Comments for page: Eico Cortina 3070

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Posted by Les L. January 18, 2016 - 08:25 am
Hi Steve, I really appreciate your answers which I am re-reading several times and thinking about. Here are a few more, if you don't mind . . .

0. Something I must be missing. I don't get VBase Q301 as 1.5V. I get 1.84V by 20Vx6K8||28K / (27+27+6.8||(27+1))K. I then get Vo = 65.7mV, not 54mV!! I see the DC circuit here as 20V on R302 + R303 + parallel R304 with R312+R16.

1. Emitter Resistors R12, R14 1Ω; make sense but I am puzzled by the two Bridging Diodes CR2, CR4. At 1Ω;, they'll switch on ~0.6 V = 0.6 A through R12, R14. I'm not sure about I Quiescent through Q2 Q3 - Sedra & Smith have it based on the diode equation plus relative junction areas diodes and emitters. Now adjustable.
If Quiescent exceeds 0.6 A, they are on and the voltage across R12 R13 is Vf ~ 0.6V →;Voltage across R309 ~ 1.2 V. Analogue Electronics - What actually drives what!!

2. I assume no overcurrent protection circuit against shorting the outputs is needed because of C6 - right? Whereas a +ve to -ve Power Supply would not need C6 and thus have at least 2 - 4 active overcurrent elements to protect Q2, Q4? BUT - why not remove the Protection Circuit and put in an output capacitor? Very simple.

3. R16 1KΩ; outputs mean the output never drives into an open circuit. Does that matter? No resistor means no ac current.

4. Since this circuit is based on late 1950's technology (PNP's none too good then), would a Full Complementary Output Stage be better? IE: Q303 / Q2 become a Sziklai Pair with say 220Ω; on Collector of Q303 off +Vcc, also Q302 / Q4 as is and R311 4K7Ω; is removed / shorted? This then becomes symmetrical.

Posted by Steve L. January 18, 2016 - 08:12 am
Hi Les,
EIC split this into 2 x 27K - R302 & R303 with C302, presumably to remove this as an ac signal path back into Q301 base, right? --- Yes.

How it is determined that it is 1/2 and 1/2 rather than some other ratio?
--- Given that the other ends of R302 and R303 are low resistance compared to these resistors, choosing them equal maximizes the source-R that is driving C302. For a given cutoff frequency target, that will minimize the value of C302, saving cost. Conversely, for a given value of C302, it will minimize the cutoff frequency, improving filtering.

I gather Q301 DC Bias = ratio of R304:R302+R303, off the centre rail that settles to +Vcc/2. Why not take a resistor divide from +Vcc to ground for Q301 bias? Not so good?
--- R302/303 provides DC negative feedback which stabilizes the DC voltage at Eico's node-7 (circled). Providing fixed bias from Vcc to Q301 would make it difficult to set the DC voltage at node-7 accurately.

To effect bootstrapping across R306, R305 is required. How is the 1K decided on please?
--- Another excellent question! The total of R305+R306 sets the operating current of Q301 because the voltage drop across them will be roughly Vcc/2. So we have a fixed value for the total. As R305 is reduced, the load resistance that C303 is driving drops, requiring that C303 be larger for a given low frequency corner. Below that corner frequency, bootstrapping isn't very effective, lowering loop gain and increasing distortion, etc. As R305 is increased, R306 must be lowered to maintain a constant sum. But R306 is the resistance that we're trying to bootstrap to a high value. The bootstrap factor multiplies the values of R306, so decreasing that resistance undercuts the advantage. Hence, the designer may want to keep R305 fairly small compared to R306.

This is partly a tradeoff between better performance at low versus high frequencies. Lowering R305 hurts the low end while lowering R306 affects the full band. If it's economical to make C303 fairly large (as it is), R305 can be kept small relative to R306 and therefore the full band performance won't be reduced.

I assume the 50V 47 mF on the +Vcc rail removes ac from the DC supply, right?
--- Well yes, but another way of looking at that is that it lowers the AC impedance on the Vcc rail, close to the output stage. Transistor amps like this tend to go unstable if the power supply impedance isn't kept low at high frequency. Hence, when employing opamps (which have a lot in common with transistor power amp designs), it's standard practice to bypass the supply rails to ground. Interesting discussion!

Posted by Les L. January 18, 2016 - 08:02 am
Hi Steve, Thanks for the answer and I have another 3 questions please if you don't mind.

1. Dr. H. C. Lin's power amp. circuit (1959) has R3 as 47K, no capacitor C302. EIC split this into 2 x 27K - R302 & R303 with C302, presumably to remove this as an ac signal path back into Q301 base, right?

How it is determined that it is 1/2 and 1/2 rather than some other ratio?

2. I gather Q301 DC Bias = ratio of R304:R302+R303, off the centre rail that settles to +Vcc/2. Why not take a resistor divide from +Vcc to ground for Q301 bias? Not so good?

3. To effect bootstrapping across R306, R305 is required. How is the 1K decided on please?

I assume the 50V 47 mF on the +Vcc rail removes ac from the DC supply, right?

Posted by Steve L. January 18, 2016 - 07:54 am
Hi Les, Thank you for your kind message and insightful question. You are absolutely correct, that there will be some DC feedforward current through R312 to the output. However, it will be very small in practice. Without a speaker connected, about 54mV will appear across R16, due to the divide ratio between it and R312. If a connected "8ohm" speaker has, say, 6ohms DC resistance, the DC voltage at the output will drop to around 0.3mV. Hence, the DC is insignificant. It's an interesting observation though.

Congratulations on your self-education program. From your question, it seems to be working! :) Best wishes with your studies in EM!

Posted by Les L. January 18, 2016 - 07:51 am
Hi Steve,

May I say what an excellent description of the E.I.C. equipment circuits you have provided.

I myself, out of writing software and industrial consulting, decided to buy > 50 textbooks /workbooks on electronics, electrical engineering and calculus (including GE's GlowLamp Manual 1963 - I like negative resistance devices) and have been teaching myself electronics / elec.eng. for the past few years.

As a test of skill and before I move onto Electromagnetics, I have been studying your description of the EICO Cortina Amplifier circuits, particularly the power amp. stage, along with Dr. H. C. Lin's RCA 10 Watt power amp.

If I may ask . . . Re "The DC output voltage (ideally half the supply voltage) is set by feedback through R302 and R303". Isn't a DC voltage put across R16 load via the feedback resistor R312? = feedforward?
As a result of the DC Bias on Q301, set up by the ratio of R304 to R302+R303 x centrerail DC voltage?

Regards ... Les​, Melbourne, Australia​;

Posted by josaih nimako August 17, 2015 - 04:34 pm
i like the diagram


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