Current source in Princeton Electrometer repair

[charles.ouweland]

I'm currently in the repair of a Princeton Applied Research Model 136 Electrometer. The input amplifier is not working. Unfortunately I could not find a schematic so I'm reverse engineering the whole thing. One of the problems is the noobtainium quality of some of the components, so it takes some thinking how the circuit was originally intended. But I'm coming a whole way. One of the things I don't understand is what appears to be a current source, the schematic of which is attached to this post. I have measured it and it delivers a consistent 0.16mA out of the collector of T11 over the whole range of -9V to +7V (the emitters of T9 and T10 are connected to nothing). That's great but can anyone explain how this circuit works? Question number two: how is the current set and/or is it possible to change the current? TIA

[perieanuo]

that schematic is incomplete, those npn's wont open like that, so something is missing. or you have shorted transistor (the one you say it outputs cc)

[charles.ouweland]

That's what I was thinking too, so I double checked. But then again, the bc-junction of T10 will conduct no problem and for T9: The 2N3565 specifies a Collector Cutoff Current (ICBO) of max 50nA under test conditions IE=0, VCB=25V and T=25ºC. That multiplied by T11's hFE of 200 gives 10µA, nowhere near 0.16mA. Remains a mystery to me.

[charles.ouweland]

And here are the photos of the component side and solder side (mirrored) of the board, where you can clearly see that the top leg of both T9 and T10 are not connected to anything

[fzabkar]

Measure the voltages. For a start, the base of T11 should be sitting at 1.8V below the 9V supply.

[charles.ouweland]

I'll measure next week

[Alex Nikitin]

The circuit would make sense if pin 3 of  T9 and T10 is an emitter. A reverse biased emitter junction can operate as a very low current capable zener at about 6-7V.

Cheers

Alex

[fzabkar]

The circuit would make sense if pin 3 of  T9 and T10 is emitter. A reverse biased emitter junction can operate as a very low current capable zener at about 6-7V.

http://njsemi.com/datasheets/2N3565.pdf

Emitter to Base Breakdown Voltage (BVEBO) = 6.0V (min) at IC = 0, IE = 10uA

[charles.ouweland]

I'll double-check T9's connections next week when I'm back. This particular package is not in that datasheet (?). Maybe I have indeed swapped c and e

[charles.ouweland]

My 2N3565's are in a different package, I think TO-106, which is not in that datasheet. I desoldered T9 and put it in a transistor tester, which showed 123=EBC (as in the datasheet), but the pin numbering is different (top view CCW, starting from the flat side), which is opposite from the TO-92 numbering. All very confusing. Anyway, this led to the conclusion that I had e and c mixed up in my drawing.

[charles.ouweland]

The circuit would make sense if pin 3 of  T9 and T10 is an emitter. A reverse biased emitter junction can operate as a very low current capable zener at about 6-7V.
Alex

This is an excellent reply. I desoldered one of the 2N3565's and put it in a transistor tester. I double checked everything and you are right, I had the c and e of T9 and T10 mixed up.
I tried to measured T9's I_B but my meter showed 0.0µA.  A very low current indeed

Measure the voltages. For a start, the base of T11 should be sitting at 1.8V below the 9V supply.
Emitter to Base Breakdown Voltage (BVEBO) = 6.0V (min) at IC = 0, IE = 10uA

See attached revised schema with voltages. T9's V_EB is only 5.0V and there is actually almost no T9.I_B current running, so I'm still a bit puzzled as to how this is supposed to work.
I measured the current coming out of T11's collector and it's 0.37mA (not .16mA) regardless of when the collector is connected to 0V or to -9V

[fzabkar]

There must be around 1 or 2 uA flowing in the base circuit. T9 and T10 set the voltage on the base of T11, and this voltage should be reasonably constant. This then determines the current flowing in the 7K5 resistor.

This is the kind of circuit I'm used to seeing: