audio3000 said:
I think it will work. The key is that at start-up C4 is uncharged and at zero volts.
Then, as C4 charges (time constant C4*R10+R11) the junction of R10/11 essentially floats up to Vsupply and the protection will be off, unless there is trouble at any of the other sense points. R11 is there to limit the current pulled from C4 in case of activation to protect IC1A's output stage and should not be too big as it also delays the activation. A few 100 ohms should do it. Then chose C4 and R10 for the required switch-on delay. I would chose R12/13 for half Vsupply.
Jan Didden
On the surface, it's supposed to work that way. C4 got charged. When the + and - inputs of the comparator is the same, it flips and turns on the relay. Isn't it? But if you look deeper, the circuit component values used does not allow this happen. Let 's draw that portion of the the circuit to discuss.
+---------------+ +15V
| |
R10=470K R12=15K
| |
+---[+] +------[-] = 15*(1.5)/(15+1.5) = 1.36V
| |
R11=120K R13=1.5K
| | At power up, volt across C4 = 0V
C4 | ==>[+] = 15*(120)/(120+470) = 3.05V
| | See that at power up, [+] is already
gnd gnd greater than [-], the relay will be on
without delay. As time goes by, volt
across C4 will just getting higer.
Kam: the above diagram changed when shown on the forum. All space between the vertical stroke disappearred. I don't know how to make it easier to read at this time...:-( It's a simple voltage bridge.
I would suggest that [-] should be set at about half of the V+ rail. This way, you can make better use of the cap by getting longer delay with the same size of cap.
There's another problem with this design. If you momentarily turn off the power supply of the amp, C4 will still be fully charged, so there will be no delay right after the power is on again. To fix this problem, we normally put a diode between the +side of C4(anode leg) and +15V(cathode leg). This way, C4 will be discharged faster to get ready to serve the next delay.
So some redesign work is needed to make it work as claimed.
Hope this helps,
Kam
+---------------+ +15V
| |
R10=470K R12=15K
| |
+---[+] +------[-] = 15*(1.5)/(15+1.5) = 1.36V
| |
R11=120K R13=1.5K
| | At power up, volt across C4 = 0V
C4 | ==>[+] = 15*(120)/(120+470) = 3.05V
| | See that at power up, [+] is already
gnd gnd greater than [-], the relay will be on
without delay. As time goes by, volt
across C4 will just getting higer.
Kam: the above diagram changed when shown on the forum. All space between the vertical stroke disappearred. I don't know how to make it easier to read at this time...:-( It's a simple voltage bridge.
I would suggest that [-] should be set at about half of the V+ rail. This way, you can make better use of the cap by getting longer delay with the same size of cap.
There's another problem with this design. If you momentarily turn off the power supply of the amp, C4 will still be fully charged, so there will be no delay right after the power is on again. To fix this problem, we normally put a diode between the +side of C4(anode leg) and +15V(cathode leg). This way, C4 will be discharged faster to get ready to serve the next delay.
So some redesign work is needed to make it work as claimed.
Hope this helps,
Kam
audio3000 said:
Yes, as I said, the midpoint of R12/13 I would set at mid supply. If I understand you correctly, you say that the cap charges to 3.05V? That is not the case, C4 will charge all the way up to Vsupply. After the charge time there is no longer any current through R10/11 and therefore no voltage drop across them.
The diode to discharge C4 on power down is a good idea.
Anyway, I am sure this circuit works, but the resistor values may need to be selected. I didn't see the values (the circuit in post # 1 had no values IIRC.
Jan Didden
No, I said that at the very first moment of power on, voltage across C4 is zero. However, the voltage at the junction of the charging resistors is already at 3.05V. Therefore, the relay will be turned on right away given there's no other problem detected.
Yes, I believe the circuit topology works fine just some component values adjustment is needed to make it works the way it was described.
Thanks
I've finish the AV400 and the ETI-466 modules. I'm trying to put them into a decent home, that's why I have to get the protection circuits. I'll be going after the super-leach, PA300 and Hafler200 if I can find all the parts. Some of them are either hard to get or over charged for shipment (cause I my buying small quantities at very high shipping rates) I'll have to make a longer purchase list before I can justify to actually getting them.
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.