cathode follower

What is the voltage at the junction of the 1M, 1k and 47k resistors? I would have said with 219v across the 1k + 47k that you should get about 206 volts at the junction. Either your meter is taking current through the 1M or the valve is... If your meter has input resistance of 1M then you could expect to see 1/2 the actual value.

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Geo

Remove the .01 and the .047 caps. This will isolate the tube from any external influence. The bias for the grid is developed across the series 1k and 47k resistors. Seeing as the grid does not draw current, the 1 meg resistor drops no voltage. So the grid sits at whatever voltage is at the junction of 1k and 47k. The cathode draws current and will be positve with respect to ground. If there needs to be 139 volts from the junction of

1k/47k, then there needs to be about 3ma of current flowing thru the 47k . The actual cathode to grid bias should be about 3 volts. If the part values check out correct, then the tube is not drawing the correct current. bg

The voltage at the junction of 1k/47k should measure 205.625v . If there is no current flowing in the 1 meg resistor, and there shouldn't be, then the grid would measure 205.625 as well. Obviuosly 58 volts at the grid indicates that the grid or something is drawing current thru the 1meg resistor(see below). Perhaps the socket has carbon arcs or it is the voltmeter?? Using a simulator program with a 12ax7a and 390 volts on the plate gives me these values - cathode current = 2.18ma Voltage at 1k/47k junction = grid voltage = 102.5v Cathode voltage = 104.7v (so the actual grid bias is neg 2.2v with respect to cathode) Bear in mind that a 10 meg voltmeter connected to the grid and ground would severly upset the grid voltage. It would measure about 33 volts. The only way you can measure grid voltage here is to measure it at the junction of the 1k/47k resistors. I would assume that if you measure the voltage across the 47k resistor, use ohms law, and come up with about 3ma, the circuit is working correctly. Measure the resistance from grid to ground and it should be 1meg + 47k. Measure the resistance from grid to plate and it should be infinite.

I take it you do not know vacuum tube functions at all, that's what I grew up on. No knocks, that's just the way things happen.

Any tube will conduct current until the grid is at a somewhat lower voltage than the cathode, ie the grid is negative with respect to the cathode. Any meter measuring a high resistance/impedance circuit will load the circuit and distort the readings. The meter should have an input resistance of at least 10 times the highest resistance in the circuit. SInce you circuit shows resistances of 1 megohm, your meter should have an input resistance of at least 10 megohms on anay scale youuse to measure the voltage. 20x would be even better. Pin 2 should have a somewhat lower voltage than pin 3, by the ratio of

47/48. The voltage at the junction of the 1K and 47 K resistors in the cathode should be the same as the voltage measured at Pin 2 unless the tube is defective or more likely, the voltmeter impedance is changing the grid voltage.

I take it you do not know vacuum tube functions at all, that's what I grew up on. No knocks, that's just the way things happen.

Any tube will conduct current until the grid is at a somewhat lower voltage than the cathode, ie the grid is negative with respect to the cathode. Any meter measuring a high resistance/impedance circuit will load the circuit and distort the readings. The meter should have an input resistance of at least 10 times the highest resistance in the circuit. SInce you circuit shows resistances of 1 megohm, your meter should have an input resistance of at least 10 megohms on anay scale youuse to measure the voltage. 20x would be even better. Pin 2 should have a somewhat lower voltage than pin 3, by the ratio of

47/48. The voltage at the junction of the 1K and 47 K resistors in the cathode should be the same as the voltage measured at Pin 2 unless the tube is defective or more likely, the voltmeter impedance is changing the grid voltage.

I agree with the above, but these old high value resistors have a tendency to go way up in value and that would just exasperate the measurement problem. It may be worth measuring the 1 meg resistor to see if it still is anything close to 1 meg.

David

Uh well yes but just not with a cathode follower/splitter.

The meter is a Fluke 77 so I would expect there to be at least 1 meg on the probe without checking to be sure.

Yeah the resistors are all within +- 10%.

Think: Depletion mode MOSFET. :)

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Now you have to explain that to me, my higher education on hardware stuff stopped with PNPs and NPNs. Then I worked EMC issues for 40 years before retriring from Bell Labs.

Bob,

FET = Field Effect Transistor JFET = Junction FET. These are depletion mode so a Gate-Source voltage of the opposite polarity as the Drain cuts off current just like a tube. MOSFET = Metal Oxide Semiconductor FET. There come in enhancement mode where a Gate-Source voltage toward the drain turns it on. Depletion mode MOSFETs operate more like a JFET.

David

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A tube and JFET/Depletion mode Mos Fet, are naturally biased in the on state. These devices need to be turned off to control them instead of On like you would logically think.

With a Tube and like the N channel fet, the voltage on the grid/gate must be lower than that of the cathode/Source to enter the pinch off state of the devices.

The pinch off voltage is most likely not the same spec as that of the original tube.. The pinch off voltage = the voltage on the grid that is lower than the voltage on the cathode, which turns off the tube.

In that circuit, it is most likely acceptable for that value to differ from the original spec and have it still work with in reason.

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