Leader LAG-120A thermistor replacement

orbanp

Member
2007-07-25 2:36 pm
Hello Everyone,


This is the reply to RACBs question about the burned out thermistor in the LAG-120A generator.

His question seemed to hijack the original thread about requesting a manual.

The generator is a Wien-bridge circuit, theoretically it needs a gain of 3, or about 9.6dB.
Looking at the schematics, the gain set by the inverse of the feedback, meaning the ratio of TH101+R130 over R119.
For a 1:3 voltage division TH101+R119 should be about the double of R119, or about 1.3kOhm.
Simulation of the circuit also bears this out.

For proper operation the thermistor should have the smallest thermal inertia possible. Glass-bead thermistors have this feature.
Looking around the web, this place produces such thermistors:
NTC Thermistors - Glass BR Series | Thermometrics Inc.
Do download the datasheet.
The BR11 series has the fastest response, the thermal time constant is 0.8sec. It is also produced in the required 1.3k value.
I also noticed others suggesting thermistors from DigiKey, the time constant of that thermistor is less than 7sec!
Where to get that glass-bead thermistor - good question!
Could ask for samples!

Now onto how to bring up the system if you got that glass-bead thermistor.
The output from the generator is about 3Vpp (if I interpret the manual correctly). The output amp has a gain of just about 1, so the output from the oscillator is also about 3Vpp.
The max power rating of that BR11 thermistor is 7mW, if the oscillator runs away, e.g. the output gets to 30Vpp or close (the supply voltage is 40V, it could happen!), it will burn out the thermistor.
The output voltage is set by the adjustment of the feedback in the oscillator amp.
Other Wien-bridge oscillators (like from HP) have means of adjusting the feedback. There the equivalent of the R119 resistor is made up from a fixed resistor and a trimmer pot.
I suggest you do the same here, e.g. use a 500Ohm resistor in series with a 500Ohm trim-pot. Start out from the trim-pot having 500Ohm value and reduce it slowly while measuring the oscillator output voltage. First there will be no output as the feedback would be too much. When the output appears, reduce the the trim-pot while get to 3Vpp output. If I am correct HP suggests to use 1kHz frequency for such adjustment.

As there is no feedback adjustment in the original LAG-120A oscillator output could have ran away, and probably that is what killed the thermistor.
The VR103, 4.7k trim-pot, is for setting the DC voltage at the output, it should be half of the supply voltage at the junction of R128 and R129 resistors.
Adjusting the feedback also changes the output DC voltage, so do this adjustment after getting the 3Vpp sinusoidal signal right.

Hope this helps.
Regards, Peter
 

BSST

Member
2018-07-07 2:21 am
I'm pleased you were able to get a service manual.

I agree with most of the comments and advice in the post from Peter above, but I do take exception to the assertion that the thermistor should have the minimum possible thermal inertia. Said differently, it would have a very fast time constant and would respond very quickly to temperature. Taken to the limit of 0 inertia, the thermistor would respond instantaneously to instantaneous power within the course of a single cycle of the generator's output. It's resistance would drop in response to increasing voltage; i.e. it would be a non-linear resistor and disastrous for low distortion. The oscillator would have very strong odd harmonic content.

For best distortion, the thermistor would have constant resistance over the course of a single cycle, but it must respond to average power in order to find a stable output amplitude. I know that lamp (or thermistor) fluctuation is responsible for worsening distortion as operating frequency is lowered. Of course very long time constants would lead to objectionably long settling time, so compromise is necessary.

To help choose a thermistor, I suggest trying to compare the physical size and surface area of the failed thermistor with catalog offerings. You don't need precision---this is not a temperature measurement application. The original Leader part surely wasn't costly. Remember that the 25C resistance must be higher than 1.3K to ensure the oscillator can start.

Good luck!