Killing signal generator outputs

[Matador]

I have a few signal generators that I use for equipment troubleshooting. I've killed both of them, so I'm wondering if there's a better way to do testing.

What I've done: the generators use BNC outputs, so I made up a cable that passes the output signal through a hefty 4.7uF/1000V film cap (it's large, and gives me something to hold on to). The output of the signal generator goes to one lead of the cap, and I can use the other end to inject signals into equipment I'm testing or troubleshooting.

For example, I'll inject a test sine wave through the grids of a tube amp, looking to see when a signal stops, becomes distorted, etc. In general, grids are ground referenced, however some grids, like power tube grids in a fixed bias amp, or cathode followers, have grids at large negative or positive voltages. Most of the time, injecting even there through the cap is fine, as the cap should protect the (50 ohm) output...

However it obviously doesn't, which probably means the initial contact between the cap lead and the (powered) device causes a large undershoot and/or overshoot on the output which flows through to devices that are driving the output signal. I also tried opening the generator and adding clamp diodes to the rails and the output, however I've found that even if the outputs don't exceed the rails, that distortion is added to the signal. Maybe terminate the 'driven' side of the cap with a mid-ish resistor to ground, to limit the slew rate of the driven side? I don't care about voltage accuracy, it's just used to listen for presence of the test signal, and I can dial up or down the magnitude if needed.

Is there a better way to do this?

 

[Audio1Man]

Brian

Try adding a 1K resistor in series to the cap (Resistor is the probe)
Duke

 

[Newmarket]

I think a schematic or sketch would really help here ?

 

[metalockpick]

There should be a load resistor at the output of the sig gen, before the cap. That way, the cap will provide protection. Without the resistor, the cap is blocking DC, but allowing pulses of up to the voltage (grid bias, etc) on what you're trying to run it into.
The load resistor value is the rated load of the sig gen.

 

[Matador]

I think a schematic or sketch would really help here ?

It's just a cap in series with the signal generator, between the signal generator and the test points that are receiving the signal.

 

[Matador]

The load resistor value is the rated load of the sig gen.

It's a 50 ohm output: so you mean a 50 ohm resistor on the output to ground (aka. a -3db resistive divider)?

 

[Newmarket]

It's just a cap in series with the signal generator, between the signal generator and the test points that are receiving the signal.

Understood. But you also mention diodes etc. Your choice and no problem but imo a graphical explanation always beats a textual one. Picture = 1000 Words etc. Although I might be channeling Telly Savalas at this point
As it happens I'm interested as I have a Sig gen that I haven't yet used. Initial thoughts are to stick a transformer on the output to give a galvanically isolated balanced/floating output.

 

[Audio1Man]

By adding a 1K series resistor you limit the charge/discharge current. If you use a TX the input pulse and the disharge or flyback will kill the generator output stage.
Duke

 

[JohnRoberts]

the sine wave output was in the bottom right hand corner.... note clamp diodes to the PS rails, AFAIK this was pretty reliable.

JR

 

[metalockpick]

It's a 50 ohm output: so you mean a 50 ohm resistor on the output to ground (aka. a -3db resistive divider)?

Yes, but seeing the schematic, 100-500 Ohms will do it.
For now, Check those output op amps and output diodes!

 

[abbey road d enfer]

There should be a load resistor at the output of the sig gen, before the cap. That way, the cap will provide protection.

Or not.

Without the resistor, the cap is blocking DC, but allowing pulses of up to the voltage (grid bias, etc) on what you're trying to run it into.

A resistor there may or may not reduce the amplitude of the pulse. If the point that is probed is low impedance, the pulse will be the same.

The load resistor value is the rated load of the sig gen.

That would just reduce the output of the generator by 6dB, not much else.

 

[metalockpick]

Yes, IF the probed point point is high voltage and LOW IMPEDANCE, there could still be an issue.
However, if you probe a high impedance point with more than 40 VDC without the load resistor, there WILL be an issue.
Adding the load resistor means any voltage that gets through the capacitor gets divided; the source impedance of the voltage source, plus the impedance of the capacitor @ the voltage pulse frequency (near DC) divided by the load resistor. For most things in audio, the capacitor plus the load resistor should save the test equipment, as it greatly increases the impedance of the capacitor at very low frequencies. The amount of energy the Sig Gen is getting from the probe hitting DC will be greatly reduced, and then the protection diodes can do their job.

 

[abbey road d enfer]

However, if you probe a high impedance point with more than 40 VDC without the load resistor, there WILL be an issue.

Why 40V?

Adding the load resistor means any voltage that gets through the capacitor gets divided; the source impedance of the voltage source, plus the impedance of the capacitor @ the voltage pulse frequency (near DC) divided by the load resistor.

If you accidentally probe a B+ capacitor, the contact is immediate and results in a very high slew-rate pulse, to which the capacitor acts as a short, so nothing is divided. Only a series resistor can limit current.

For most things in audio, the capacitor plus the load resistor should save the test equipment, as it greatly increases the impedance of the capacitor at very low frequencies.

You cannot neglect the fact taht when you probe, you have a transient. The capacitor start with zero volt across it so the whole DC voltage goes to the generator. Series resistance is needed to protect.