| john-china |
i want to build 200w aleph-x. i collect a lot of irfp250 and irf9610.
i worry about too high input capacitance of irfp250 which will affect high frequency. so i want to increase the current of 9610 to 50ma ( how much is enough?). but i dont know whether or not
the modification of original sch is correct. my aleph-x will parallel
6 banks of irfp250 for each side.
if i increase the current , the volt of 392OHM resistor will not be about 4v , so i must resolve this problem , but i dont like capacitor-decouple. then i must keep the joint of 392OHM and irfp250 stay about -26v because rail to rail of my aleph-x is +30v to -30v. so negative supply of 9610 will be -26v-50ma*392OHM=
-45v. then i must separate power supply of 9610 from irfp250.
finally my aleph-x will use +-45v for front-end(irf9610) and +-30v for back-end(irfp250). can i do it so? will this aleph-x work?
i dont know if above modification is correct. pls tell me how to do.
maybe some simple way i dont know. give me more suggestion. thanks |
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| Nelson Pass |
You can crank up the front bias all you want in the front end.
Just adjust the resistors on the bias and off the Drains of the
diff pairs. |
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| john-china |
this is really a good way to resolve my question.
so i will adjust 392(OHM) resistor to 80(OHM) for 50ma current of front-end. but i have another question. is 80(OHM) too low ?,
will this affect gain of diff pair of irf9610? |
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| Nelson Pass |
| Yes, it will lower it. |
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