# Low-Voltage ECL : part 2, Current sinks : here we go again !

A project log for Germanium ECL

How fast can germanium transistors compute ? And how much current will that draw ?

Yann Guidon / YGDES 05/25/2020 at 04:583 Comments

For the newcomers, have a look at the log 2. moving forward of a related project. The comments are great too :-)

So I want to make ECL gates. I want to make them fast and if possible reduce the amplitude and frequency of the PSU's "digital noise". Oh and I would love to have an even lower supply voltage, because about half of the power is burned by the common resistor, that wiggles between 0.8 and 1.2V. If I could win maybe 0.5V there, the PSU would have to supply only 1.5 or 1.6V and save 20 to 25% power. Furthermore I might want to "stack" layers of ECL circuits later...

According to @Julian, reducing the current below 2mA starves the circuit and the speed is reduced, so I'll stick to 2mA for now and try to get the equivalent of the 470 ohms I now have.

What that would bring me :

• The ability to change the settings and working points of the whole circuit with current mirrors spread around the PCB and trim the values with one pot,
• reducing the operating voltage (and power), or increasing the current at will to see the effect on the speed
• reduce power noise and decoupling

There is not much margin though : 0.8V is already low, and a degeneration resistor will already eat 0.1 or 0.2V and the transistor's Vcesat can impose limits. The complexity of the system would be overkill while a simple resistor already works rather well.

But there is something else : the lower the resistor, the lower the swing at the common node, and the lower the sensitivity of the inputs... Maybe a transistor could help there too ?

Looking at https://wiki.analog.com/university/courses/electronics/text/chapter-11, I find this interesting circuit :

but this is a current mirror : this means that to get 2mA in the output (which can be replicated/multiple) there must be another place where 2mA is sunk. Which is not great for power savings. Things start to get complicated from there, if I want to reduce it to 200µA for example...

R2 would be about 50 ohms and drop 0.1V at 2mA. Energy saving means I should get the lowest Vce, so it must be (almost) saturated. Which means : pumping significant current in the base. Which means that the R2 drop would also increase because of that base current and might reduce its effectiveness. The transistor's gain should thus be high to reduce base current and Vce, but all those counter-effects conspire to make the transistor current sink overly complicated.

After all : looking at Motorola's MECL Data Book, no transistor is used in the major old families:

• MECL I : 1.24K
• MECL II : 1.18K
• MECL III : 365
• MECL 10K : 779

The high-side resistors however are lower than my values : 100 (MECL III) to 300 (MECL I & II). I could investigate reducing my values but this comes at the prince of increasing the current...

Do I really need a sink transistor ?

## Discussions

Tim wrote 05/25/2020 at 08:41 point

If you have a reference voltage somewhere (by using a bandgap IC like TL431 maybe? :) ) you can easily transform that into a current using a precision low TC resistor.

Keep in mind that a lot of compexities of on-chip references arise because you actually have to build your voltage reference first and all resistor options have horrible tolerances (+- 20%) and temperature coefficients. In discrete implementations one could take some short cuts.

One thing that should be more critical in discretes though is matching of transistors both in hfe and temperature.

Are you sure? yes | no

Yann Guidon / YGDES wrote 05/25/2020 at 08:47 point

I was considering building a discrete bandgap reference, since at one time I needed 1.2V.

The resistor might be the right solution for now... or I'll lose disproportionate amounts of time and energy on that current sink thing that would not bring any benefit for now.

Are you sure? yes | no

Yann Guidon / YGDES wrote 12/01/2020 at 18:00 point

I see that you have gone full LED with #LED Coupled Logic (LCL) :-D

Are you sure? yes | no