I fear I cannot understand how this circuit provides temperature stabilization for Q2. My primal idea was that an increasing heat leads to increased collector currents, which will decrease the collector voltage at Q1 by a little and this will reduce both base currents and eventualy the collector current of Q2. Everything seems fine for now, but here comes the confusion: the reduced Ib1 will decrease Ic1 and Vc will be the same as before. I can understand that Ib1 is stable, duo to the clearly visual collector feedback, but what about Ib2? When Vc is back to normal(as it was before the variation in Hfe) i see no reason for Ib2 not to increase again. Apparently this one is missing something and he will be glad if someone helps him.
PS: please don't send me to other threads in this forum or in other forums, i've tried and nothing helps. It will be great if you can explain it here. Thanks!

 

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I fear I cannot understand how this circuit provides temperature stabilization for Q2. My primal idea was that an increasing heat leads to increased collector currents, which will decrease the collector voltage at Q1 by a little and this will reduce both base currents and eventualy the collector current of Q2. Everything seems fine for now, but here comes the confusion: the reduced Ib1 will decrease Ic1 and Vc will be the same as before. I can understand that Ib1 is stable, duo to the clearly visual collector feedback, but what about Ib2? When Vc is back to normal(as it was before the variation in Hfe) i see no reason for Ib2 not to increase again. Apparently this one is missing something and he will be glad if someone helps him.
PS: please don't send me to other threads in this forum or in other forums, i've tried and nothing helps. It will be great if you can explain it here. Thanks!

I don't think it does anything for HFE. It only compensates for Vbe with temperature.

 

This circuit depends on Q1 and Q2 being identical, as they would be in an IC, and at the same junction temperature. Q1 is forced to have a collector current equal to 1 mA (the 20V supply minus one Vbe voltage drop, divided by the 20 kΩ collector resistor), and since Q1's and Q2's bases are fed through identical resistances from the same point (Q1's collector), Q2 will have a collector current of 1 mA, also. Since Q2's collector resistor is 10 kΩ and drops 10V, that leaves Q2's collector biased at +20V - 10V = +10V. Since the transistors are matched and at the same temperature, the bias point will be stable against temperature variations.

(Paraphrased from the description given on pp. 97-98 of The Art of Electronics, 3rd Edition)

The circuit was designed and patented by Bob Widlar, the guy who designed the LM101 op amp.

 

A long, long time ago one manufacturer placed two transistors inside one case to insure proper temperature tracking. This was long before integrated circuits. Another company made clips that would hold two TO5 cases together for heat sharing. Some were side by side and another style was head to head.

 

This circuit depends on Q1 and Q2 being identical, as they would be in an IC, and at the same junction temperature. Q1 is forced to have a collector current equal to 1 mA (the 20V supply minus one Vbe voltage drop, divided by the 20 kΩ collector resistor), and since Q1's and Q2's bases are fed through identical resistances from the same point (Q1's collector), Q2 will have a collector current of 1 mA, also. Since Q2's collector resistor is 10 kΩ and drops 10V, that leaves Q2's collector biased at +20V - 10V = +10V. Since the transistors are matched and at the same temperature, the bias point will be stable against temperature variations.

(Paraphrased from the description given on pp. 97-98 of The Art of Electronics, 3rd Edition)

The circuit was designed and patented by Bob Widlar, the guy who designed the LM101 op amp.

But why, or how the bias point will be stable against temperature variations?
By the way, I also read The art of electronics, but mine is Second edition

 

But why, or how the bias point will be stable against temperature variations?

The two transistors are identical: they have exactly the same Vbe voltage drop at any given collector current and temperature, and exactly the same current gain at any given Vbe and temperature. They are both biased from the same point in the circuit (Q1's collector), through equal value resistors. Therefore, they will have the same collector current. And that current will be roughly equal to [(20V - Vbe) / 20 kΩ].

The above holds true regardless of temperature: both Vbe and current gain will change with temperature, but it will be the same in both transistors. Therefore, Q2's collector voltage will be constant over temperature.

 

But why,

This circuit depends on Q1 and Q2 being identical, as they would be in an IC,

two transistors inside one case to insure proper temperature tracking.

The two transistors are identical:

The critical point is two transistors be identical and held at the same temperature. This circuit sucks in real life because you are never going to get two transistors in separate packages to be identical and track temperature. The only way it works is to buy a dual transistor in one package.
http://www.mouser.com/Semiconductor...r-Transistors-BJT/_/N-ax1sh?P=1z0y4diZ1z0xz4p

 

The only way it works is to buy a dual transistor in one package.

Gotta keep in mind that "dual transistor" doesn't always mean dual matched transistor.

 

Gotta keep in mind that "dual transistor" doesn't always mean dual matched transistor.

Really? I never suspected a designer would make a dual that isn't matched.
Wassup? They design them the same and then sort the batch for the ones that actually match?

 

Really? I never suspected a designer would make a dual that isn't matched.
Wassup? They design them the same and then sort the batch for the ones that actually match?

Yup, really. Ordinary dual transistors might be matched well enough to do the job in this rather crude circuit, but units that are sold as a "matched pair", such as the NST45011, are usually matched closely enough that they can function as the differential input stage in precision amplifiers.

 

If you can't even trust a batch of monolythic duals to match, it really puts a point on the fact that two separate transistors in different packages don't have a tinker's chance of tracking in the circuit of post #1

 

If you can't even trust a batch of monolythic duals to match, it really puts a point on the fact that two separate transistors in different packages don't have a tinker's chance of tracking in the circuit of post #1

Agreed. I wouldn't expect them to work well at all.

 

Thanks, guys, I think I understand now. When temperature rises, since both transistors are identical, in both the beta will increase, but also the Vbe will decrease. This decrease in Vbe will increase Ib1 which will increase Ic1 and the current through the 20k resistor as a whole. This will reduce the voltage at the collector node of Q1 and this will counteract the rising of the base currents in both transistors.