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I am confused by the LM386. In the data sheet, multiple examples connect the inverting input directly to ground.

simple LM386 circuit

To me, that feels weird. How can the IC amplify if the input goes below GND?

The data sheet goes on: when left unconnected (or AC coupled) either input will float to 12.5 mV. That is certainly not enough to allow for the full 2Vrms output (500 mW into 8 Ohms) at a gain of 20.

Does the LM386 actually allow the input to go past the lower rail? How does it do that? Are the input transistor pairs all in saturation, i.e. \$V_C>V_B\$ ?

LM386 equivalent circuit

polwel
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  • Vbe is the answer : with the input pin at 0, the input stage emitter is at +Vbe or about +0.6V. That's just enough +ve bias for all the rest to work. By -0.4V, the input transistor starts to saturate producing unacceptable distortion; hence the input specification. –  Apr 04 '21 at 13:06

3 Answers3

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Section 6.3 in the data sheet says that an acceptable input voltage is +/- 0.4 V. That is just below a single diode drop, so it would confirm your theory.

enter image description here

Yes, that confirms what I thought about the PNP BJT input stage. It will sail close to BJT saturation with an instantaneous value at -0.4 volts but, it'll still work: -

enter image description here enter image description here

As can be seen, the input signal falls below 0 volts quite significantly (-200 mV) without any problem or distortion in the output waveform (connected to an 8 Ω load). Signal is 1 kHz and transistor models are BC557 (PNP) and BC547 (NPN). Current source is 5 mA.

Simple answers are the best

Simple answers are usually the best but, not at the expense of truth.
Andy aka
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Note: This answer and the one by Andy aka were posted independently and within minutes of each other. The content is virtually identical. I appreciate Andy's effort, so I accepted his answer.

With Andy aka confirming my suspicion that the input is indeed allowed to go negative, I rigged up the equivalent circuit in LTspice.

LTspice scmeatic

trace

For a supply voltage of 18V, the circuit deals with an 0.3Vp input with remarkably little distortion (50dB). Increasing the input further, the output starts to be clipped by the supply rails.

I will need to do some more digging to actually fully grok this circuit.

Q5 is an emitter follower, that is in saturation whenever the input goes negative. It still feeds a current waveform into Q6 that accurately tracks the input. Q6 has its collector pinned to 0.7V by Q10, so Q6 is in saturation almost all the time.

polwel
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4

When differential inputs are PNP the device may sense just below 0V=Vee.

When Ib=250nA (LM386) use Vbe=500mV ( for any BJT at 25’C) then when Vb=Vin=-0.2V (max in spec) Ve = 300mV which is reduced hFE but NOT “saturated” since Vc=0.

“Saturation” is when hFE is about < 20% of hFE or derated to 10% at rated Ic(sat) , Vce(sat) { or Ic/Ib=10 !} here Vce(sat) would be Vce(sat)<<100 mV (more like <50mV) so Vce=3xx mV at this low Ic is not saturated.

So the simple explanation is iff the output is in the linear range to pull up the input PNP emitter with Vce=300 mV when Vbe=500mV and Vb(in)=-200 mV, you can use the base input below Vee!

Tony Stewart EE75
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