# Help understanding amplifier

1. Jun 3, 2013

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.

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
V=IR

3. 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!!!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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2. Jun 4, 2013

### Staff: Mentor

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.

3. Jun 4, 2013

### rude man

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

4. Jun 4, 2013

### 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.

5. Jun 4, 2013

Thanks for the welcome Berkeman! I will definately 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.

6. Jun 4, 2013

### Staff: Mentor

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".

7. Jun 6, 2013

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 lets 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??

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8. Jun 6, 2013

### Basic_Physics

Last edited: Jun 6, 2013
9. Jun 6, 2013

### rude man

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.

10. Jun 6, 2013

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 [Broken]

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

Last edited by a moderator: May 6, 2017
11. Jun 6, 2013

### CWatters

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.

12. Jun 6, 2013

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....

13. Jun 6, 2013

### 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.

14. Jun 6, 2013

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?

15. Jun 6, 2013

### 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.

Last edited: Jun 6, 2013
16. Jun 6, 2013

### CWatters

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.

17. Jun 6, 2013

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!

18. Jun 6, 2013

Sorry I forgot to attach the schematic!

#### Attached Files:

• ###### built amp.jpg
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19. Jun 6, 2013

### rude man

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

Last edited: Jun 6, 2013
20. Jun 6, 2013