dual transistor common emitter amplifier - Page 1
Forum section: electronics / projects

author: robse
author_rank: Contributor
timestamp: June 02, 2017, 06:26:06 pm
content: Hi,I'm trying to implement a dual transistor amplifier for low signals. I'm a newbie at RF, but after many tries and some math I achieved a good result on multisim.It's a common emitter ampifier, I just need voltage gain (up to 100Khz is fine, input will be some mV. I set the gain at -10).Here is the circuit:Now I would like to add a second stage with unity gain (-1) just to invert the phase of the signal (Ideally I need both an amplified signal in phase with the input and an amplified signal in counterphase with the input). I tried to add another 2n2222 but it screws everything.Could someone help with this ?Thank you,robse

author: edavid
author_rank: Super Contributor
timestamp: June 02, 2017, 06:43:00 pm
content: Take the amplifier you have and add a phase splitterhttps://en.wikipedia.org/wiki/Phase_splitter(BTW, I don't think your amp actually has a gain of -10, because of the bypassed emitter resistor. What does the simulator say?)

author: robse
author_rank: Contributor
timestamp: June 02, 2017, 06:57:29 pm
content: I tried adding a phase inverter without luck, i'm probably doing something wrongCould you provide a schematic ?Multisim says that the circuit has a gain of -10ThanksRobse

author: Zero999
author_rank: Super Contributor
timestamp: June 02, 2017, 08:09:56 pm
content: The amplifier will have a gain of around 55, at low frequencies. It wouldn't surprise me if it has a gain of 10 at 100kHz, due to the Miller effect.What bandwidth do you require? Is it just a narrow range of frequencies around 100kHz you're interested in?

author: robse
author_rank: Contributor
timestamp: June 02, 2017, 08:24:12 pm
content: 10Hz-100khz would be ideal.

author: edavid
author_rank: Super Contributor
timestamp: June 02, 2017, 08:42:22 pm
content: Quote from: robse on June 02, 2017, 06:57:29 pmLet's see what you tried.QuoteThat's at DC.

author: rstofer
author_rank: Super Contributor
timestamp: June 02, 2017, 08:43:51 pm
content: What is the output impedance of the first stage?How does that work with the input impedance of the second stage?In rough numbers, the output impedance is RCor 10k and you should be able to simulate an input impedance to see what happens.

author: robse
author_rank: Contributor
timestamp: June 02, 2017, 09:10:38 pm
content: edavid: I know that bypassing Re the gain should be different, but this is what the oscilloscope says:Consider that A Channel (red) is at 1 mV/Div while B (green) is at 10mV/DivAs per the second stage, I tried adding something like this, that doesn't work as I expect, ie as a phase splitter (It must be really wrong, beware):The oscilloscope reaches more than 2V on both sides and the waveforms are clipped and distortedrstofer: I'm not skilled on impedance matching, I don't know how to simulate that yet

author: edavid
author_rank: Super Contributor
timestamp: June 02, 2017, 09:53:07 pm
content: Bottom of R8 needs to connect to ground, not input.You might need to do something about DC bias for Q2, but give it a try.

author: robse
author_rank: Contributor
timestamp: June 02, 2017, 10:03:40 pm
content: I fixed the R8 error but it doesn't workThanks

author: edavid
author_rank: Super Contributor
timestamp: June 03, 2017, 12:05:02 am
content: What is the DC level at Q2's base?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 03, 2017, 03:09:23 am
content: First lesson:If all you need is voltage gain, use a transformer (or matching network)!If you actually need power gain, only then do you need an amplifier.Of course, transformers and networks can be more expensive than transistors, so this is not a strict requirement! But if you're talking RF, it should be the first thing you think about, andthenyou weasel your way out of it based on specs like how much noise / SNR, dynamic range and distortion your application needs, and of course, cost and availability.Transformers have impedance, so this also gets you thinking immediately about impedance. RF is all about impedance matching, conservation of energy. Once you know that, everything else falls into place: how to choose L and C (roughly), wideband transformer design, which amplifier types to choose, etc.Tim

author: Audioguru
author_rank: Super Contributor
timestamp: June 03, 2017, 03:19:22 am
content: Your scope is useless since it does not show DC voltage levels like 0V and +5V.I calculate that the first transistor is biased wrong so it is saturated and cannot ampllfy anything unless the input signal level is huge.I show your circuit and my modification of the biasing:

author: robse
author_rank: Contributor
timestamp: June 03, 2017, 11:42:05 am
content: Audioguru: could you also suggest the second stage with the phase splitter ?Huge thanks,Robse

author: rstofer
author_rank: Super Contributor
timestamp: June 03, 2017, 02:11:41 pm
content: The bias changes above incorporate one subtle change that is shown but not mentioned: the AC coupling capacitor. This allows the bias resistors to set the operating point of the amplifier and blocks the DC component of the input. This concept is probably even more important on the input to the second stage because we know that the collector of the first stage can only swing between 0V and 5V. It's average value will be around 2.5V. In fact, when we set up the Dc bias resistors, that's what we're aiming for: Vcc/2 at the collector (give or take).The problem with capacitors is that they are frequency sensitive. To couple low frequencies, they need to be fairly large. For audio circuits, it isn't unusual for these capacitors to be 100 ufd or so.https://coefs.uncc.edu/dlsharer/files/2012/04/D7.pdfsee note where 'all C -> infinity'You can search Google for 'ac coupled small signal amplifier' to find other references.

author: Audioguru
author_rank: Super Contributor
timestamp: June 03, 2017, 03:05:40 pm
content: The first transistor is biased so that its collector can swing equally up to +5V and down to about +0.6V then it idles at about +2.8V.I do not know why you want a phase splitter or the impedances it must drive but I made one that has 1k output resistors.I needed to use a voltage divider with R7 and R8 for the widest output swings.

author: robse
author_rank: Contributor
timestamp: June 03, 2017, 03:10:45 pm
content: I made a capacitor coupled scheme, It works but I have a gain of about 5 at all frequencies ... can someone explain why ? Should be R3/R4 = (-)10 ?Thanks

author: robse
author_rank: Contributor
timestamp: June 03, 2017, 03:20:42 pm
content: Audioguru: Thank you, I achieved a similar result with the capacitor coupled scheme posted above, I never saw the way you placed R8.Can I adjust the impedance matching of the load by just changing r5/r6 (say 10k/10k for a 10K load ?)rstofer: very well made document, thanks

author: rstofer
author_rank: Super Contributor
timestamp: June 03, 2017, 09:51:55 pm
content: Quote from: robse on June 03, 2017, 03:10:45 pmThe output impedance of the first stage is about 10k (the collector resistor) and the input impedance to the second stage is similar (82k in parallel with 15k). So, there's a voltage divider that cuts the first stage gain down to about 5 as far as the base of the second stage is concerned. The gain of the second stage is just 1 so, there you have it: an overall gain of about 5.You might try scaling up the bias transistors of the second stage or scaling down the collector and emitter resistors of the first stage. Try multiplying the second stage bias resistors by 2 or 5. Try dividing the first stage collector and emitter transistors by 2 or 5. It is the RATIO of the resistors than determine gain. Make sure you still get full swing on the transistors.It's easy to simulate so go for it. You already have a circuit that works even though the overall gain is low.

author: MarkyMark
author_rank: Newbie
timestamp: June 04, 2017, 02:05:10 pm
content: Hi,you could try this circuit (the attached file).The circuit gain is somewhere around -23 but it can be easily changed by modifying some resistor values. I would increase the value of Re1 and decrease the value of Re2 to obtain a gain of -10. Re1+Re2 should be 1.1kohm for the bias point not to change.

author: robse
author_rank: Contributor
timestamp: June 04, 2017, 05:39:32 pm
content: Quote from: rstofer on June 03, 2017, 09:51:55 pmI ended up with this:I have a gain of about 20 and the phase splitter working correctly. I adjusted r7/r8 for impedance matchingIs r4 value too low ?Now another recipe: would it be possible to replace both c3 and c2 with resistors ?Audioguru: changing r4 your voltage divider gets ruined, could you update it ? (maybe telling me how to calculate it too ?)Thank you guys,Robse

author: rstofer
author_rank: Super Contributor
timestamp: June 04, 2017, 06:10:23 pm
content: Quote from: robse on June 04, 2017, 05:39:32 pmThe way I read your scope, you have an input of 20 uV and an output of 1.1 mV which seems like a gain of 55 (all numbers rounded).Consider R3, R5 and R6 as all being in parallel and working as Rc in the gain equation of Rc/Re so about 5.6k / 100 gives the gain figure of something around 56.No, you can't replace the capacitors with resistors and expect to be able to pass only the AC signal without passing the DC component which upsets the biasing. In theory, the DC operating point of both transistors is such that Vcc/2 appears at each collector due to biasing with no signal applied. The AC component of the signal just wiggles the base through the capacitor to vary the output around this operating point.

author: robse
author_rank: Contributor
timestamp: June 04, 2017, 06:38:01 pm
content: Quote from: rstofer on June 04, 2017, 06:10:23 pmThis is it, T2: Channel A (red, input) 991.925uV, Channel B (green output 1) 17.982mV, Channel C (yellow, output 2) -18.103 mVThe previous image showed osc values from the very beginning of the recording, so I think they were wrong.Let me know,Robse

author: akis
author_rank: Frequent Contributor
timestamp: June 04, 2017, 07:05:42 pm
content: There will typically be some voltage between R1 and R2 you may want to turn C3 over.For the number of components have you thought of using an op-amp? eg the LM6172. It will allow you to mess with Av (voltage amplification) without screwing up the bias every time you make a small change.

author: rstofer
author_rank: Super Contributor
timestamp: June 04, 2017, 09:47:47 pm
content: When I simulate this on LTSpice, I get really bad results. Probably my fault but...Let's think about Q1:The bias resistors would like to hold the base at 0.9V (that's 5V/100k * 18k). If that's the case, the emitter is about 0.7V lower or 0.2V. Given the 100 Ohm resistor, that's a current of 0.2V / 100 Ohms or 2 mA. But 2 mA times 10k (collector resistor) is about 20 volts so something is hosed. We only have a 5V supply! Even if we include the effects of the 2d stage bias resistors, the collector equivalent resistance is about 5k and that would still be a 10V drop.It's been decades since I have agonized over this stuff so I suggest you look at the 10 step process given in the attachment.Note that everything starts at the output (how much current do we need to send to the load?) and works back to the bias resistors.http://www.radio-electronics.com/info/circuits/transistor/common-emitter-amplifier-design.phpThere's a reason everybody is using op amps!Also, I doubt that the inductor is going to help in the real world. You might try probing the 5V rail and see what you get.Base current = collector current / beta.Bias current (ignoring the base current) is base current * 10. This is to make the bias network sufficiently 'stiff' that the base stays centered around the bias point.

author: Audioguru
author_rank: Super Contributor
timestamp: June 04, 2017, 10:26:19 pm
content: You need to learn how to bias a transistor.You reduced the value of the emitter resistor without re-biasing the base then the emitter and collector currents are much higher causing severe clipping distortion because the transistor is turned on too much.In my earlier circuit I did not bias the second transistor with a voltage divider from the supply voltage and use a coupling capacitor. I simply biased the second transistor from a divided down voltage from the output of the first transistor.Here is what happens when you reduce the value of the emitter resistor without re-biasing the base:

author: Audioguru
author_rank: Super Contributor
timestamp: June 04, 2017, 10:31:42 pm
content: Why is your input signal level in micro-volts which is almost nothing??Don't you want a real input level so you can see if the output can swing equally up and down?

author: robse
author_rank: Contributor
timestamp: June 05, 2017, 04:35:44 pm
content: Quote from: Audioguru on June 04, 2017, 10:26:19 pmYes, I still don't understand how to see the dc level on multisim. Can you suggest the proper biasing (r1/r2) when r4 is say 220ohm ?Thank you,Robse

author: Audioguru
author_rank: Super Contributor
timestamp: June 06, 2017, 03:20:29 am
content: Multisim is used by old school teachers. Most of us use modern LTspice simulation software that is free.

author: Audioguru
author_rank: Super Contributor
timestamp: June 06, 2017, 04:01:49 am
content: Quote from: robse on June 05, 2017, 04:35:44 pmThe DC gain will be very high so the resistors, transistor and temperature must be very accurate. Why not make its AC gain high but its DC gain low with a capacitor and another emitter resistor in series with it?Instead of using your transistors circuit which is 50 years old why not use a modern opamp?Here is one of your transistors with a 220 ohm emitter resistor and it is biased accurately for the maximum output swing:

author: rstofer
author_rank: Super Contributor
timestamp: June 06, 2017, 02:16:54 pm
content: Quote from: Audioguru on June 06, 2017, 03:20:29 amAnd then we set up the simulation command to .op to get the operating point. In building the schematic, it is helpful to name the nodes because the default naming is seldom helpful. Q1B, Q1E, Q1C are examples.