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PSU theory question..

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patroclus

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In this schematic, there's a psu which outputs are 5,6V, 6,6V and 13.1V(variable). It uses simple LM voltage regulators, but.. why at the output of LM7805 are 5,6V and not 5V ??? Also in LM7806.

http://home.quicknet.com.au/andrewm/eprom1/psul.gif

The author says that the circuit must be feed by 12 V AC.. wouldn't it be the same if you supply, let's say, 15 V DC??

Thanks for the help.
 

Nigel Goodwin

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Most Helpful Member
patroclus said:
In this schematic, there's a psu which outputs are 5,6V, 6,6V and 13.1V(variable). It uses simple LM voltage regulators, but.. why at the output of LM7805 are 5,6V and not 5V ??? Also in LM7806.
Because the ground connection of the7805 doesn't go to ground, it goes to the top of D2 - adding an extra 0.6V to the output of the regulator, as it does to the 6V regulator as well. The 5.6V supply is then fed through a rectifier to return it to 5V, presumably it's to block a reverse feed from a higher voltage?.

The author says that the circuit must be feed by 12 V AC.. wouldn't it be the same if you supply, let's say, 15 V DC??
15V is really too low, you want 18V DC or so to give the LM317 decent headroom.
 

patroclus

New Member
So, 15V DC is too low but 12V AC isn't??

This circuit is used to feed this :

http://home.quicknet.com.au/andrewm/eprom1/schel.gif
(lower left corner)

There's some transistor theory I also can't understand... why when you put the base of Q1 to 5V and the base of Q3 to 0V, you get 12.5V +/- between Q2 and Q3 collectors??

I really apreciate some help cause I spent some hours trying to undertand this..
 

Nigel Goodwin

Super Moderator
Most Helpful Member
patroclus said:
So, 15V DC is too low but 12V AC isn't??
Yes, 12V AC will give about 18V DC when it's rectified and smoothed.

There's some transistor theory I also can't understand... why when you put the base of Q1 to 5V and the base of Q3 to 0V, you get 12.5V +/- between Q2 and Q3 collectors??
Well looking at that circuit, there looks to be an error in the drawing around Q1, R3 shouldn't connect to the emitter - it should just carry on straight past it.

Q2 and Q3 are just used as switches, they both require their bases taking low to turn them on. Q3 can simply be switched by a logic gate, as it only feeds off 5.6V - Q2 however feeds off 13.1V, and a logic gate wouldn't be able to turn the transistor off - it needs an open-collector drive. Q1 is simply used to provide an open-collector drive, and also inverts the signal (presumably this is taken care of in the software?).

Q4, Q5 and Q6 work in exactly the same way, and R7 again shouldn't be connected to the emitter of Q4.
 

patroclus

New Member
Thank you very much :)
I alomost got all my questions answered
Anyway, if you want to get 12.5 V +/- between both transistors, you should have to enabled Q2 and disable Q3, right??
But that's not the way it work, cause the software puts the base of Q1 to 5V and the base of Q3 to 0V, so, If I'm not wrong, bot transistors are working...
This is the main problem I have. I can't understand why...
 

Russlk

New Member
When Q1 is turned on, Q2 turns on and pulls Q3 base up to 5 volts, so Q3 is turned off regardless of what the input signal is. There is nothing to limit the current from Q2 to the 5.6 volts source, so Q2 is in danger of burning out.
 

patroclus

New Member
I don't think there's danger, because this circuits is tested.
anyway, what you say is that Q2 is on, so in its collector there's about 12.5 V, and that does not allow Q3 to turn on?? is that it??
but, there's no path from Q3 to 5,6V is there? isn't it supose to flow current from 5,6V source to emiter and then to Q3 colector??
 

Nigel Goodwin

Super Moderator
Most Helpful Member
If Q2 is on, as you say it's have 12.5V on it's collector, as will the collector of Q3. Current can't flow from the 5.6V rail to the collectors because the collector is already at 12.5V - current only flows from a high voltage to a low voltage.

I wouldn't expect both transistors to ever be turned on together, all they are doing is switching either voltage to the EPROM.
 
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