What's wrong with my amplifier? - Page 1
Forum section: electronics / beginners

author: DuanerD
author_rank: Newbie
timestamp: September 14, 2016, 10:56:12 pm
content: Hi everybody,I'm a physic student who try to learn electronic for fun, to try to see how equations work in the real world!For the moment i'm 100% old school, I try to work with basic components only, no IC.My goal is to take two signals (a signal and a carrier) and to modulate them, AM and FM.But i'm far from it. For the moment i have a signal of 14MHz, Vcc=0,4V; created by a Collpitts Oscillator. I want to amplify it, to learn how amplifier worked.I attached the type of amplifier,common emitter, i made, very basic, found it in a book for newbie like me. I also attached a picture of my circuit on my breadboard, with in the left the oscillator and in the middle the amplifier. They are powered each one by one DC 5V generator (I had just one powering them both, but since it didn't work, i tried that), and both grounds are linked.So here what i understand. The AC signal leave the oscillator (f=14MHz, Vcc=0,4V) and enter the amplifier. When I take the voltage of R2, the sine signal is still here, but since he is powered by the 5V generator, he is at 2V. Which it's normal (i think) because of the voltage divisor. When i take the voltage base-emitter of the transistor it show 0,750V. So my transistor is "on", and I try to take back my amplified signal above the collector (common emitter), after a capacitor to make it AC again.The problem is that my signal disappeared. I just have a 4V DC voltage, completely flat.I tried a few things (changed values of resistors, changed transistor, checked 100 times for stupids mistakes), but no result. I must ignore something important here, certainly about transistor. I checked the datasheet of the two transistor i used (2N2222A and BC547) to see if there is a limit of the frequency, and 14Mhz don't seem to be a problem, but i could be completely wrong, i'm not a expert yet to understand all the characteristics... Ho I also tried with a 2KHz sine signal, and no result either...Thanks for any help and suggestions!

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 14, 2016, 11:40:51 pm
content: Check transistor pinout? Check emitter-ground voltage? Check resistor values?14MHz is a little more challenging to amplify, as the transistor capacitance (Ccb) acts in parallel with the collector load resistor, reducing gain significantly. If Ccb = 10pF and Rc = 2.2kohms, then the cutoff frequency will be 7.2MHz. At 14MHz, you will have only a little gain.Miller effect also causes the input capacitance to be multiplied. If your source is 50 ohms (such as a signal generator), it should be able to "push" it around well enough. If the source impedance is higher, perhaps direct from an oscillator, it may cease oscillation entirely!FYI: starting with a carrier, and introducing FM to it, is quite challenging. Rather, it is preferable to vary the oscillator itself. If this is what you meant, then that's alright. I just wanted to make note that: while you can generate AM that way (i.e., with a separate carrier oscillator, signal source, and modulator), not all modulation methods can be applied this way, and the reasons for this can be fascinating.Tim

author: Audioguru
author_rank: Super Contributor
timestamp: September 15, 2016, 12:47:18 am
content: You are very confusing. You say the supply voltage is 5V but your schematic shows 4.5V. Which voltage is correct?You say the 10k and 3.9k voltage divider produces 2V on the base of the transistor which is impossible. The base will be about 1.2V to 1.4V depending on which supply voltage.Where did you measure 4VDC? The collector should be about 2.5VDC, not 4VDC.

author: JS
author_rank: Frequent Contributor
timestamp: September 15, 2016, 01:26:02 am
content: I'd start with a lower frequency to start playing around, maybe 1kHz, low enough not to have any issues with parasitic capacitance and high enough not to mess with the bias or coupling caps.You should draw the voltages you are measuring in the schematic and upload that pic. All referred to ground, maybe an extra BE voltage, but probably not over the schematic. 0.75V is probably too much for the BC547 as it should be in saturation in your circuit. Also, what transistors are you using? That doesn't look like a BC547. You seem to have a choke and a 2nd can in there, what are those doing?JS

author: alsetalokin4017
author_rank: Super Contributor
timestamp: September 15, 2016, 02:31:21 am
content: Is there any signal at the collector of the transistor, before the 10uF output capacitor?And why is there a steady 4 VDC measured on the output side of that capacitor?If the transistor in your photo is a 2n2222a, you may have it connected wrong. The "tab" on these TO-18 metal can transistors indicates the Emitter.It's hard to tell for sure from your photo; you may have the Base lead bent under the transistor so it appears to come out under the tab. Please confirm that you have the transistor pins connected correctly.

author: alsetalokin4017
author_rank: Super Contributor
timestamp: September 15, 2016, 04:19:10 am
content: You are right, that circuit does not work as you have it on the breadboard. Try it with the C3 capacitor installed.This circuit works:... but with only a gain of about 4 or 5 at 14 MHz with a 5 volt supply and 2n2222a.

author: Zero999
author_rank: Super Contributor
timestamp: September 15, 2016, 07:58:59 am
content: Presumably it's a fairly narrow bandwidth 14MHz signal? If so, replace the collector resistor with an LC circuit tuned to 14MHz.

author: DuanerD
author_rank: Newbie
timestamp: September 15, 2016, 09:43:58 am
content: Thx for all yours replies!Quote from: Audioguru on September 15, 2016, 12:47:18 amSorry about the confusion, i'm not use to talk "electronic" at all. My book where i took the amplifier is old, and the supply voltage is 4.5V to match a standard battery used at the time. I use a 5V power supply instead. The values of the resistor i used change a little also. For exemple, i used a 4,7k ohms instead of 3,9k in the book, I thought that it would be ok. About the 4VDC, my mistake it was the voltage on R3, i missed use my oscilloscope... The collector-ground is about 1,125VDC. The emitter-ground is 0,874VDC.Quote from: alsetalokin4017 on September 15, 2016, 02:31:21 amSorry about the mistake of the 4VDC on the output of the 10uF capacitor.There is no signal at the collector before the 10uF output capacitor.I used the BC547 but at the time I took the picture, i changed it for the 2N2222A to see if any changes. Yes i bend the lead base connector under the transistor, it's pin out correctly on the photo.Quote from: alsetalokin4017 on September 15, 2016, 04:19:10 amI will try that today, thanks!Thanks T3sl4co1l, JS and Hero999 for yours answers. I'm not technical enough to really understand what you saidbut I will try with lower frequency and with more precision. I'm waiting for a function generator that should make experiments easier.I'll try today and post my results.

author: CJay
author_rank: Super Contributor
timestamp: September 15, 2016, 10:07:57 am
content: Check your input signals, be sure you're using 2KHz as you stated.14MHz on a breadboard is unlikely to work well.

author: danadak
author_rank: Super Contributor
timestamp: September 15, 2016, 01:04:50 pm
content: Sim of your amp, attached.

author: alsetalokin4017
author_rank: Super Contributor
timestamp: September 15, 2016, 07:11:33 pm
content: And another sim:

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 15, 2016, 07:47:55 pm
content: Note that the source is zero ohms -- rather hard to come by at 14MHz. Now try it with, say, 1kohm source resistance?Tim

author: Ammar
author_rank: Regular Contributor
timestamp: September 15, 2016, 11:37:54 pm
content: I saw 14MHz and breadboard in the same paragraph. Maybe you could try connecting everything up dead-bug style?

author: danadak
author_rank: Super Contributor
timestamp: September 16, 2016, 01:20:05 am
content: Source = 100 ohms, attached. If 1K quite significant reduction. Then you wouldconsider a CE - CB compound, cascode, if you want to stay with discretes.Note your design is not a high gm design, your Icbias, to get the Ft up, that youcould experiment with w/o using cascode, while still evaluating effects of sourceRE. Your current Icbias is ~ 1 mA, could try it using 10 mA. gm ~= Ic / 26mV.CE gain is ~http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/npncegain.htmlhttp://people.seas.harvard.edu/~jones/es154/lectures/lecture_3/bjt_amps/bjt_amps.htmlhttps://en.wikipedia.org/wiki/Common_basehttps://en.wikipedia.org/wiki/CascodeRegards, Dana.

author: Audioguru
author_rank: Super Contributor
timestamp: September 16, 2016, 03:12:01 am
content: Why didn't anybody simulate the original circuit, then the circuit with the wrong value for R2? The original circuit should work fairly well and the one with the wrong R2 should work but with a limited output level before clipping its bottom.

author: alsetalokin4017
author_rank: Super Contributor
timestamp: September 16, 2016, 03:15:42 am
content: Quote from: Ammar on September 15, 2016, 11:37:54 pmThe circuit I posted works at 14 MHz on a breadboard, with a sine wave input from a cheapo DDS FG and a supply voltage of 5 to 12 VDC. Here it is with a 10V supply and about 120 mV p-p input sinusoid:

author: alsetalokin4017
author_rank: Super Contributor
timestamp: September 16, 2016, 03:35:02 am
content: Quote from: Audioguru on September 16, 2016, 03:12:01 amThe original circuit did work for me in the QUCS simulation but I could not get it to work on the breadboard.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 16, 2016, 06:57:05 am
content: What is your voltage source?

author: alsetalokin4017
author_rank: Super Contributor
timestamp: September 16, 2016, 09:25:27 am
content: Quote from: T3sl4co1l on September 16, 2016, 06:57:05 amWho, me? For the breadboard I'm using an old HP721A small regulated PS. For the Qucs simulation I'm using the "ideal DC voltage source" component.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 16, 2016, 09:57:30 am
content: Sorry, not DC voltage source, the AC one!Tim

author: alsetalokin4017
author_rank: Super Contributor
timestamp: September 16, 2016, 11:13:48 am
content: Quote from: T3sl4co1l on September 16, 2016, 09:57:30 amFor the breadboard I'm using a cheap DDS FG, the MingHo MHS-5200A 25MHz version, set to Sine, 150mV (the lowest it will go) at 14 MHz, which shows up as about 120 mV on the Rigol. But as I said I couldn't get the original circuit to work on the breadboard.For the Qucs sim above I'm using the "ideal AC voltage source" set to 400 mV, 14 MHz, phase 0 and theta 0. I also just tried it at 120 mV and it gives an even cleaner output waveform.

author: Audioguru
author_rank: Super Contributor
timestamp: September 16, 2016, 12:32:28 pm
content: My simulation runs the original circuit at 1MHz so that phase shift does not mess up the waveform much. It is shown biased wrong (with the value of R2) and biased correctly.

author: DuanerD
author_rank: Newbie
timestamp: September 16, 2016, 12:44:32 pm
content: I tried the circuit you gave me and it worked!I made few measurements based on this :Quote from: alsetalokin4017 on September 15, 2016, 04:19:10 amWith the 14Mhz signal.I put a resistor of 1kat the output of the AC source before C1, likeT3sl4co1lsuggested. I'll call it the resistor R0.I made the measurements trying to understand the role of R0, C1, and Ce. All the voltages i put are the peak voltages, Vp, not the peak to peak. I don't know the common way to express voltage of alternative signal, so I prefer to precise it. Voltages are not very accurate, i measured them with my oscilloscope (Hameg HM604 60Mhz):With R0,C1 and Ce : input signal Vp = 0,2Voutput signal Vp = 0,6VWithout R0, with C1 and Ce : input signal Vp= 0,5Voutput signal Vp=0V (flat)With R0, Ce and Without C1 : input signal Vp=0,2Voutput signal Vp=0,6VWith R0, C1 and Without Ce : input signal Vp=0,2Voutput signal Vp=0,1VSo the most important component to make it work with the amplifier is R0, C1 is not necessary but if I just remove R0 and take the signal directly, there is almost no difference with the amplifier. So the amplifier is not a great help.First i must say that i don't really understand the meaning of the value of the resistors and capacitors here (different from my book), to make my signal passed through the amplifier or not.The role played by Ce is also something i don't really understand. The capacitor Ce get charged when the transistor let pass the current, and when he doesn't, the capacitor discharged through Re i think, but what role it play on the signal i try to amplify?I gonna resume what I seem to understand (with my basic words) : we use the transistor on the range of possible voltages he could amplify before he get saturated. So the goal is to "squeeze" the signal between the lower voltage where the transistor let the current passed and the voltage where he get saturated. On that range, the voltage collector-ground "follow the shape" of the voltage base-ground and we can take back our signal amplify by the 12V power supply, then we make it passed to a capacitor to filter the DC current. I don't really see the meaning of RL also.If i'm totally wrong, please tell me! Thanks for all you help, and i'm definitely gonna use a simulation software! If you have any to suggest for linux/ubuntu?

author: Audioguru
author_rank: Super Contributor
timestamp: September 16, 2016, 02:24:04 pm
content: An input coupling capacitor (C1 in your circuit) is very important for an AC amplifier because it blocks DC but passes the AC signal. If the output DC voltage of the signal source is 0V like most, then without C1 the base of the transistor is forced to 0VDC and it rectifies instead of amplifying.CE increases the gain of the transistor because it bypasses CE then CE is shorted for AC signals and the gain of the transistor is a little less than (RC parallel with RL)/RS or about 16 times at low frequencies.This amplifier is not a great help because its output impedance is high then the stray capacitance between the wires and rows of contacts on your breadboard cause a lowpass filter that cuts the level of high frequencies. When an amplifier has a low output impedance then stray capacitance does not affect it much. Also, your coupling capacitors have a very high capacitance that is not needed at high frequencies. 1uf and 10uF are usually electrolytic type that has inductance that does not pass high frequencies. You need to learn the simple formula for calculating the capacitance of a coupling capacitor and use a ceramic capacitor at high frequencies.I tried LTspiceXVII to sim your latest schematic and it created a serious error twice. There is something wrong with this XVII version.

author: danadak
author_rank: Super Contributor
timestamp: November 07, 2016, 09:43:36 pm
content: I found this gents presentation excellent, learned a thing or two.Regards, Dana.

author: Bryan
author_rank: Frequent Contributor
timestamp: November 11, 2016, 08:26:53 pm
content: W2AEW has a excellent video on common emitter, collector, base amplifiers. It's a good primer to watch first.https://youtu.be/zXh5gMc6kyU

author: vk6zgo
author_rank: Super Contributor
timestamp: November 12, 2016, 02:02:13 am
content: QuoteI started to write out a lot of stuff from a historical perspective,but after I read the first few paragraphs,ny eyes started to glaze over,so I'll try to make it a bit simpler.Going back to your original circuit,imagine replacing the transistorT1 with a resistor Rt from collector to emitter,& R4 with a short circuit ( the base connection now goes nowhere).You now have a voltage divider consisting of R3 & Rt across your 5v supply.The voltage ( with respect to the negative PSU terminal) at the junction of R3 & Rt (pt "c") is now +5( Rt/(Rt +R3) volts.Now make Rt a variable resistor.You can see that you can vary the voltage at point (c) by varying the value of Rt.Now,imagine you have superpowers & can look at a signal incoming & move the knob of the variable resistor fast enough to keep up with it.You will now have a dc voltage at point (c),varying at the signal rate.Thisac componentwill pass through C2,& appear at the "sortie" terminals.After a while,you will discover (assuming Rt is a real-world component) that for best distortion of the output signal,at expense of voltage level,you will not use all of the adjustment range of the variable resistor.The transfer function of Rt is not completely linear across the full range,so you have applied a mechanical "bias " to get around this.(we'll come back to this)Now restore R4 --the voltage at (c) now becomes:- +5( Rt+R4/(Rt +R3+R4)Varying Rt now varies the total value of the "bottom" part of the divider by a lesser proportion,& hence the level of variation of the current through the divider is less,the voltage at (c) has a smaller ac component,& the ac output at the "sortie" terminals is less.At this point,Microsoft has revoked your superpowers,& you have to restore T1.T1 needs a dc bias to place it in the linear portion of its "transfer function",just like Rt needed a mechanical bias.In the case of a BJT,this is a current biasR1,R2,R4 are all part of the circuit to provide this bias.So we are pretty much stuck with R4,but in this circuit,the gain is lousy.The formula I learnt back in the Dark Ages for CE amplifiers with unbiased emitter resistors was A =gmRc/1+gm Regm is usually >>1,so this is near as dammit A= Rc/Re,which,for the original circuit ,is a gain of about 4.There is,a "fiddle" though--if you use C3,for the ac component of emitter current it looks like a short,so Re disappears from the gain formulaA= gmRc/1 ( as gm times zeroiszero),or just,gm Rc.