Some trouble with Mojotone FX loop

[Yoda]

On Q1 and 2 what dc voltages to you read when you prob the side of the transistor with one tab on it?

I get 17.6v on Q1 and 1.15v on Q2.

[Stevem]

You took a chance and bought a dud board it looks like.

[Yoda]

Any idea what those voltages should have been at Q1/Q2?

[sluckey]

How good are your soldering skills? For under $5 you could replace Q1, Q2, and C6. Chances are really good that the FX loop would work good as new. Worth the gamble IMO, unless you do a botched soldering job.

[R.G.]

Any idea what those voltages should have been at Q1/Q2?

Maybe we can guess.

Te LND150 is a depletion mode device, meaning it conducts if the gate to source voltage is not negative. It's a MOS device, so its gate caannot conduct any current. The datasheet says that the cutoff voltage from gate to source will be -1V minimum, -3V maximum.

Since the gate cannot conduct current, the gate of Q! will be held at half of the supply voltage after the dropping resistor. This is likely to be in the 100V to 20oV range, so we can ignore it.
The channel of Q1 will let current through to the source and this current through the 47K resistor will raise the source voltage until it gets into the range of 1 to 3 V higher than the gate. So the 47K source resistor will have something like half the power supply voltage across it.
The current through Q1 will be Vsupply/2*47K. For a 200V supply to the unit, this will be very roughly 200/2*47000= 2.12 mA. For a 400V supply, this would be 4.2ma.
Q2 is a little harder, more guesswork. The gate is held at ground by resistors, so current rises until the 1.5K source resistor is from 1 to 3v above ground and this starts shutting off the transistor. The current in this device will be between 1V/1500 and 3V/1500. That's between 666uA and 2ma. if we guess that most devices will be in the middle of the range, the most probable current is 1ma.
1ma through the drain of Q2 will pull down the 150k drain resistor by 1ma* 150k = 150V. So the drain of Q2 will sit 150V lower than the supply voltage after the 10K dropping resistor.
This has been ignoring the drop in the 10K dropping resistor. The current the transistors pull can be between 3-ish ma minimum and 6.2ma. So the 10K will drop between 10K*0.003 and 10K*0.0062 volts, 30 to 62V.

The estimation process can be run again, and get closer approximations to the voltage drop for the supply. You can eventually get really close for any starting guess on the high voltage dc fed to the board. However, it's fair to just pick off some estimates after only one pass.

The 10K resistor should have no more than 65V or so across it. If it does, one of the transistors is pulling too much current or a capacitor is failed, something like that.
The drain of Q1 will be at the supply voltage after the 10K. The gate of Q1 will be at half that, and so will the source, within a volt or two.
Q2'gate will be at 0V. Q2 source will be one to three volts, and Q2's drain will be 50V to a big fraction of the supply voltage after the 10K, depending on its threshold voltage and transconductance.

[Ten Over]

Maybe we can guess.

Te LND150 is a depletion mode device, meaning it conducts if the gate to source voltage is not negative. It's a MOS device, so its gate caannot conduct any current. The datasheet says that the cutoff voltage from gate to source will be -1V minimum, -3V maximum.

Since the gate cannot conduct current, the gate of Q! will be held at half of the supply voltage after the dropping resistor. This is likely to be in the 100V to 20oV range, so we can ignore it.
The channel of Q1 will let current through to the source and this current through the 47K resistor will raise the source voltage until it gets into the range of 1 to 3 V higher than the gate. So the 47K source resistor will have something like half the power supply voltage across it.
The current through Q1 will be Vsupply/2*47K. For a 200V supply to the unit, this will be very roughly 200/2*47000= 2.12 mA. For a 400V supply, this would be 4.2ma.
Q2 is a little harder, more guesswork. The gate is held at ground by resistors, so current rises until the 1.5K source resistor is from 1 to 3v above ground and this starts shutting off the transistor. The current in this device will be between 1V/1500 and 3V/1500. That's between 666uA and 2ma. if we guess that most devices will be in the middle of the range, the most probable current is 1ma.
1ma through the drain of Q2 will pull down the 150k drain resistor by 1ma* 150k = 150V. So the drain of Q2 will sit 150V lower than the supply voltage after the 10K dropping resistor.
This has been ignoring the drop in the 10K dropping resistor. The current the transistors pull can be between 3-ish ma minimum and 6.2ma. So the 10K will drop between 10K*0.003 and 10K*0.0062 volts, 30 to 62V.

The estimation process can be run again, and get closer approximations to the voltage drop for the supply. You can eventually get really close for any starting guess on the high voltage dc fed to the board. However, it's fair to just pick off some estimates after only one pass.

The 10K resistor should have no more than 65V or so across it. If it does, one of the transistors is pulling too much current or a capacitor is failed, something like that.
The drain of Q1 will be at the supply voltage after the 10K. The gate of Q1 will be at half that, and so will the source, within a volt or two.
Q2'gate will be at 0V. Q2 source will be one to three volts, and Q2's drain will be 50V to a big fraction of the supply voltage after the 10K, depending on its threshold voltage and transconductance.

A depletion mode device will conduct with a negative gate-to-source voltage.

LND150 Average Characteristics copy.jpg

[Yoda]

How good are your soldering skills? For under $5 you could replace Q1, Q2, and C6. Chances are really good that the FX loop would work good as new. Worth the gamble IMO, unless you do a botched soldering job.

They’re good enough to give it a shot if I end up stuck with the board, then I have nothing to lose.

[Yoda]

Any idea what those voltages should have been at Q1/Q2?

Maybe we can guess.

Te LND150 is a depletion mode device, meaning it conducts if the gate to source voltage is not negative. It's a MOS device, so its gate caannot conduct any current. The datasheet says that the cutoff voltage from gate to source will be -1V minimum, -3V maximum.

Since the gate cannot conduct current, the gate of Q! will be held at half of the supply voltage after the dropping resistor. This is likely to be in the 100V to 20oV range, so we can ignore it.
The channel of Q1 will let current through to the source and this current through the 47K resistor will raise the source voltage until it gets into the range of 1 to 3 V higher than the gate. So the 47K source resistor will have something like half the power supply voltage across it.
The current through Q1 will be Vsupply/2*47K. For a 200V supply to the unit, this will be very roughly 200/2*47000= 2.12 mA. For a 400V supply, this would be 4.2ma.
Q2 is a little harder, more guesswork. The gate is held at ground by resistors, so current rises until the 1.5K source resistor is from 1 to 3v above ground and this starts shutting off the transistor. The current in this device will be between 1V/1500 and 3V/1500. That's between 666uA and 2ma. if we guess that most devices will be in the middle of the range, the most probable current is 1ma.
1ma through the drain of Q2 will pull down the 150k drain resistor by 1ma* 150k = 150V. So the drain of Q2 will sit 150V lower than the supply voltage after the 10K dropping resistor.
This has been ignoring the drop in the 10K dropping resistor. The current the transistors pull can be between 3-ish ma minimum and 6.2ma. So the 10K will drop between 10K*0.003 and 10K*0.0062 volts, 30 to 62V.

The estimation process can be run again, and get closer approximations to the voltage drop for the supply. You can eventually get really close for any starting guess on the high voltage dc fed to the board. However, it's fair to just pick off some estimates after only one pass.

The 10K resistor should have no more than 65V or so across it. If it does, one of the transistors is pulling too much current or a capacitor is failed, something like that.
The drain of Q1 will be at the supply voltage after the 10K. The gate of Q1 will be at half that, and so will the source, within a volt or two.
Q2'gate will be at 0V. Q2 source will be one to three volts, and Q2's drain will be 50V to a big fraction of the supply voltage after the 10K, depending on its threshold voltage and transconductance.

That is certainly a lot of info, I’ll admit I don’t understand all of it but what I do understand is very helpful so I appreciate it. Mojotone did get back to me that there may have been some incorrect resistors installed on some of those boards so I will have to confirm what is going on when I get home.

[Ten Over]

The channel of Q1 will let current through to the source and this current through the 47K resistor will raise the source voltage until it gets into the range of 1 to 3 V higher than the gate. So the 47K source resistor will have something like half the power supply voltage across it.
The current through Q1 will be Vsupply/2*47K. For a 200V supply to the unit, this will be very roughly 200/2*47000= 2.12 mA. For a 400V supply, this would be 4.2ma.

According to the Output Characteristics curve in the data sheet, the Source voltage would have to be lower than the Gate voltage at those currents (i.e. Vgs is positive). The Vgs would typically be less than +1V. The range of 1V to 3V has nothing to do with it.

[Ten Over]

Q2 is a little harder, more guesswork. The gate is held at ground by resistors, so current rises until the 1.5K source resistor is from 1 to 3v above ground and this starts shutting off the transistor. The current in this device will be between 1V/1500 and 3V/1500. That's between 666uA and 2ma. if we guess that most devices will be in the middle of the range, the most probable current is 1ma.

Why would the Source resistor be from 1V to 3V above ground? Gain stages like that don't idle at cut-off, unless the Source resistor is ridiculously large. They strike an equilibrium where the Drain current causes a voltage across the Source resistor that creates a Gate-to-Source voltage that causes that particular Drain current. A 1.5k Source resistor typically results in a Drain current of 0.43mA and a Source voltage to ground of 0.65V.

[Ten Over]

...and Q2's drain will be 50V to a big fraction of the supply voltage after the 10K, depending on its threshold voltage and transconductance.

The Drain current of Q2 is determined almost exclusively by the 1.5k resistance of the Source resistor. The voltage drop across the 150k Drain resistor is typically 65Vdc and this is independent of the supply voltage. Again, the threshold voltage has nothing to do with it.

[Yoda]

...and Q2's drain will be 50V to a big fraction of the supply voltage after the 10K, depending on its threshold voltage and transconductance.

The Drain current of Q2 is determined almost exclusively by the 1.5k resistance of the Source resistor. The voltage drop across the 150k Drain resistor is typically 65Vdc and this is independent of the supply voltage. Again, the threshold voltage has nothing to do with it.

Well funny you mention the 1.5K resistor. After talking to Mojotone it seems some of these boards shipped out with the 47K and 1.5 K resistors erroneously installed in the wrong places and sure enough this board has a 47k where the 1.5k should be and vice versa so we’ll see if swapping them fixes anything.

[R.G.]

The channel of Q1 will let current through to the source and this current through the 47K resistor will raise the source voltage until it gets into the range of 1 to 3 V higher than the gate. So the 47K source resistor will have something like half the power supply voltage across it.
The current through Q1 will be Vsupply/2*47K. For a 200V supply to the unit, this will be very roughly 200/2*47000= 2.12 mA. For a 400V supply, this would be 4.2ma.

According to the Output Characteristics curve in the data sheet, the Source voltage would have to be lower than the Gate voltage at those currents (i.e. Vgs is positive). The Vgs would typically be less than +1V. The range of 1V to 3V has nothing to do with it.

In this setup, if the power supply is suddenly connected, the gate goes to the mid-power-supply voltage. Current flows in the channel and the source voltage rises until it gets to a point where it starts cutting back on the ability of the channel to conduct by backing down the Vgs. We know that if the source gets somewhere between -1 and -3V, depending on which LND150 we got, the channel current will drop to zero, so the source voltage will not rise to more than 1 to 3V higher than the gate, or it will turn off the current making it rise.
The chart only shows Vgs for -1.2V to +1.4V, it being a typical device. the source voltage will rise until it hits a gate voltage between -1.2V and +1.4V, because -1.2V causes 0 drain current. For this one device, it won't need to get higher than +1.4V because that gets you about 5ma of current. As an aside, it probably gets the device really near it's power maximum, too, that hyperbola in the chart.
Given that the gate is sitting at half the power supply, the source must rise to that same voltage, give or take a volt or two. So you're going to get about half the 200-ish to 350-ish power supply at the source, as long as the LND150 can conduct that much current and not burn up. The source will beat the gate voltage +/- a volt or two.

The cutoff voltage is important in this case because it sets a hard limit on how low the source resistor's voltage can go.

[R.G.]

Q2 is a little harder, more guesswork. The gate is held at ground by resistors, so current rises until the 1.5K source resistor is from 1 to 3v above ground and this starts shutting off the transistor. The current in this device will be between 1V/1500 and 3V/1500. That's between 666uA and 2ma. if we guess that most devices will be in the middle of the range, the most probable current is 1ma.

Why would the Source resistor be from 1V to 3V above ground? Gain stages like that don't idle at cut-off, unless the Source resistor is ridiculously large. They strike an equilibrium where the Drain current causes a voltage across the Source resistor that creates a Gate-to-Source voltage that causes that particular Drain current. A 1.5k Source resistor typically results in a Drain current of 0.43mA and a Source voltage to ground of 0.65V.

No, they don't idle at cutoff. They idle at the voltage where there is just enough voltage away from cutoff to make the tranconductance let enough current through above cutoff to support that gate and source voltage.

Using the cutoff range of -1 to -3 accounts for the range of devices. If you have a device that has a Vgs off of -1.2V, close to the device in the typical chart, the gate-source will not be 1.2V, but some fraction above that. That fraction is enough so that the particular device's transconductance - current increase per Vgs increase in tiny intervals - raise the channel current enough to steady the device at an idle point.

I think I see some of the confusion. The OP wanted debug data, what to measure and order-of-magnitude target voltages. Without a few hours of equations or some simulator time, the process I outlined was where to start debugging. It gets you very close, a volt or two in 100s of volts circuits. A full design pass would need a very much larger amount of information, not least of which would be the range and sigma of the Vgs and transconductance for likely commercial quantities of LND150. A design with only typical voltages can be significantly off when you put in real devices with variation.

So OK - the source might be 0.65V for a solidly typical device. Sounds reasonable. It's not ten, or fifty. If you happened to have the bad luck to get a device with a real -3V Vgsoff, then it could easily be 1 to 2.5V, depending on the transconductance that went with that particular device. Probing a known-not-working board, knowing the source is between 1 and 3 V is probably good enough for debugging.

[R.G.]

...and Q2's drain will be 50V to a big fraction of the supply voltage after the 10K, depending on its threshold voltage and transconductance.

The Drain current of Q2 is determined almost exclusively by the 1.5k resistance of the Source resistor. The voltage drop across the 150k Drain resistor is typically 65Vdc and this is independent of the supply voltage. Again, the threshold voltage has nothing to do with it.

Well, maybe. the threshold voltage sets limits, even if soggy ones, for what the 1.5K resistor current will be. It sets a hard limit on what it cannot be. The voltage in that resistor can't for instance, be bigger than the threshold voltage, because that would mean the channel current is zero, and clearly it can't be for a good device.

This lets us estimate things even with >> zero << device knowledge other than the cutoff voltage, which is what I did. I didn't get to see the handy chart until after I posted the estimate.

The voltage drop on the 150K resistor may well be typically 65V, for typical values of LND150s having typical values of threshold voltage and typical transconductances. In fact, the voltage on the 150K is something I only estimated based on the current that flows in the 1.5K resistor. The fact that the channel current flows in both resistors means that the voltages will be in the ratio of the resistances - whatever voltage appears in the 1.5K, a 100 times larger voltage appears on the 150K, since it's the same exact current.

I agree that the drain current is determined by the 1.5K resistor - and the transconductance, and the cutoff voltage for that particular device.