Line Signal Muting

Line Signal Muting

While studying my Sony TC-K61 cassette deck, I was intrigued how the muting transistor action works? After sketching out a similar circuit of my own to analyse and assemble, I think I may now understand how this muting is achieved.

Here is Sony's muting circuit for the TC-K61 cassette deck for both recording and at Line Out - marked in green.

After sketching out my own generalised circuit, I had some components easy-to-hand, and so put this simple circuit together.

Here we have - an external signal generator shown to the left with its own internal resistance of around 600 ohms. Then I just added a simple, non-specific RC network to simulate general external circuitry. After that there is a load resistance of 20,000ohms. An oscilloscope was connected across the load RL.

The muting transistor here is simply a S9014 NPN which is about to be switched 'on'. A sine wave signal voltage of 1Khz across RL was set to 2 volts peak-to-peak with no interference from the S9014. Once I had switch the S9014 'on', the output completely muted. 

The results ..

VL (before muting) = 2v.
VL (during muting) = 5mv.

This gave me an effective reduction of 20Log(0.005/2) or -52dB.

So why does this work?
Firstly, note that - the transmission line is effectively dc decoupled. Secondly, the S9014 is not experiencing any external one-directional electric field (hence voltage) to motivate charges across the collector-base-emitter junctions. 

So how does forward biasing the base-emitter junction result in an efficient -52dB muting reduction?

My theory: Both base-emitter, and base-collector junctions are forced into forward biasing modes.

There is no, or very little dc current running through RL, but since both forward biased junctions are now offering a conduction path, and in particular to an ac signal - it is this ac signal voltage component that gets shunted through both junctions as both junction slope resistances 𝚫V/𝚫I (thanks to biasing) are very small.

Effectively, the source ac signal (component only) voltage drop, almost entirely occurs across the 600Ω, and 670Ω+j/𝛚C impedances since the slope resistances (collector-base, base-emitter) are so low in comparison. It's basically potential divider law, the ac signal voltage at the two junctions is 'lost' earlier across the said impedances. *

(To understand slope resistance 𝚫V/𝚫I or slope conductance 𝚫I/𝚫V in this context, you'll need to study transistor NPN junction characteristics, and in particular the Ic vs Vbe (or even Ie vs Vbe) curve which are not always illustrated in datasheets) *

Shunting of the ac signal works in both directions, only one direction shown above.

Can we expect a forward biased base-emitter junction to offer a better conduction path than that of a forward biased base-collector junction?, probably, but I haven't investigated this.

I noted that during this electrical state, there was a very small dc potential across RL (collector-emitter) of 6.9mV, and again 6.9mV when I completely removed the circuit to the left of the S9014. This suggests that while one junction (probably base-emitter) was approximately 0.7v (forward biased), the other PN junction was at around 0.7v-0.0069v; effectively the same!

One final note - efficient muting was also realised when I reversed both emitter and collector.

23/03/2020.

2nd revision, 24/3/2020.
** 3rd revision: 20/03/2024

 

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I also later simulated the circuit in https://www.falstad.com/circuit/circuitjs.html

The results were different concerning muting ability - this simulation was less efficient for some reason?, but the remaining analysis was reasonably accurate. You can see that both PN junctions were effectively forward biased and the slope resistance very low.

 
25/03/2020.