# How a constant current of 23 mA flows from the capacitor when the transisitor turned on?

simulate this circuit – Schematic created using CircuitLab

The following is a circuit I got from a research paper and I am trying to analyze it. The inverter is a high speed inverting buffer whose output is always one. The circuit starts functioning when the inverter sends an output of zero. I will quote the sentences in the paper about the idle conditions (When inverter output is one)

"The output of inverter biases Q1 Such that its emitter Voltage is higher than the base voltage of Q3, thus cutting off the collector current of Q3. Capacitor C6 will have been charged to 5 volts through R17 with a time constant of about 5 us"

When the inverter ouput is zero,

"The inverter generates a fast falling edge and quickly turns off Q1 thus turning on Q2.The collector of Q2 draws a constant 23 ma from C1, thus discharging it at the uniform rate of 0.1 volts per nanoSecond"

My doubts :

1) How turning off Q1 can turn on Q2?

2) When Q2 is turned on, how the collector become constant taking 23 mA from C1?

• 1 & 2/ The circuit is a so called "long tailed pair" I suggest you read up on that. Also the discharge statement is only partly true because at some point the discharge will stop. Dec 9 '19 at 5:47
• @Oldfart But how can I calculate it to be 23 mA? Even if it is 23 mA, how can the capacitor current be constant? I can't see any circumstance I the circuit where linear discharging of capacitor is possible.. !! Dec 9 '19 at 5:56
• why is V3 labeled as -5 V? ... is the polarity correctly oriented? Dec 9 '19 at 6:05