Surely it will: When VDD is switched on (even if it rises slowly), C initially provides the necessary startUp current. The limiting R isn't necessary, IMHO, because the pmos limits the current anyway.
Surely it will: When VDD is switched on (even if it rises slowly), C initially provides the necessary startUp current. The limiting R isn't necessary, IMHO, because the pmos limits the current anyway.
Surely it will: When VDD is switched on (even if it rises slowly), C initially provides the necessary startUp current. The limiting R isn't necessary, IMHO, because the pmos limits the current anyway.
Oops, you are totally right. If the node is zero, the P type transistors are on and if the node is higher than zero, the cap can provide the current. I have spoken too fast.
Right: If you want to switch on the circuit several seconds after switching off, you should spend a (very) hi-ohmic resistor in parallel to the cap to discharge it.
I have used junction capacitors as startup circuits in bipolar PTAT & bandgap designs and they work fine provided you ensure there is a discharge path for the capacitor. I cannot remember the exact mechanism in the designs I did but there was a distinct, predictable discharge path when the power was removed (possibly through one of the wells being forward biased when the power collapsed). The discharge could also be seen in simulations.
I think the value depends on the possible leakage you need to overcome and the rate of rise of the power supply. Normally when designing such circuits I deliberately add in an excessive amount of leakage & check the circuit doesn't work. Then add the startup circuit with the slowest power supply rise you expect and make sure it then works.
I think the value depends on the possible leakage you need to overcome and the rate of rise of the power supply. Normally when designing such circuits I deliberately add in an excessive amount of leakage & check the circuit doesn't work. Then add the startup circuit with the slowest power supply rise you expect and make sure it then works.
Because that is when leakage will be highest and so when the circuit is most likely to fail. However, do also check at low temperatures - on theory is that the higher leakage of the larger transistor will be sufficient to ensure starting, but that is not a method I have relied on, even though simulations may show it to be the case. It is a method which caused some controversial discussion, I think.