Phase Splitter - Learning The Art Of Electronics - Page 1
Forum section: electronics / beginners

author: SewingYard
author_rank: Contributor
timestamp: April 26, 2016, 10:56:07 pm
content: Hi,Another newbie question I'd love some advice on if someone can spare a moment.I have been working my way through this book (as per topic) and I'm not just making the circuits, I want to genuinely understand how they work, which is how the author intends it.So, I have arrived at the phase splitter, I designed the circuit as per the book and it works as intended which I expected.I don't however have and understanding of how they arrive at the values for some components, the first one I have struggled with for example, the R Collector is 4.7k and can't work out why.I have posted the pages from the book (I'm not sure if I'm allowed to under copyright but I'm sure I'll be moderated if I've done wrong)

author: SewingYard
author_rank: Contributor
timestamp: April 26, 2016, 10:58:39 pm
content: Another page

author: danadak
author_rank: Super Contributor
timestamp: April 26, 2016, 11:07:48 pm
content: The idea in a phase splitter, in this case a 180 degree splitter,one would want not only the correct phase but essentially sameamplitude, pk-pk voltage swing. So for a high beta, gm transistorIc =~ Ie, so by using Re = Rc you get the same amplitude out ofthe outputs.The design should also have a nominal bias Vrc = Vre = Vceto maximize dynamic range.Regards, Dana.

author: Marco
author_rank: Super Contributor
timestamp: April 26, 2016, 11:10:11 pm
content: First of all Rc it sets the DC current through the transistor together with the base bias voltage, which determines power consumption (can also determine the useful bandwidth, but that's not really relevant here). Most importantly though it sets the output impedance of your circuit, which you want to be significantly smaller than the input impedance of the next stage so you can design them separately (we don't care about getting max power out of the circuit, so we're not looking for 1:1 impedance matching).PS. actually strictly speaking it only sets the output impedance for outc, if the input impedances attached to outc and oute are "significantly" mismatched you have a problem.

author: SewingYard
author_rank: Contributor
timestamp: April 26, 2016, 11:18:42 pm
content: Hi Again, thanks for the replies, I think if i type out the sentence i don't comprehend it might help matters."Page 182 : Therefore, we should put Vc.quiescent at the midpoint of the range available to it. That range is 10-20V, so the quiescent voltage should be 15V. To get this result, we want to drop 5V. So, RA = 5k. The nearest 10% value is 4.7k"Hope this helps

author: Marco
author_rank: Super Contributor
timestamp: April 26, 2016, 11:31:02 pm
content: Quote from: SewingYard on April 26, 2016, 11:18:42 pmWith due respect, the reasoning in the book is bullshit. The base bias voltage determines the quiescent voltages at the outputs pretty much 100%. You can change Rc/Re to almost anything and it will not change. So to say you chose the resistance to set it is putting the cart before the horse.It's not a logical approach.

author: danadak
author_rank: Super Contributor
timestamp: April 27, 2016, 12:00:22 am
content: Here is a good summary of various biasing techniques.http://www.electronics-tutorials.ws/amplifier/transistor-biasing.htmlRegards, Dana.

author: rstofer
author_rank: Super Contributor
timestamp: April 27, 2016, 01:04:38 am
content: Quote from: SewingYard on April 26, 2016, 11:18:42 pmIf we short out the transistor, Vc and Ve are approx equal at 10V. So, Vc can vary between 10V when the transistor is shorted and 20V when the transistor is open. Ve, OTOH, can vary between 10V when the transistor is shorted and 0V when it is open. This gives both signals approx equal range of 10V. That's why the resistors match.To get mid-range output (center), we want Ve to be 5V and Vc to be 15V. That implies that 10V is dropped across the transistor. That should make sense because we still have to account for 10V. 5V dropped in Vc, 5V dropped in Ve, 10V dropped across the transistor. Just what we want. In this scenario, we want approx 1 mA flowing (give or take selected values). Given Beta = 100, say, we need 10 uA of Ib. Vb needs to be somwhere around 0.6V higher than Ve and Ve is 5V. So, we need to work out the resistors to bias the base to get the proper voltage. We need to get the real Beta from the datasheet, just assuming 100 won't get the same answers as the book.My circuit theory is a bit rusty, I'll let somebody else play with the bias resistors.

author: mikerj
author_rank: Super Contributor
timestamp: April 27, 2016, 09:18:45 am
content: Quote from: Marco on April 26, 2016, 11:31:02 pmIt's a completely logical approach if you actually read the entire section. The collector resistor is chosen based on the desired quiescent current of ~1mA combined with the ideal bias voltage to give maximum signal swing.How would you have designed it?

author: LvW
author_rank: Frequent Contributor
timestamp: April 27, 2016, 10:41:54 am
content: Quote from: Marco on April 26, 2016, 11:10:11 pmNo - Rc does not "set the DC current". This current is determined by the base-emitter voltage only.Remember: The BJT is a voltage-controlled current source and Rc acts as a load only.Quote from: Marco on April 26, 2016, 11:31:02 pmThe quiescent voltages are determined by the collector current (set by Vbe) and the resistors, of course. Changing the resistor values changes the DC voltage drop accordingly (Ohms law).The proposed dimensioning is fully OK. The collector voltage (against ground) is +15V and the emitter voltage is Ve=5V.Hence, both output nodes can swing to both sides by app. 5 V amplitude. This is the only alternative for symmetrical behavior at Ic=1mA

author: orolo
author_rank: Frequent Contributor
timestamp: April 27, 2016, 02:08:23 pm
content: For the circuit to behave as a phase splitter, you need Rc = Re, because then the gain as CE amplifier will be -Rc/Re = -1.The design starts with this requirement: Ic = 1mA. Also, Vcc = 20V. This for itself doesn't determine the amplifier: you can design an infinite number of phase splitters with Ic = 1mA at this voltage. You still have a design choice remaining: where does the collector voltage set when there is not signal (ie, the quiescent voltage). You can put this voltage almost anywhere from 10V to 20V, and that position, together with Ic=1mA, will determine Re = Rc.The books does a very smart choice: since a phase splitter drives two equal (in amplitude) waves, one at the emitter and other at the collector, the 20V rail is divided in two zones: from 0 to 10V, where the emitter wave will swing, and from 10V to 20V, where the collector wave will swing. So when quiescent, the emitter voltage should be at 5V (middle of its zone), and the collector should be at 15V (middle of its zone).Now it's time to use Ohm's Law: if Ic = 1mA, and the quiescent collector is at 15V, that means that Rc drops 5V, so: Rc * Ic = 5 -> Rc * 1e-3 = 5 -> Rc = 5k, nearest typical value Rc = 4.7K. The very same applies to Re which, anyway, must be equal to Rc.Now that we have Re, and we know the quiescent emitter is at 5V, we deduce that the quiescent base is at 5+0.6 = 5.6V (assuming the transistor is active, which is a safe assumption, though it is always best to check for it after finishing the design). From Vb = 5.6, and Ib = Ic /beta, one uses that the base divider must be 'stiff enough' to drive the transistor, meaning that Rb1 || Rb2 = Re*beta/10. This gives us an equation. Another equation comes from the fact that Vcc * Rb1 / (Rb1 + Rb2) = Vbe (the base divider gives the base voltage computed before). We solve that easy system of equations and we get Rb1 and Rb2. Finally check if the solution you got is consistent with an active transistor (sometimes there are surprises, especially at very high beta or temperature). And you are done.

author: Marco
author_rank: Super Contributor
timestamp: April 27, 2016, 05:19:08 pm
content: Quote from: LvW on April 27, 2016, 10:41:54 amRc==ReVe~=Vb-0.6Vc~=Vcc-(Vb-0.6)ic~=ie~=(Vb-0.6)/RcVb is chosen to maximize available signal swing which it almost solely determines, pretty much regardless of Rc/Re. Thus there is no freedom in choosing Vb, hence Rc as the only free variable determines the DC current (and output impedance and potentially bandwidth).QuoteThe problem is not the dimensioning, the problem is the order in which the book pretends they are chosen. You begin with determining the base voltage because that has a single optimal value. After that you chose Rc/Re.

author: LvW
author_rank: Frequent Contributor
timestamp: April 27, 2016, 05:43:27 pm
content: Marco - you wrote:First of all Rc it sets the DC currentAnd this is wrong - at least, it sounds very confusing. I hope, you did not mean "Rc controls the DC current" (what is the meaning of "to set"?).The classical sequence of steps is first to select a proper collector current, choose resistors Rc and Re in accordance with the voltage supply, and lastly design the corresponding bias circuitry at the base. That``s all I wated to clarify. ("Rc sets the DC current" is very misleading).More than that, can you please explain the following sentence from your reply#5"You can change Rc/Re to almost anything and it will not change. So to say you chose the resistance to set it is putting the cart before the horse."

author: BobsURuncle
author_rank: Regular Contributor
timestamp: April 27, 2016, 07:34:07 pm
content: Quote from: SewingYard on April 26, 2016, 10:56:07 pmIt's not nearly as complicated as people are making it.Rc sets the quiescent voltage for OUTc, i.e. 15V. (that is the voltage you want when the AC signal = 0V) It was given to you that you will design the circuit for a 1mA collector quiescent current. Given this, you need to drop the 20V positive rail by 5V down to 15V at Outc. 5V/1ma = 5K. The nearest standard resistor is 4.7K. Do you understand why 15V Outc and 5V OutE quiescent voltages were chosen?

author: Audioguru
author_rank: Super Contributor
timestamp: April 28, 2016, 01:31:42 am
content: I was trying to think of a circuit that uses an old fashioned phase-splitter so I looked in Google and found many circuits that used vacuum tubes a long time ago. Then I remembered this circuit from Philips that was used in a transistor portable radio 61 years ago:

author: Marco
author_rank: Super Contributor
timestamp: April 28, 2016, 02:54:50 am
content: Quote from: BobsURuncle on April 27, 2016, 07:34:07 pmWith the assumption that you change the rest of the circuit along with it to maintain a given quiescent current ... which to me just feels unnatural.If you change Rc (and Re along with it) in the physical circuit without touching the rest of it, the quiescent voltage does not change ... the quiescent current on the other hand does. That's why I say Rc sets the current, no assumption needed apart from Rc==Re.

author: BobsURuncle
author_rank: Regular Contributor
timestamp: April 28, 2016, 05:14:02 am
content: Quote from: Marco on April 28, 2016, 02:54:50 amI see your point but his question was how Rc was chosen given the stated problem, not how one might want to approach the design of such a circuit.Also Re and the bias voltage on the base set the current, not Rc. If you change Rc to 10K then Ie and Ic will still be 1ma as long as Re is still 5K - but the quiescent voltage at Outc would be poorly centered.

author: LvW
author_rank: Frequent Contributor
timestamp: April 28, 2016, 08:33:31 am
content: Quote from: BobsURuncle on April 28, 2016, 05:14:02 amYes - full agreement.It is, of course, the emitter resistance which "sets" the current (together with the base bias). Choosing Rc=Re is another design step which (practically) does not influence Ic.I consider this clarification as important because I know (and have experienced) that some beginners/newbies think that Rc would "limit" the current Ic (they don`t realize that the BJT works as a current source).

author: SewingYard
author_rank: Contributor
timestamp: April 28, 2016, 07:40:52 pm
content: Quote from: BobsURuncle on April 27, 2016, 07:34:07 pmYou just cracked it for me, I knew it was simple but my brain disagreed, I really appreciate you putting it across in a way my grey matter could comprehend itI couldn't understand where the 5k came from even with the book but now the penny has dropped and the light bulb hath illuminated!