# Hand Calculations of Emitter Coupled Logic circuits

I am having trouble approaching doing hand calculations for the MECL shown below. I know since Qa, Qb, Q1 all have common emitter voltages so they cannot all operate in the same mode if there base voltages are different. If VA = -0.6 V and VB = -1.5 V, then at least two of the three BJTs must be operating in cutoff mode. My question is how do you determine which ones are in cutoff mode and which one is on or to approach any type of problem like this Thank You Image Taken from EE 307 at Cal Poly

My question is how do you determine which ones are in cutoff mode and which one is on

When you are first learning, use the method you were taught for analyzing any non-linear circuit: Guess and check.

First guess the state of each nonlinear element. Then analyze the circuit with those assumptions. Then check that there's no logical contradiction (for example if you guessed that one of the BJT's is in saturation, then make sure you didn't calculate its $$\V_{ce}\$$ to be greater than 0.2 V) in the result.

Once you've done a few problems like that you'll quickly see patterns and have a good chance to "guess" the correct states on the first try.

Like in this case, since $$\V_A > V_B\$$, and I know that ECL logic gets its high speed from not driving BJTs into saturation, I would guess that QA is forward active and QB is cut-off.

• Could you explain your logic why Va is in Forward active. You mentioned that since Va is at a higher potential with respect to Vb, then Va must be on. That is the correct answer but I'm not sure why guessing that Vb is on isn't a valid guess too. – Aaron Mar 11 at 15:53
• Think of the b-e junctions of QA and QB as two diodes in a diode OR circuit (like this). Then use the same logic you'd use to solve the diode circuit. – The Photon Mar 11 at 15:56
• ok then I'm guessing since the voltage across Va base to Vee is larger than the voltage across Vb. Then Va must be on, which implies that Vb is off. Is it a rule of thumb that BJT that has the most voltage across (assuming symmetry except the base voltage lik in this case) will be the BJT that is operating? – Aaron Mar 11 at 16:15
• Since the two emitters are connected together, you don't have to worry about VEE. You just know that if you pull one base higher than the other (by more than a few mV) it will shut the other b-e junction off. (Again, like the diode logic) – The Photon Mar 11 at 16:17
• Ok that makes sense. Thank you – Aaron Mar 11 at 16:19

1. Block diagram.

2. Operation. Here is how I have visualized ECL operation in three steps:

Input logical '0'.

Transition 0 -> 1.

Input logical '1'.