24-02-2017, 04:39 PM
(24-02-2017, 02:40 PM)Mark Hennessy Wrote: Just another point. It is hopefully "stating the obvious", as it applies equally to valve circuits, but just in case...
DC conditions first. Then add signal.
In other words, you must always bias your valve or transistor before it can amplifier the way you want. The AC signal basically "wobbles" the circuit either side of the quiescent DC condition.
The signal at the base-emitter junction of a transistor consists of a DC offset and and AC signal. Learning to mentally separate these two components so you can consider each separately is probably the biggest favour you can do for yourself.
That DC offset is likely to be close to 0.2V for a germanium transistor (0.6V for silicon). Give or take...
If there is an emitter resistance, it'll have a DC offset on it. More on that later if needed.
"The AC signal basically "wobbles" the circuit either side of the quiescent DC condition."
I sort of had some issues lately with signal voltage and DC bias. Here is where I got curious: In a valve radio, you will know the signal that hits the antenna is voltage-wise very weak. Millivolts. So, the first stage is to amplify the signal. In the very old TRF's of course the first valve would be the HF amplification. The modern sets just use the Mixer Triode Hexode. At any rate, to get to the point we wind up with an IF that passes through the IF transformer as, say, 455 KHz. Normally the IF transformer is shown as primary winding between HT and the anode of the Mixer (a hefty voltage). The secondary will connect between the detector grid and either AVG (or is it AVC) or chassis. Here is the point: The detector grid is DC biased at less than zero volts. Thus you only need very small amplitudes of AC signal to drive the grid into more positive values. In AARL, the example used is a grid at - 5 and a signal input of 2 volts+. We're told the grid will swing to - 3 on positive waves or - 7 on the negative amplitudes. For every positive swing, current is going to increase. What had me foxed is a very simple observation. If the actual IF amplitudes are at high voltage, how do we wind up with a signal on the grid at just, say, 2 volts AC? My only guess is that the sheer speed of the alternating cycle (1000's per second) are too fast to literally push the grid over the zero bias. So maybe it's not a good idea to try and apply graphs too literally? Of course, then we have grid leak which also balances out the bias. Also, possibly the capacitors may attenuate the signal. Even so, no textbook seems to have addressed this and seems to assume it's too obvious to highlight. However, it bugged me. So, your remarks about not mixing up DC and AC remind me of that. What I do know is there's a voltage drop across the diode load resistance that responds to the grid swings as current wobbles about. So, we eventually get the audio decoded. Now with transistors, the principle I imagine is the same but the holes and so forth, plus the smaller voltages required are no doubt a science in itself.
Agreed, though, none of this is needed in repair work.
