Understanding HV Shutdown Circuit for HPGe Detector using Thermistor Input

[hjerald1]

Hi all,

I am hoping that someone here on the Forum can provide this chemist with some electronics assistance .
I am working on setting up my newly acquired HPGe detector with my MCA system
The only thing I am missing is a HV bias supply shutdown in the event of a detector warm up.
I have obtained a circuit diagram (attached) for a HV shutdown, which I believe, senses a thermistor in the HPGe/Preamp.

Some questions:
First---Using a thermistor input (pin 8), how does this circuit function?

Second---What is the output voltage of the circuit assuming a thermistor value of
greater than 1.5K at cold temperature. What is the output voltage assuming
a thermistor value of less than 1K at warm temps.

Third--- On the diagram, what do WP? and PTR signify?
Thanks,
Jerry

 

[vk6kro]

The sensor is in a bridge circuit.

If the pot is set to maximum resistance, I calculate there will be 1.136 volts at the bottom of the 820 ohm resistor.
ie 12 * 5000 / (5000 + 820 + 47000)

At the low input point, the voltage with a 1 K thermistor would be 12 * 1000 / ( 1000 + 47000) or 0.25 volts.
With 1.5 K it would be 0.37 volts.

The opamp acts to compare these voltages. This works OK, but it can cause rapid on-off switching when the input voltage is very near the switch-over point. So, we could talk about hysteresis later if you like.

So, you would have to adjust the pot until the upper voltage (on pin 3) is between 0.25 volts and 0.37 volts.
The temperature would then determine whether the relay is pulled in or not.

The 820 ohm resistor seems unnecessary.
Changing to a 1 K pot would be a bad idea with the thermistor you suggest. It would not give you enough voltage on pin 3.
Third--- On the diagram, what do WP? and PTR signify?
PTR could mean Positive temperature coefficient thermistor, I guess.

If pin 2 is more positive than pin 3 then the opamp output will be about -10 volts. This will turn on the PNP transistor, 2N4403, and pull in the relay and turn on the red LED.
The relay could then be arranged to turn off the power to the power supply.

As it stands, the power would be restored to the power supply when the thermistor cools down. I think you would really need to disable the power supply until it can be inspected or else it will cycle on and off.

 

[hjerald1]

Hi VK6KRO,
Thanks much for the explanation! I believe I am now starting to understand the circuit.

The purpose of the circuit for my application is to shut down the HV supply to the HPGe detector if the temperature warms above about -160 deg C. Lower than this temperature (thermistor >1.5K) the circuit should supply a signal to allow the HV to be on. Higher than this temperature (thermistor <1.0K) the circuit should turn off the HV supply. Since the required cold temperature is supplied by liquid nitrogen, a warm condition would be caused by evaporation of coolant. Therefore I believe hysteresis is not necessary.

For an HV enable condition (cold HPGe detector, thermistor > 1.5K) the signal needed is an open circuit or active high greater than or equal to +1.2V to +24V. The inhibit signal condition (warm detector, thermistor < 1.0K) is -24V to < +1.2V.
It is now not clear to me how I adjust the circuit to provide these needed voltage ranges. Can you clarify for me?
Thanks again,
Jerry

 

[vk6kro]

Could you accept +12 V and - 12 V as your outputs? These seem to be in the ranges you specified.

If so, then you could have a single pole double throw set of contacts on your relay and select one of your power supply lines as the output.

I would modify the bridge circuit a little, though.

If you substitute 10 K resistors for the 47 K resistors in the diagram (and omit the 820 ohms) then you will get a range of 1.09 V to 1.56 V for the thermistor changing from 1 K to 1.5 K.

Previously, it was a change of 0.25 volts to 0.37 volts, so you can get better control with the 10 Ks.

It will need to be adjusted with actual operation, though. Opamps are not perfect and you could get an offset voltage that upsets the operation, so you need to be able to compensate for it.
You can do this just using the LEDs before you start cutting off the power supply.

 

[hjerald1]

Thanks again VK6KRO,
Yes, +12V output for HV enable and -12V output for HV inhibit (off) would work. The output lead with these enable or inhibit signals would feed via a BNC connector directly into the back of my multichannel analyzer, which apparently has internal circuitry to switch the HV bias voltage on or off depending on this signal.

That said, I must beg addition clarification from you on exactly how to implement "a single pole double throw set of contacts on your (my) relay"...new relay with this feature? Can you supply a wiring diagram incorporating this and with specific suggested parts?

I do understand the resistor substitutions you suggested.

I also appreciate your willingness to tolerate my limited knowledge of electronics!
Jerry

 

[vk6kro]

You would need to get a relay that does the switching needed.

Here is a diagram of it:
[PLAIN]http://dl.dropbox.com/u/4222062/bridge%20cct.PNG

I included a diagram of the bridge circuit. The 1.2 K could possibly be a fixed 1.2 K resistor (you can buy them anywhere) or a pot that covers this resistance. You can also get multiturn pots that give you better control.