Electronic – the maximum negative output voltage of an emitter follower with a split +10v/-10v supply

voltage

I'm perplexed by a passage in Horowitz and Hill's "Art of Electronics" (1st ed., p. 54). They present the reader with a simple emitter follower circuit. The input voltage is a 10v amplitude sine-wave. The transistor itself is fed +10v directly to the collector. There is a 1k resistor on the emitter, which is supplied with -10v. According to the authors, the sine wave input will be preserved at the output. The positive peak will look just like the input at almost full amplitude but the negative peak will clip at -4.4v. I can't for the life of me figure out where they got -4.4v. If the power supply is split +10/-10v, wouldn't the output clip at -10v, not -4.4v like the text says?

I tested the circuit on EveryCircuit and the output clipped as expected at -10v. Here is the circuit I built to try to figure this out for any EveryCircuit users.
http://everycircuit.com/circuit/6493906064375808
I'm a novice here, and having an impossible time figuring out if I've simply misunderstood something or it's a misprint.

Best Answer

The emitter and load resistors form a voltage divider for the \$V_{EE}\$ supply. When the transistor goes into cutoff each resistor drops \$5V\$ across them.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

Related Solutions

Negative voltage

Someone may have better words to explain this than me, but the big thing you have to remember is voltage is a potential difference. In most cases the "difference" part is a difference between some potential and ground potential. When someone says -5v, they're saying that you are below ground.

You also need to keep in mind that voltage is relative. So like I mentioned before, most people reference to "ground"; but what is ground? You can say ground is earth ground, but what about the case when you have a battery powered device that has no contact to ground. In this situation we have to treat some arbitrary point as "ground". Usually the negative terminal on the battery is what we consider from this reference.

Now consider the case that you have 2 batteries in series. If both were 5 volts, then you would say you would have 10 volts total.

But the assumption that you get 0/+10 is based off of "ground" as being the negative terminal on the battery that isn't touching the other battery and then 10V as being the location of the positive terminal that isn't touching the other battery. In this situation we can make the decision that we want to make the connection between the 2 batteries be our "ground" reference. This would then result in +5v on one end and -5v on the other end.

Here is what I was trying to explain:

+10v   +++   +5v
       | |
       | | < Battery
       | |
+5v    ---   0v
       +++
       | |
       | | < Another Battery
       | |
0v     ---   -5v

Electronic – Adding LFO Modulation to VCA in synthesizer

The crux of your problem seems to be that you want to multiply the LFO signal with the A/R signal, and to do this multiplication requires circuitry that is effectively a second VCA.

One way of making a VCA is by making use of pulse-width modulation technique as Andy aka suggests. The classical 'analog synth' method of doing so is by altering the bias current on a transistor differential pair -- usually within a transconductance op-amp. See pages 8-10 or thereabouts of the LM13700 OTA datasheet.

The above talk of multiplication and signal VCAs assumes that the input to your existing VCA is linear (which I believe is the usual case). If the input is in fact logarithmic, you simply need to sum the A/R signal and LFO signal (e.g. using an op-amp mixer) before sending it to the VCA.

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