Comments for page: Eico Cortina 3070

<<first - <previous - 1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 - 10 - next> - last>>

Posted by Mike S. September 24, 2020 - 06:28 pm
Thanks so much for your article!

I just finished restoring my 3070, which I purchased for $15 on eBay. After a complete (electrolytic) recap, resistor check, new power supply diodes and cleaning, I turned it on and it worked great! I checked the bias current at idle. One side was 50mv (across R11/12/13/14), and the other was 110mv(!). I presume that both should be around 25mv. I was going to use trimpots per your article. Instead, I paralleled higher-value resistors across the 220 ohm R307 until the bias reached 25mv (more or less; tested for 15 minutes). Idle bias is now ~25mv both sides. Amp still works great, and the output transistors stay cool. I tightened every screw I could find. I think this improved grounding and improved the heat sinking of the output transistors. I ordered (10) 2N5089s from China (eBay). Cost: $3 including shipping. I also bought a Kuman Mega 328 Graphic Transistor Tester for $15 on Amazon for transistor testing, identification, and matching. Amazing device for so little money! This is my first "solid state" project. I mostly work on tube amps.

Anyway, thanks for your great advice on the E3070. I've learned a lot.

Posted by Steve L. July 19, 2016 - 02:44 pm
[For benefit of others, Google translates msdava's message from Turkish as, "helpful; one page".] Hi msdava, I guess this is a compliment---thanks.

Posted by msdava July 19, 2016 - 12:37 pm
faydalę; bir sayfa

Posted by Steve L. January 18, 2016 - 11:14 am
Hi Les,
0. Something I must be missing. I don't get VBase Q301 as 1.5V. --- In the Cortina 3070 manual voltage chart, Eico gives the reading as 1.6V so that accounts for part of the discrepancy. [Eico was never known for the dependability of their voltage readings.] The other factor is the loading of Q301 base current. The 1966 RCA Transistor Manual shows beta of the 40317 as 40-200 at Ic=10mA. From the 0.9V emitter voltage, we have it running at 3.3mA, which isn't too far away from the spec point. Taking the geometric mean of beta at about 90, then Ib=37uA. I get the Thevenin source resistance driving the base as 4.9K, so the base current drops the base voltage by 0.18V. That reduces your calculation from 1.84 to 1.66V, which is close to the 1.6V given in the manual.

1. Emitter Resistors R12, R14 1Ω; make sense but I am puzzled by the two Bridging Diodes CR2, CR4. --- R12/14 are there to make quiescent current (Iq) in Q2/4 more stable and predictable. If R12 and R14 were each shorted, the voltage drop across R307, CR301/302 would be working against only the sum of Vbe's for Q303/2/3/302. As we know, Ic versus Vbe rises very steeply, which would lead to unpredictable bias current. With R12/14 in place, CR301/302 are supposed to match the Vbe's, leaving the drop across R307 to fall across R12/14. (There should have been three diodes, not two!) CR2/4 don't come into play with just Iq present. The purpose of those is limit the loss across R12/14 when high current is being delivered to the load. At 19Wrms into 4ohms the peak output current is about 3A, so without the diodes, losses would be high. With the diodes total losses are less than 1W or so.

2. I assume no overcurrent protection circuit against shorting the outputs is needed because of C6 - right? --- C6 could protect against DC or VLF currents but not against AC or transient overcurrent. The sad fact (discussed in the article) is that the poor Cortina isn't short-circuit protected (SCP) and will die in a millisecond or so if the output is shorted while output is present (and in other scenarios too). After all, the five-transistor power amp design was all about simplicity and SCP would have increased the number of transistors in the power amp by 40%.

3. R16 1KΩ at the outputs mean the output never drives into an open circuit. Does that matter? No resistor means no ac current. --- R16 is there to pull the DC output to near zero when no load is attached. Without it, when a speaker is attached, you might experience a loud POP. Moreover, if you shorted the output with that DC present, even without an AC signal, one or more output transistors would surely die. Now, cost-cutters at Eico may well have pointed-out that R312 will serve much the same purpose. The problem with that is the time constant is 54-seconds, which is plenty for bad things to happen. (Not to mention, it would still leave ~1.6V there.)

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? --- Yes, that would be more symmetrical as you point out and one would expect better even harmonic distortion. It was a cost-saving measure at the time but even so, Eico was able to achieve 0.25% distortion across the band at half output voltage. (See "Summary of the Test Results".)

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!


<<first - <previous - 1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 - 10 - next> - last>>