Input impedance

[Mark Hennessy]

Thinking some more, as there's no emitter decoupling capacitor, the Miller effect will be based on a gain of RL/RE, or approximately x8.2. Of course, the gain will be a little less in practice because of the non-infinite open-loop gain (circa 400, ignoring the next stage).

So 5M6 reduced by 9 (Av+1) is 650k, give or take. In parallel with the 400k, that's around 240k. Much more plausible.

Back to the fencing (thanks, Doris!). Now, as earlier, I'm procrastinating

[Murphyv310]

Easy to fall into a trap like I do when I don't study a circuit in depth.

[Hartley118]

It's certainly not immediately obvious what's going on.

I too was inclined at first to dismiss that RC feedback network (Ca1 & Ra2) as just a bit of frequency response correction. However, on a second look, it seems to me to be central to the whole basis of the circuit because Ca1 at 2200 pf is probably of similar order of capacitance to the crystal pickup, and what's more will be 'amplified' in value by the gain of the stage (Miller effect). Not sure what Ra2 is for because its resistance value only becomes significant compared with the reactance of Ca1 in the tens of kHz. Perhaps it went unstable without it! So forget its presence for a moment and what do we have? - an integrator.

So I look at it as a capacitor with gain. Roughly its value will be 2200pF x 18k/2.2k or 18nF. As Mark has calculated, its voltage gain will be 18k/2.2k, or just over 8.

Why was it designed this way? Well, a crystal pickup has a capacitive output impedance,so loading it with a capacitor reduces its output voltage, but also reduces the output load resistance needed to preserve low frequency performance. Quite a useful way to match it into a bipolar transistor input.

Martin

[Mark Hennessy]

Hi Martin,

Your last paragraph fills in the gap in my knowledge - the behaviour of crystal cartridges into differing loads. Thanks

I wouldn't mind betting that there was a fair bit of "cut and try" at the design stage

[Hartley118]

I reckon so!

M

[Crackle]

In case you wanted to discuss the circuit further, here is the whole of the circuit and a description taken from the manufacturers service sheets.
The record deck was a BSR MA65 and the cartridge is described as a stereo C1 with diamond stylus.

I also found this information

C-1
C1  C 1  c1  c-1  c 1
Stereo Ceramic Piezo

• Cartridge Type: Ceramic Piezo.
  
• Stylus Shape: Conical.
  
• Stylus Type: Conical 0.7 mil. Diamond for LP & 3.0 mil. Sapphire for 78 RPM. (Flip-Over)
  
• Frequency Response: 20 Hz. - 20 kHz.
  
• Output Voltage: 0.43 V at 1 kHz, 5 cm/sec.
  
• Dynamic Compliance: 8 x 10-6 cm/dyne.
  
• Vertical Tracking Force Range (Grams): 1.5 grams - 6.0 grams.
  
• Recommended Load Impedance: 47,000 Ohms.
  
• Number of Terminals: 4.
  
• Net Mass: 4.5 grams.
  
• Replacement Stylus: ST-8, ST-9, ST-10, ST-11, ST-12 & ST-23.
  
• OEM Part Numbers: BSR C-1, C1-ST3.
  

Mike

[Mark Hennessy]

Hi,

Mike and I have been discussing this via email - he'd like me to remove some of his earlier posts. The question was basically "what value resistor would do the same job as the transistor", and the answer to that is indeed 1.3M. In other words, a potential divider of 5.6M and 1.3M would indeed give 1.9V at its junction when fed from 10M.

We all thought that he was asking about the AC input impedance - our "bad" for not thinking harder about the numbers he posted. But the discussion that ensued was interesting and worthwhile, so we'll leave all that here.

After the tidy-up, please keep the discussion going

Mark

[Murphyv310]

Hi.
Thanks for the info. Seeing the cartridge load impedance is 47K that makes more sense, some of the older crystal types were far higher.
I also remember the old Philips Electronic Engineer kits with the AF116 & AC126, these used direct collector to base biassing with either a 680K or 330K resistor. Once I did my C&G and got more theory I understood the reason they did it that way rather than having two resistors between supply and ground to give a pre-determined bias point.
Snag for me is it's a long time since I did all the learning, involved with servicing and some health issues a lot has gone out of the window. At least it seems to be coming back....... well a little bit.

[Mark Hennessy]

Thanks for the extra info on the cartridge, Mike.

The obvious thing that leaps out is the recommended load impedance of 47k. Now, that's more like what I'd expect for a magnetic MM cartridge.

In light of earlier comments, and to help me relax after battling the fence, let's reach for the breadboard!

The first problem I had was finding a BC109 with a hfe of 200 (I was hoping to match practice with our earlier calculations). Most of mine are 400 or more, but I found one that was almost exactly 300. OK, so we'll have to run the numbers again.

Let's recalculate the input impedance: without the 5M6, we've got 300 times (re + 2k2). re is still 45 ohms as gm hasn't changed. So that gives us 675k. Quite respectable.

What about the Miller effect on the the resistor? Well, assuming the voltage gain is x8.2, then it'll look like 610k. So...

However since the input impedance is largely defined by beta x Re, and badly defined at that, the miller effect isn't going to make much difference.

Miller effect is more than halving the input impedance to 320k, so I'd say that's quite an affect. Nothing like what I erroneously suggested initially, but plenty. On paper, at least. The 2n2 cap changes all that. More on that story later...

On breadboard, fed from a 50 ohm source, and with no load beyond the 1M of the 'scope, the built circuit has a voltage gain of 7.64. As expected, it's not as high as the 8.2 suggested by the simple RL/RE approximation.

Next, I measured Zin using the added resistance method, using an old RS decade box in series with the input, and adjusting for half the output amplitude. I got 310k. Not bad!

For fun, I decided to eliminate Miller effect. To do that, I simply split the 5M6 in half, and grounded the mid-point via a capacitor. This obviously maintains the DC conditions, and the AC gain didn't change because of the relatively low open-loop gain.

Now Zin is 500k. Not as high as 675k, but to be honest, I wasn't surprised at that result, given that we're using simplified models and grotty breadboards. Also note that the 5M6 is being "Millered" slightly less in reality (because of the reduced gain) to 650k. But happily these "practical realities" converge nicely to give an overall input impedance that is within 10k (3%) of the prediction. That's pretty good, really...

However, on a second look,  it seems to me to be central to the whole basis of the circuit because Ca1 at 2200 pf is probably of similar order of capacitance to the crystal pickup, and what's more will be 'amplified' in value by the gain of the stage (Miller effect).  Not sure what Ra2 is for because its resistance value only becomes significant compared with the reactance of Ca1 in the tens of kHz.  Perhaps it went unstable without it!  So forget its presence for a moment and what do we have? - an integrator.

Yes, I initially saw 2k2 and 2n2 and thought "33kHz" and ignored it. But 5M6 and 2n2 screams much louder! We can't ignore it after all.

I added the 2k2 and 2n2 cap. I've put the single 5M6 back in, so the circuit looks exactly like the original.

Fed from 50 ohms, nothing much changes at 1kHz. The -3dB point of the circuit has reduced (not surprisingly!) from ~70kHz to 4.3kHz.

But obviously the cartridge won't have a 50 ohm source impedance. That isn't specified for this model. So all we can sensibly do is plot Zin against frequency:

Code:

16Hz  - 290k
31Hz  - 247k
62Hz  - 175k
125Hz - 102k
250Hz - 54.8k
500Hz - 27.5k
1kHz  - 13.8k
2kHz  - 6.9k
4kHz  - 4k
8kHz  - 2.88k
16kHz - 1.63k

Quite a range!

For the record (groan!), I did check Zin vs frequency without the 2n2/2k2 network, and found that it falls to half of 310k at about 20kHz. Much less dramatic.

So for a given source impedance, the frequency response will change considerably as a result of this.

What does the 2k2 resistor do? I found that shorting it added overshoot to a square wave response. That's good enough for me - I didn't explore further.

So there we are. An apparently simple circuit that hides a wealth of intrigue within its many layers. And like the proverbial onion, the more you attempt to peel them back, the more you want to cry!

So when poor Mike innocently asked the simple question about input impedance, who knew it would be this involved?

Hope this is of interest!

Mark

[Hartley118]

Thanks for the extra info on the cartridge, Mike.

The obvious thing that leaps out is the recommended load impedance of 47k. ............

Code:

16Hz  - 290k
31Hz  - 247k
62Hz  - 175k
125Hz - 102k
250Hz - 54.8k
500Hz - 27.5k
1kHz  - 13.8k
2kHz  - 6.9k
4kHz  - 4k
8kHz  - 2.88k
16kHz - 1.63k

Quite a range!
.................

So when poor Mike innocently asked the simple question about input impedance, who knew it would be this involved?

Hope this is of interest!

Mark

It's always good to test these things by experiment!

So that input impedance characteristic does, as expected,  look like a capacitor, though an imperfect one because of the limited gain of that first stage amplifier.

That specified cartridge optimum load impedance of 47k  is a surprise for a ceramic cartridge - one would expect something more like 1Mohm.  A 47k load would typically turn its usual constant amplitude frequency response curve into a constant velocity curve, approximating the characteristic of a magnetic cartridge.  That would then need an RIAA equaliser in the amp just like a magnetic cartridge.  Maybe that's the intention, but there's no sign of an RIAA EQ in this record player circuit.  I guess that it relies on that idiosyncratic capacitive input impedance to get a flat response from the cartridge.

Mark - if you set exam questions for your BBC students, this example would surely get them thinking! 

Martin