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loudspeakar protection from DC w/ softstart

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andrew24

Joined Aug 20, 2008
76
Hallo, i have some trouble with this circuit,hope you can help me.


First part of the circuit should provide a delay of a few seconds before connecting the relay contacts (speaker to amp. line). It basically consists of 47k resistor anr 470u capacitor. When i power on the circuit, nothing happens. I measured all the voltages in reference to ground, i think that should help solving the problem.

The transistor seems to be cut off, and there is only 0,25V on the 470uF capacitor. There should be approx. 12V when it's fully charged. It seems like the 2SD1266 transistor prevents capacitor from charging up.. When i connect the RC circuit alone to the GND and +12V and measure voltage drop on the capacitor it starts from 0 and gradualy reaches 12V, but when connected ir to the base of a transistor, capacitor's voltage ends up with 0,25V..
Thanks for reading this, Andrew
 
beenthere

beenthere

Joined Apr 20, 2004
15,819
If the 2SD1266 were short, the relay would pull in immediately. The problem is more likely to be with one of the output protection transistor above - the 2BD547 or the 2BC557. I wonder where the 9.94 volts on the cathode of the 1N4004 comes from?

I would remove and check those transistors and the diodes.
 
Ron H

Ron H

Joined Apr 14, 2005
7,063
It looks to me like the BC547 is saturating, clamping the base of the 2SD1266 to Vce(sat), which is why you are measuring 0.25V.
Also, 48k is much to large to allow the 2SD1266 to saturate and turn on the relay, even if the base were not clamped.
Where did you get this design?
 
eblc1388

eblc1388

Joined Nov 28, 2008
1,542
@andrew24,

I would say the circuit is working correctly, because you have 5.14V at the amplifier output. This is what the purpose of the protection circuit is for.

Why would you think the circuit should energize the relay to connect the speakers if there are voltages at the amplifier's output?

Test the circuit again by:

1. disconnects the ends of the two 22K from amplifier's output and connects them to 0V. The relay should then energize after a delay.

2. disconnects one end of one 22K from 0V, the relay should stay energized.

3. use a 1.5V battery and connect its terminals to end of the 22K in step(2) and 0V. The relay should release immediately and disconnect the speaker.

4. confirm step(3) occurs using opposite polarity of the 1.5V battery in step(3).
 
A

Thread Starter

andrew24

Joined Aug 20, 2008
76
eblc1388 said:
@andrew24,

I would say the circuit is working correctly, because you have 5.14V at the amplifier output. This is what the purpose of the protection circuit is for.

Why would you think the circuit should energize the relay to connect the speakers if there are voltages at the amplifier's output?

Test the circuit again by:

1. disconnects the ends of the two 22K from amplifier's output and connects them to 0V. The relay should then energize after a delay.

2. disconnects one end of one 22K from 0V, the relay should stay energized.

3. use a 1.5V battery and connect its terminals to end of the 22K in step(2) and 0V. The relay should release immediately and disconnect the speaker.

4. confirm step(3) occurs using opposite polarity of the 1.5V battery in step(3).
Click to expand...
Thanks for a reply. I tested the protection circuit not connected it to power amplifier, so i have those 5.14V from protection circuit itself!
 
eblc1388

eblc1388

Joined Nov 28, 2008
1,542
andrew24 said:
Thanks for a reply. I tested the protection circuit not connected it to power amplifier, so i have those 5.14V from protection circuit itself!
Click to expand...
If that is the case, then check your circuit for bad connections and measure the voltages again.

At the moment 90% of those readings are not making sense. :confused:
 
A

Thread Starter

andrew24

Joined Aug 20, 2008
76
I assembled the soft start circuit on the breadbord, changed 1K resistor to 90 ohms. Now the base current is sufficient to turn 2sd1266 On. The delay is also provided.
This relay on the breadboard has 276 ohms coil resistance. On my soldered circuitt i used a bit diferent relay, i think it has 170 ohms coil resistance. would this affect the biasing of the transistor? do i need to change something?

 
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A

Thread Starter

andrew24

Joined Aug 20, 2008
76
Ron H said:
It looks to me like the BC547 is saturating, clamping the base of the 2SD1266 to Vce(sat), which is why you are measuring 0.25V.
Also, 48k is much to large to allow the 2SD1266 to saturate and turn on the relay, even if the base were not clamped.
Where did you get this design?
Click to expand...
can you tell me what type is the connection of those two transistors BC547 and 557? looks like diff amp, but stange one ..
 
Ron H

Ron H

Joined Apr 14, 2005
7,063
andrew24 said:
can you tell me what type is the connection of those two transistors BC547 and 557? looks like diff amp, but stange one ..
Click to expand...
It's not a diff amp. Consider the voltage at the junction of the two 22k resistors. If the DC level at that node (audio is filtered out by the series 47uF caps) exceeds approximately ±1.5V, one or the other of those two transistors will be on, diverting base current from the base of the 2SD1266, preventing it from turning the relay on.
 
A

Thread Starter

andrew24

Joined Aug 20, 2008
76
Ron H said:
It's not a diff amp. Consider the voltage at the junction of the two 22k resistors. If the DC level at that node (audio is filtered out by the series 47uF caps) exceeds approximately ±1.5V, one or the other of those two transistors will be on, diverting base current from the base of the 2SD1266, preventing it from turning the relay on.
Click to expand...
So if the voltage is positive 1,5V, the NPN transistor BC547 turns on?
Could you explain, how it turns 2SD1266's base to 0V ? thanks :)
 
Ron H

Ron H

Joined Apr 14, 2005
7,063
andrew24 said:
So if the voltage is positive 1,5V, the NPN transistor BC547 turns on?
Could you explain, how it turns 2SD1266's base to 0V ? thanks :)
Click to expand...
When an NPN turns on, it's collector-emitter voltage ,Vce(sat), is typically less than 100mV at low current, such as this application, where the collector current will be about 250uA.
 
A

Thread Starter

andrew24

Joined Aug 20, 2008
76
i got it. Now one thing left unclear: why i get about 0,6V on the capacitor's + (the one in the foto i made it's the delay RC circuit)? i thought when the capacitor is finished charging, it should have 12V dropped..
 
Ron H

Ron H

Joined Apr 14, 2005
7,063
andrew24 said:
i got it. Now one thing left unclear: why i get about 0,6V on the capacitor's + (the one in the foto i made it's the delay RC circuit)? i thought when the capacitor is finished charging, it should have 12V dropped..
Click to expand...
Because current is flowing through the 47k and the 1k (or whatever you changed it to), into either the base of the 2SD1266, or into one of the clamp transistors, if one of them is turned on due to excessive DC on the amplifier outputs.
 
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Thread Starter

andrew24

Joined Aug 20, 2008
76
ok..now i have another problem.. most of relays have about 200ohms resistance.
because of low resistance on the collector path, there is low voltage drop on the relay, so it doesn't energize.
the only way to energize a relay is to saturate a 2SD1266 transistor by extremely reducing its base and RC delay circuit resistance. Everythink is ok, but.. there is no delay, because RC constant has changed!

Can someone suggest, how to provide a delay, and at the same time make a transistor saturated? maybe use a FET instead of BJT?
 
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Thread Starter

andrew24

Joined Aug 20, 2008
76
I used field effect transistor BUZ21 instead of 2SD1266.Now the delay is privided, relay activating.

only one problem left .. :D when i attach DC detector's transistors(557,547) to FET (with the orange wire), BUZ21 gate gets 0V and becomes cut off.
I dont realize from where, but BC557 transistor's base(on the left, where the diode's anode is connected) gets 0,6V from somewhere.. it makes it saturated. PLease note that i didn't introduce any DC in the circuit intentionaly,nor i connected power amp to the circuit!

where could it go wrong? maybe the circuit needs to be completely redisigned?

 
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R!f@@

R!f@@

Joined Apr 2, 2009
10,004
Your circuit is quite simple. When it comes to Speaker protection u'll need faster detection from both supply rails.
I have one that I built, one is with a IC and other is with using transistors and is quite reliable and fast acting.
Which one do you like?

Rifaa
 
A

Thread Starter

andrew24

Joined Aug 20, 2008
76
R!f@@ said:
Your circuit is quite simple. When it comes to Speaker protection u'll need faster detection from both supply rails.
I have one that I built, one is with a IC and other is with using transistors and is quite reliable and fast acting.
Which one do you like?

Rifaa
Click to expand...
this simple circuit is rather for educational purposes, to learn how to control relays with transistors and etc. For serious amplifier i'll use better project.
I'd like to take a look at the one using transistors :)
 
Ron H

Ron H

Joined Apr 14, 2005
7,063
As you pointed out, the 47k is too large to saturate the 2SD1266 (I told you that in my first post). If you change to a MOSFET, the PNP, being an emitter follower, will never turn off completely. Even with the inputs floating, there is a sneak current path through the PNP emitter, the two 100k's and 1N4004's, and the base on the NPN, to ground. This leaves the PNP emitter at about 1.5V, which is too low to turn on the MOSFET. If you connect the inputs (amp outputs) to ground, it gets worse.
Below is a circuit which works in simulation. You might want to try it. It switches the speakers off when the amp outputs get above about 0.6V, or below about -1.3V. Not great, but as good or better than the original circuit would have done if it had worked (like if the 2SD1266 had very high beta).
 

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Thread Starter

andrew24

Joined Aug 20, 2008
76
Ron H said:
As you pointed out, the 47k is too large to saturate the 2SD1266 (I told you that in my first post). If you change to a MOSFET, the PNP, being an emitter follower, will never turn off completely. Even with the inputs floating, there is a sneak current path through the PNP emitter, the two 100k's and 1N4004's, and the base on the NPN, to ground. This leaves the PNP emitter at about 1.5V, which is too low to turn on the MOSFET. If you connect the inputs (amp outputs) to ground, it gets worse.
Below is a circuit which works in simulation. You might want to try it. It switches the speakers off when the amp outputs get above about 0.6V, or below about -1.3V. Not great, but as good or better than the original circuit would have done if it had worked (like if the 2SD1266 had very high beta).
Click to expand...
thanks a lot for the circuit you designed for me! I see one major difference that you used both NPN transistors. Is that a darlington pair ?
 
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