Input impedance

[Crackle]

Following on from a previous thread, I thought I would set myself a little exercise to work out the input impedance of this pre-amplifier stage of a ITT-KB KP038 record player.

   

Calling the input impedance "I" my reasoning was; (and I have to admit I struggled over this, as I discovered I had forgot all my algebra)

5.6   x 10
--------          = 8.1
5.6 + I

so

56
----      =8.1
5.6 +I

56
-------   = 5.6 +I
8.1

so then;

6.913 - 5.6 = I

1.313 = I or in other words input impedance is 1.3M

This seems to be about right, am I correct

thanks

Mike

[ppppenguin]

Roughly Hfe x emitter resistor in parallel with 5M6. Since Hfe is highly variable let's say 200 for a BC109, so 4M or so in parallel with 5M6. Say 2.2M with a pretty wide variation due to Hfe.

Haven't taken into account the CR network which is presumably for getting rid of RF or possibly for tone correction. At higher frequencies the impedance will be lower.

[Murphyv310]

Hi.
I'm with Jeffrey with the calcs. One clue is the 1nf coupling caps to the base of the transistor even with out any form of calculation, these are low values for audio coupling so would need to "see" a high impedance load to have any low frequency response.

[Mark Hennessy]

Mike,

You've picked an unfortunate circuit. I'm afraid that this is a non-trivial exercise.

The amplifier is a shunt feedback amplifier.

In theory, if the transistor offered infinity gain and Ra4 was bypassed with a large capacitor, the input impedance would be zero. But the gain isn't infinity, and that resistor isn't bypassed.

What about Ra2 and Ca1? They should be working outside the audio band, so we'll ignore them.

Always break a problem down.

If we ignore the effect of feedback first of all, the "raw" input impedance of the transistor is (re+RE) times hfe. "re" is the intrinsic emitter resistance (1/gm), and RE is Ra4.

("Little re" will be small, so we could ignore it. But we might as well work it out, as we'll need gm later.)

At room temperature, gm is Ic times 40. About 22. So re is 45 ohms. Let's assume the BC109 has a hfe of 200, so 200 times (45 + 2.2k) is 449k. Call it 500k.

So before the effect of negative feedback, Zin is 500k. Next would be to consider the effect of NFB on the value of the feedback resistor.

This is Miller effect. You put a small positive signal voltage on one end of the 5M6 and expect a certain current to flow. But because the other end of the 5M6 is not tied to ground - it has moved negative, so more current than you expect will flow, and you'll "see" a lower resistance than you think. It's like "anti-bootstrapping". Roughly, the resistance is reduced by the voltage gain of the amplifier.

The amplifier should have a voltage gain of approx 400 (gm times Rl). So that makes the 5M6 resistor look like 14k. This is in parallel with the 500k of the transistor.

However, that will be wrong because we don't know what comes next. When I worked out the gain, I used 18k for Rl, but Rl is in parallel with the next stage. Whatever that is, it'll reduce the open-loop gain of the input stage, and that'll mean that the 5M6 is reduced less by Miller effect. Post the whole diagram and I'll try to guess at it.

All of the above is approximate in a "back of the envelope" sense. However, the numbers aren't important - but trying to understand what's happening before doing the maths is. And there are many ways to approach this problem, but what I've said here is the most intuitive, least mathematical approach.

Of course, you can measure it really easily. Feed in a sine wave via a variable resistance in series with the input. Set resistance to zero and, monitoring the output, increase the resistance until the signal has fallen by half. Measure the resistance with a DMM. Viola

I'm afraid I couldn't follow your working - I got lost at the first line because I couldn't see where 10 came from. Sorry

[Craig]

Roughly Hfe x emitter resistor in parallel with 5M6. Since Hfe is highly variable let's say 200 for a BC109, so 4M or so in parallel with 5M6. Say 2.2M with a pretty wide variation due to Hfe.

Haven't taken into account the CR network which is presumably for getting rid of RF or possibly for tone correction. At higher frequencies the impedance will be lower.

beta x emitter resistor is 200 x 2.2k = 440k ~ 400k not 4M. So input R ~ 370k absolutely dominated by the poorly defined value of beta.

Craig

[Murphyv310]

The resultant input impedance though has to be high as the coupling capacitors and xtal cartridge would result In poor audio quality.

[Mark Hennessy]

Don't forget Miller effect on the 5M6!

I've found the schematic on RM.org. I won't upload it, but Mike could as he put it on RM.org originally

The output of this stage feeds the tone control network and volume control. So the impedance they present - in parallel with RL (Ra3) - is highly dependant on frequency and the settings of the user controls. That makes it near-impossible to determine the open-loop gain of the first stage, and hence near-impossible to predict the Miller effect on Ra1.

The input impedance of the following stage is rather low as well.

My 14k does seem unreasonably low, and as I said, it won't be that low in practice because of the above. But it will vary according to the position of the user controls. Bad design, but presumably "good enough", especially if the gain structure is such that the volume control can never get near to "max" before gross clipping sets in...

Trevor makes a good point about the 1n input capacitor. But don't forget the source impedance of the cartridge, which is in series with the capacitor. If that's 1M, then the LF rolloff will be in the order of 100Hz. The DIN socket is worrying, because signals from that also go through the 1n caps, but the source impedance from that might well be 100s of k.

To get a 100Hz rolloff with 1n in circuit, the total impedance "seen" by that 1n cap would need to be 1.6M. As described, the "best-case" input impedance be ~500k, before it's been lowered by Miller effect.

[ppppenguin]

I managed to get the order of magnitude worong in a calculation and negelct the miller effect. 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.

[Mark Hennessy]

It's easy to look at this circuit and think how simple it appears - more simple than the one I presented to NM in the other thread. However, nothing could be further from the truth. This version is popular because it uses one less resistor, but it's a lot harder to analyse. IIRC, you need to revert to simultaneous equations to determine the collector voltage. The collector voltage depends on hfe, so Vc will vary in production. It's OK for small-signal stuff, but where you definitely want Vc in the middle for maximum output swing, you'd pick something else.

Thank goodness for op-amps

[Craig]

In 1961, Dinsdale showed a single germanium transistor RIAA input stage for use with a moving magnet and crystal cartridges, tape EQ and flat http://69.195.158.203/Archive-Wireless-W...961-12.pdf . Of course transistors were new fangled devices back then and were very expensive, so there was a powerful incentive to minimise the number of them in the circuit.

The old two-transistor RIAA stage stage was also originally devised by Dinsdale in 1965 http://69.195.158.203/Archive-Wireless-W...965-01.pdf had both a DC feedback path to stabilise the operating point, and an AC path to set the gain at 1kHz and apply RIAA correction, and a bootstrapped first transistor. Used OC44's. The performance of that can be made quite acceptable using silicon transistors, adding an emitter follower buffer to drive the RIAA network, changing the bias conditions to ensure symmetric clipping, and upping the supply voltage (typically <0.01% distortion).

But like Mark said - thank goodness for op-amps! Although they used to be expensive; when I built the Texan back in the early 70's, 741 and 748 op-amps were 60p each (about £6 in today's money). Forking out the equivalent of £36 for six opamps was a big deal for a 16 year old.