How do I "Un-AC Supply" this circuit ?

[ceolwynn]

hi, I need this circuits AC Supply related parts
removed, and replaced with DC Supply parts ?



I already fried one trying it, ooops ....

I have ordered 2 more that will be here friday or
monday. since it is a MCU Circuit I know it's
heart is Using DC primarily.

regards !!! Tom

 

[ke5frf]

What sort of load is the TRIAC driving?

I'm not familiar with TRIAC circuits that don't have an inductive (AC) load.
If the load requires AC, then you don't have any choice here, I don't think. TRIACS are dependant on the rise and fall of AC current for their switching characteristics.

 

[MikeMl]

...
If the load requires AC, then you don't have any choice here, I don't think. TRIACS are dependant on the rise and fall of AC current for their switching characteristics.

TRIACs turn off during the AC line zero crossing (polarity reversal). If connected to a DC circuit, they will trigger once, and stay triggered until the DC power is disconnected.

 

[ke5frf]

TRIACs turn off during the AC line zero crossing (polarity reversal). If connected to a DC circuit, they will trigger once, and stay triggered until the DC power is disconnected.

But as for his intended load? The schematic implies that the load is inductive, so this wouldn't apply.

I think I've read what you described, but I am only familiar with TRIACS firing heater coils and AC motors in personal experience.

 

[ceolwynn]

I guess I was too specific for this situation ...

I am not going to use the triac.

on The MPU ...
Pin 3 Emitting through R7 is the end of the
part I am going to use. beyond R7 will not
exist in my application.

The triac in this pic will be relaced with a
"whatever" small transistor, pn2222 or ?

I suspect a MosFet or J-Fet will work better than
a standard BiPolar.

I will be using the 0 -to- 100% on / off curve
from pin 3's output, to gradually turn on and off
a Mosfet / Jfet. I need only a few milliamps.
and the output of pin 3 should create voltage
that spans from .5v -to- 10 v DC . that is
the ultimate goal. and...

I do understand that without changing the MPU
pin 3 output. I may need to use an OpAmp circuit
to translate the voltage range comming out if pin 3.

for example, pin 3 may be outputing say, 3 -to-
8 volts ? or 5 -to- 15 volts ? I will not know till
the other 2 kits get here soon. before tinkering
on those ones , I wll map out what the meter
says on all the junctions during it's operation first,
THEN ! I'll go Dr. Frankenstein on it ! ( evil laugh ! )

as far as I can tell, the MCU does not know it's
output is going into a Triac. it should work as
well for a Mosfet or Jfet transistor I think...

but - How do I power this with a DC supply ?

what parts are specifically the AC power circuit ?

at some point the parts of this circuit related
to it's output, are powered by DC Power I am
sure. I do not think there are any AC powered
Micro Processors ? are there ?

The original circuit designer does not have any
time for me, or anyone possibly ... but if we
get this to work we will have an awesome
programmable Soft start / soft stop affordable
circuit !!!

regards !!! // Tom \\

 

[ceolwynn]

that triac in the circuit really stole the thunder away from my question huh *grin*

 

[ke5frf]

I am more of a repair guy than a designer, but I do love analyzing circuits to figure them out.

It seems to me that the D3 and zener combination along with the parallel capacitors c2,c3 are your "power supply" and r8,9,10 are dropping the voltage to a usable level. Not sure how half wave rectification is tolerable here though.

Sice D1 is forward biased off of the LINE bus it looks like it performs dual purpose of biasing t1 and rectifying. It would seem t1 must require a pulsed base voltage, because there isnt a regulator.

actually, it almost looks like d1 and d3 form a full wave rectifier but d1 has the biasing function too.

I'm not sure about all this though, but maybe my thoughts will spur some better input.

 

[ke5frf]

at any rate, i would investigate what voltages occur here and get a datasheet for the IC to determine its function.

certainly this isnt driven by ac, im certain those diodes are rectifying and the zener is regulating....5V?

 

[ParkingLotLust]

Assuming it runs on 5v, I would remove the three resistors, two diodes, and one HV cap from the left-hand side, and apply +v to the top rail and GND to the bottom in order to run it on DC. Im assuming 5v because of the 5v6 zener.

 

[mikeReynolds]

Your AC supply is basically comprised of R8, R9, C5 R10, ZD1, D3, C3, C2, and R1. Eliminating those and supplying +5 volts DC at Pin 1 and GND at Pin 8 of IC1 will power the PIC - but your circuit will not work for a couple of reasons.

1) The PIC is programmed to drive a triac and uses zero crossing detection. The thing about a triac is that once it is turned on it will continue to conduct until the current flowing through it drops below a certain point, it then cuts off. The way the circuit works is like this. The PIC is programmed to detect when the AC voltage crosses the zero volts point and to detect which combination of SW1 switches are closed. The position of the switches most likely sets TIMER0 in the PIC with a value that determines how long the load should be on. Once the timer expires an interrupt is generated and in the interrupt service routine the gate of the triac is turned on briefly to start the triac conducting and then it is turned off (usually 1 millisecond or so). This is done to conserve power, no point in keeping the gate high once the triac is conducting. The zero cross detect is to let the PIC know when to turn on the triac, you want to turn it on as close to zero cross as possible to reduce switching noise and it also takes less current to turn on the closer it is to the zero point. The value stored in TIMER0 determines how often the triac gets re-triggered since it is only triggered in the interrupt service routine and you can only get into the ISR by the timer expiring. So the signal coming out of Pin 3 of the PIC will only be there for maybe 1 millionth of a second every second and will not be suitable for driving a mosfet or jfet.

2) Additionally the highest voltage you will get out of a PIC is 5 volts DC - with the exception of the HV series PICs that are designed to operate at higher voltages so you are not going to get 5 to 15 volts out of it.

Sorry for the long post but I would rather give you the reason why it will not work rather than just tell you it will not work.