On Tue, May 3, 2022 at 03:05 PM, Ozan wrote:
Good analysis and great idea to simulate the start up section. Your spice
netlist is missing Q1062 and the load at its emitter. For inverter driver to
function Q1062 should be able to support at least the current through R1066
and R1068 (in alternate phases) in addition to other components. R1069 is too
large to supply such current. With a single 2k from Q1062 emitter to (REF)
node there is not enough voltage across C1066 and at the emitter of Q1062 for
proper driver operation without CR1062-CR1065 working.
This is one of those Aha moments for me. Thanks for the correction. It makes perfect sense to me now.
With just the R1066 and R1068 as load, C1066 only charges to a bit less than 3V. More load (U1062-U1064) will lead to even lower voltage. So the initial push using C1072 to get C1066 charged is necessary.
It also removes another doubt for me. When Q1062 is not considered, C1066 would charge to about 40V at maximum. However, this exceeds its 35V rating. With Q1062 considered, that will not happen.
Can you help me with the following two questions? I have had difficulty figuring out an answer by myself.
1) The PWM control signal is taken from the REF line and 5V REF of TL494 through R1045 and R1046. However the REF line relative to PIN 7 of TL 494 (namely REF2 in later versions) is not a DC voltage. It is a switching voltage following the switching pattern of Q1050. Does C1035 serve as a filtering capacitor, together with the Th¨¦venin equivalent resistance R1045||R1046?
2) There is no DC feedback in the error amplifier TL494/U2. Just R1033 in series with C1033. The same happens with U4 - no feedback at all. Therefore the output voltage V2 and V4 should be either LOW or HIGH, but not in the middle, as Figure 3-11 seem to suggest. It can work like this in a bang-bang way (100% or 0% duty cycle). But is this how the circuit is supposed to work?
Thanks so much.
Luca
