Vbe multiplier with Si for output stage with Ge

[Citizen124032]

As a follow-up to a recent post: is it possible to apply a VBE multiplier with a Si transistor for a complementairy output stage with Ge transistors?
(See example).
R2 in the circuit must be of a very low value to avoid a melt down of the Ge's. But is it 'controllable' with a certain adjustable value of R1?
Wrapping up: what would be the lowest possible multiplied Si-Vbe?

Answers not needed:

• how Vbe multipliers work (explanations, doubts of my knowledge about this topic)
• to have things thermally connected / frequency related issues / driving or loading issues
• &c

Just: if, how, what can, what not. Nothing else.
It's theoretical.
Practical solutions from the past are appreciated, as long as they fit the scope.

Thanks Andersonix for your very single example.

 

[BSST]

I haven't been able to find a Vbe vs. temperature tempco for germanium. Any data?

 

[MarcelvdG]

For silicon, it's

(V_BE - V_BG - some correction term)/T

where V_BG is the bandgap voltage and T the absolute temperature. I guess the same holds for any other semiconductor, but with a different value for V_BG.

The base-emitter and bandgap voltages of germanium are very roughly about half that of silicon, so the temperature coefficient should also be about half that of silicon.

 

[davidsrsb]

I have seen claims that germanium is actually slightly higher than silicon at -2.5mV/C

 

[BSST]

I wouldn‘t be surprised if the tempco turned out to be about -1.1mV/C, half that of silicon. A single silicon junction might be about the right compensation for Citizen’s complimentary Ge output stage. Absolute guesswork on my part.

 

[BSST]

Just saw your post. Indeed, I guess.

 

[Elvee]

For usual current densities, the tempco is also around -2mV/°K

 

[Citizen124032]

The Ge output stage is an example where a very low value Si Vbe multiplier is needed.
So it is not about the tempco things here, but it is all about the low value Si Vbe multiplier.

Wrapping up: what would be the lowest possible multiplied Si-Vbe?

For example: is it possible to drive the bjt in saturation with R1 and R2 for a Vce of less then Vbe?
(That would make it a 'Vbe divider'!)

 

[BSST]

I think using a Ge transistor would be the simplest solution, but if you want a silicon circuit I suggest the following concept.

Add a TLV431 in series at the top, above the Q7 collector/Q11 base. Wire the shunt regulator for 1.24V and shunt it with a 1k pot.
Make R1 about 5k, adjustable. Make R2 = 3.9k and connect it to the 1k wiper rather than Q7. I think the circuit will give plenty of adjustment range.

 

[JMFahey]

I think you can not, very different materials.
To boot, even a single Si junction has higher voltage drop than two Germanium ones, which makes it all but unusable.
No voltage multipliers , multiplication factor would be less than one.

Classic Germanium bias solution was to use thermistors, which are resistive, no junction drops involved.
Look at 60's Germanium power amp schematics.

 

[EdGr]

All bipolar transistors follow the Ebers-Moll equation. The approximate form is:

Ie = Is * exp (Vbe/Vt)

The difference between germanium and silicon is a five orders of magnitude higher Is.
Ed

 

[Citizen124032]

Add a TLV431 in series...

I think you've lost me somewhere, but this will make things very complex.

 

[Elvee]

The Ge output stage is an example where a very low value Si Vbe multiplier is needed.
So it is not about the tempco things here, but it is all about the low value Si Vbe multiplier.

You need both: a relatively low absolute voltage, like 0.25 to 0.3V, and a tempco of -2mV/°K. There are known techniques to artificially increase the tempco, I think Self outlined some.
Once you have a 0.6V voltage having a tempco of -4.5mV/°K, you just need to halve it to have the correct voltage and coefficient.
Ge circuits often used a NTC to achieve compensation

 

[BSST]

I think you've lost me somewhere, but this will make things very complex.

I contend that it's not that complex--- only two more components than in the original circuit.

The tempco is ~ -(1+R2/R1)*2.2mV/C; The 1K pot position lowers the voltage across the Si transistor to trim the spreader voltage and its position has minor effect upon on tempo. (Wiper resistance varies from 0 to 250 ohm, small compared to R1+R2.)

Best regards,

Steve

 

[Citizen124032]

Many thanks BSST!!!
Very helpfull and it hints in the direction to think about it.
The application I'm busy with has nothing to do with Ge output stages (that is an example), but with low value Vbe multipliers.
I think I have my answer.
Thanks for all the contributions.

Copied your drawing for easy viewing.

 

[EdGr]

This solution does not look right.

Vt=kT/q affects silicon and germanium transistors equally. Their Vbe variation with temperature is the same when expressed as a percentage of Vbe.

This implies that one can multiply or divide Vbe without affecting the percentage variation with temperature. All bipolar transistors will track.
Ed

 

[Mark Tillotson]

I thought the Vbe variation with temperature is dependent on the difference between Vbe and the bandgap voltage. For Si this is 1.15V-0.65V = 0.6V typically, for Ge its more like 0.74 - 0.2 = 0.54V, and in particular the higher Vbe the smaller the temperature dependence.

For Si you have about 0.6V for about 300K at room temperature, so the coefficient is about -2mV/K
But for very low currents the Vbe = 0.5, the difference is 1.15-0.5 = 0.65, so 650mV/300K gives about -2.2mV/K

Have I remembered this correctly?

 

[EdGr]

Assume Is=10^-13. Ebers-Moll says:

At Vbe=0.6V, Ie=1.2mA, and tempco=-2mV/K.
At Vbe=0.5V, Ie=25uA, and tempco=-1.7mV/K.
At Vbe=0.3V, Ie=11nA, and tempco=-1mV/K.

Now, put 100,000 silicon transistors in parallel. Is is now similar to a germanium transistor.
Ed

 

[JMFahey]

I love simple practical solutions

 

[BSST]

I have no opinion on the tempo of Ge base-emitter voltage. I only the offer the circuit in post 15 as a way to set tempo and and offset trim with quasi independence, using silicon semiconductors.