if you want a swing of say +/- 4V and you only have a 9V supply.

 

[MacDaddio]

Okay sweet thanks CWatters.

I just built the amp again. I drew the schematic for it on paint, sorry for the horrible art skills...

So it ACTUALLY works! However it isn't as loud as just directly connecting the speaker to the iPod. I mean just the fact that works is amazing but it isn't "amplifying" anything. It also has static. Do I need a filter capacitor and bypass capacitor you think? I'm researching it right now, however if you have any useful advice/links that would be greatly appreciated!

 

[MacDaddio]

Sorry I forgot to attach the schematic!

 

[rude man]

Okay sweet thanks CWatters.

I just built the amp again. I drew the schematic for it on paint, sorry for the horrible art skills...

So it ACTUALLY works! However it isn't as loud as just directly connecting the speaker to the iPod. I mean just the fact that works is amazing but it isn't "amplifying" anything. It also has static. Do I need a filter capacitor and bypass capacitor you think? I'm researching it right now, however if you have any useful advice/links that would be greatly appreciated!

The output of your iPod is probably a 1V max signal at reasonably low impedance, say 1K, so actually the input of my design would be OK. Pick C = 1uF to get your low frequencies into the amplifier. Change R1 to 51K and R2 to 12K, for the same reason.

Now, your real problem is with the output. Your output Z is about 4K which is way too high to drive a basic speaker. A stand-alone speaker has impedance of only about 8 ohms. Connecting such a speaker from the collector to ground will drive the collector voltage to almost zero V dc. You gain is well below 1.

We need to know the input impedance of the speaker. Speakers for pc's for example usually have built-in amplifiers so they'll have higher input impedance. Of course, then you wouldn't need your amplifier. Have you tried to hook up a powered pc speaker pair to your iPod? Should work well.

EDIT: might look at http://www.talkingelectronics.com/projects/TheTransistorAmplifier/TheTransistorAmplifier-P1.html

 

[MacDaddio]

Okay and for C = 1uF you mean the input coupling C right?

Well my speaker is an old stereo speaker. However I'm doing this for a project, not necessarily to get better speakers. So although buying a set of powered speakers would be cool I wouldn't get a good grade XD. However this is very interesting stuff to me and I'm really thankful for all the help!

So is there anyway I can impudence match the voltages? IDK the impudence of my speaker but how would I go about testing it? I was reading that you can use a transformer, capacitor, or resistor. From those 3 the capacitor sounds like the best option but I'm not sure what value to use for it. In the link you posted it said that capacitors can act like resistors. So since my Z is 4k ohms (which not sure how you calculated that lol) I would need a capacitor that results in a resistance bringing down the current to acceptable values? I mean let's assume my speaker is an 8 ohm one. Would I need the capacitor to act like a 3992 ohm resistor?

 

[rude man]

Okay and for C = 1uF you mean the input coupling C right?

Yes.

Well my speaker is an old stereo speaker. However I'm doing this for a project, not necessarily to get better speakers. So although buying a set of powered speakers would be cool I wouldn't get a good grade XD. However this is very interesting stuff to me and I'm really thankful for all the help!

So is there anyway I can impudence match the voltages? IDK the impudence of my speaker but how would I go about testing it? I was reading that you can use a transformer, capacitor, or resistor. From those 3 the capacitor sounds like the best option but I'm not sure what value to use for it. In the link you posted it said that capacitors can act like resistors. So since my Z is 4k ohms (which not sure how you calculated that lol) I would need a capacitor that results in a resistance bringing down the current to acceptable values? I mean let's assume my speaker is an 8 ohm one. Would I need the capacitor to act like a 3992 ohm resistor?

No, that would just divide the ooutput in half and only small part of that would make it to your speaker.

The best way would be a small step-down transformer. You'd want the turns ratio to be sqrt(4000/8) or about 22:1. Connect the primary in lieu of the collector resistor and the secondary to your speaker.

Or you could just wire your speaker inlieu of the collector resistor directly. That wouldnot work nearly as well.

PS the word is 'impedance", not "impudence". Look up "impudence" in the dictionary!

 

[MacDaddio]

Okay so I went to buy one at radioshack and apparently they don't have much of a selection. And as usual don't know anything about what they are selling lol. I got the only one that wasn't enormous but I'm not sure what the specs on it are. However looking online it appears that its a 1000 ohm to 8 ohm transformer... Serial number 273-1380. Anyways can I use this? If I just wasted $3 it's no big deal just wondering.

 

[rude man]

Okay so I went to buy one at radioshack and apparently they don't have much of a selection. And as usual don't know anything about what they are selling lol. I got the only one that wasn't enormous but I'm not sure what the specs on it are. However looking online it appears that its a 1000 ohm to 8 ohm transformer... Serial number 273-1380. Anyways can I use this? If I just wasted $3 it's no big deal just wondering.

Yeah, that's a good start. Wire the primary between Vcc and collector, don't use a collector resistor.

Before hooking up the speaker to the secondary, I would get a 10 ohm resistor & hook it up instead. Check out the collector voltage with a scope when you input a nice 1000 Hz sine wave of 0.1 - 1 V.

Also, if you're using one of those small instrument 9V batteries you'd better bypass the battery with a big old 100 uF or bigger capacitor, aluminum or tantalum. Make sure it's rated for 9V. Even so, I suspect the 9V won't hold up too well if your input is too large.

 

[MacDaddio]

Okay when you talk about bypassing the battery with the capacitor do you mean to put the capacitor in parallel with Re? Thats what I've seen people do, or do you mean put the capacitor between Vcc and the battery + side?

 

[rude man]

Okay when you talk about bypassing the battery with the capacitor do you mean to put the capacitor in parallel with Re? Thats what I've seen people do, or do you mean put the capacitor between Vcc and the battery + side?

Put the capacitor across the battery. Vcc IS the battery + side.

You should never bypass Re with a capacitor unless it's a small one, because that's like using too low an Re again.

You could experiment with shunting Re with a combo R and C, e.g. R = 100 Ω and C = 47 μF. This has the effect of increasing gain without disturbing the 1 mA collector quiescent current. Adjust R while putting in a 1V sine input voltage and look at the voltage at the collector. If it distorts, increase R.

 

[MacDaddio]

Okay will do sir! Okay so let's say that I connect the ipod directly to the speaker with no amp. If the max voltage is 1v and the speaker impedance is 8 ohms then the current through the speaker is 1v=8 ohms which is 125 mA right? So power is 1*.125 = 125mW

And with the amp the output is 4 v which makes the current 4/8 = 500mA. So power would be .5*4 = 2000 mW.

Now wen I actually test it the speaker gets less loud as i run it through the amp. So the above equations obviously don't work. But how come? Where is my error?

 

[rude man]

Okay will do sir! Okay so let's say that I connect the ipod directly to the speaker with no amp. If the max voltage is 1v and the speaker impedance is 8 ohms then the current through the speaker is 1v=8 ohms which is 125 mA right? So power is 1*.125 = 125mW

And with the amp the output is 4 v which makes the current 4/8 = 500mA. So power would be .5*4 = 2000 mW.

Now wen I actually test it the speaker gets less loud as i run it through the amp. So the above equations obviously don't work. But how come? Where is my error?

Your ipod puts out 1V (I'm guessing) only if it's not loaded down too much.

A pair of earphones have impedance in the 1-2K range so they don't load the output down too much. But your speaker loads the output down by a factor of about 2K/8. You are making a voltage divider between the output impedance of the iPod and the speaker.

Also, with the amp your power output is nowhere near 125 mW. You would not get a 4V amplifier output with the speaker connected. It would be only about 8/1000 V with an input of 1V.

Did you hook up & try the transformer yet?