Preamp Current Draw

[surfsup]

When I send in a signal (sine) into the preamp and there are three gain stages...

At each stage, does the positive upswing affect stages 1&3 at the same time the negative affects stage 2? In other words, current draw will be higher for stages 1&3 while stage 2 draws less (negative swing), and the opposite on the negative signal swing for stages 1&3 as stage 2 goes positive and draws more current?

Also, if my stage is biased at -1.5 volts and I send in a signal that is 4Vpp, shouldn't I hit 0v bias and see the waveform cut off abruptly? What happens to the signal, is it a gradual compression at this point or a cutoff?

[surfsup]

Okay to clarify here is a scope shot of a 3rd gain stage (100k plate, 10k cathode)

[img:480:234]http://chicagocadcam.com/ChrisHahn/xen1 ... -t0b10.jpg[/img]

Bias is -2.7 yet a 11Vpp signal only clips on the negative side. This is due I believe due to the supply being 282 and the plate being at 252 so the 28V swing gets cut off simply because there isn't any more voltage supply, but on the other side, I always thought 2.7v bias would make the tube cut off at minimum voltage because the grid goes positive...? (due to a 10volt p-p input)

[surfsup]

Okay I understand question 2 now...load line current clipping is at 1.8mA. So with a 1.8mA current draw, through a 10k resistor, the bias voltage will shift to 18V.

I would still be interested in comments on my first question, reposting it:

When I send in a signal (sine) into the preamp and there are three gain stages, at each stage, does the positive upswing affect stages 1&3 at the same time the negative affects stage 2? In other words, current draw will be higher for stages 1&3 while stage 2 draws less (negative swing), and the opposite on the negative signal swing for stages 1&3 as stage 2 goes positive and draws more current?

[katopan]

Basically yes, but you might have the polarity around the wrong way. If you're talking about positive upswings and negative downswings on the plate signal, then less current through a stage gives you a positive upswing and more current a negative downswing. But you're right in that stages 1 & 3 will draw more current together and stage 2 less current, and viseversa when they swing the other way. But the preamp supply filter cap has the reserve to smooth this out and still provide a constant supply rail voltage. It all basically evens out over a full cycle if you're not clipping. But assymetrical clipping can mean under signal the whole preamp current draw is changed from conditions with no signal.

[jjman]

Each preamp stage could be swinging significantly different currents but they all swing very little compared to the output.

My understanding is that the cathode bias of the preamps prevents clipping at 1.5v peak voltage. This because the up-swing also pulls the cathode up. So they don't "meet" until a higher peak is reached. I think I read around 3-5v peak once but I don't really know.

[Firestorm]

When you try to drive the grid more positive than the cathode, electrons get attracted to the grid instead of the plate and effectively clamp your grid voltage preventing the signal from swinging any more positive. This accounts for the clipping on the negative side of the output signal (and is why it is sometimes said that preamp tubes clip at the input). Clipping on the positive side is basically just cut-off: the grid swings negative enough to suppress current from the cathode and the plate rises to the supply voltage, hitting the wall, as it were. You can make that happen if you really screw up the bias. But you probably don't want to clip both sides, since that typically sounds less "musical."

[rdjones]

Current flowing through the tube changes with the instantaneous voltage at the grid.
Average current drawn from the power supply is (nearly) constant assuming a properly biased class A stage, and a properly filtered supply node.
Your overly large cathode resistor will change this behavior, allowing the signal to clip only at one extreme.
This will reduce average current draw slightly when the stage is allowed to spend more time at cutoff.

Generally, a Class A stage will exhibit the same current draw at idle as at near clipping.

rd

[surfsup]

katopan, yes I had them reversed sorry was typing not thinking..

jj/Fire/rdjones, as I now understand it, grid clipping hits when voltage is zero from grid to cathode. So simply looking at the load line, we have 1.8mA flowing through the tube when Vgrid=0. So if this is the case, 1.8mA flows through the cathode resistor which is 10k, which is 18V, which should define the point at where the signal will actually clip. In merlin's book he states the signal clips but doesn't explain how everything will shift.

I'm hopefully saying the same thing you guys are saying, and thanks for the replies!

[gearhead]

Just to be esoteric, signal inversion is a function of the specific circuit hooked up to a preamp tube, and not the tube itself.

[katopan]

Also keep in mind that the valve doesn't hard-wall clip as soon as the grid reaches the cathode voltage. That diode effect has a characteristic of volts vs current and clipping is a gradual thing. It has a start threshold but the curve from then on depends on the valve and operating point.