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Old 1st June 2005, 07:10 PM   #1
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Default Yet another DC protection circuit

This should work (subject to component values being tweaked) - anyone see why it wouldn't?

It's for an amp with 35V rails, hence R6 and R10 to limit the maximum voltage to the 1st opamp. What I'm not sure of is a suitable time constant of R8 and C2 and a suitable threshold (set by R2 and R13); I'm guessing that 100mS might not be enough to fry my lovely speakers, but what voltage should be allowed?

Should I limit the amp's LF response to 10Hz to match the proposed 100mS?

Thanks in advance for any constructive advice and criticism (or indeed scathing dismissals).
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Old 1st June 2005, 08:12 PM   #2
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Link won't work for me. . .
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Old 1st June 2005, 11:31 PM   #3
ilimzn is offline ilimzn  Croatia
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Well, you seem to have a couple of problems with your circuit.
First of all, the output of the last OPamp which is used as a comparator, will be ~~positive rail as long as the + input is higher in voltage than the - input. The + input is set to roughly half the OPamp Vcc, so about 6V. Your DC detector will be switching the relay off, when it gets more than 6V DC out of the previous stage. To get 6V on the output of the previous stage, you need about 5 times that much out of the amp because of the input divider which is roughly 5:1, so that means you need 30V DC out of the amp for your circuit to detect anything. This is a MAJOR problem. More than a few V DC is capable of frying some speakers if exposed to it long enough (as in seconds), and more than some 0.5V is already high time to switch off the speaker. To make matters worse, an amp with 35V rails may never reach 30V on it's output even with a catastrophic power transistor failure - because with the full rail voltage on the load, the rail voltage will drop, and if you have source/emitter resistors in the output, you will easily have a few V drop there under maximum current.

Secondly, minor considering the above, you should not drive the transistor base with a negative voltage above a few V. Typically 8-12V will be enough to make the reverse biassed B-E junction act as a zener, which in your circuit may end up destroying the transistor. Put a reverse biassed diode (1N4148 or similar small diode) in parallel with the B-E junction of the transtor, so that it clamps the negative voltage to -0.5..-0.7V.

Thirdly, your DC detect stage must also charge C1 in order to change the voltage on the - input of the OPamp, which will take time and introduce a further delay when the speakers need to be turned off.

Fourthly, you could make this circuit with two OP-amps. You could also make a similar one that also includes power fail detect and delayed switch-on with only 3 transistors

Here is how to fix this:

First of all your comparator has it's threshold set way too high - it should have been around 0V. Just use approx 1k or so from + input to GND and keep the 1M feedback from the output of the OPamp. Also put 1M from + input to +Vcc. This will give you a threshold of about 0.2V or below on - to switch the relay on, and 0.4V or above on - to switch it off. You could even reduce this value further, by decreasing the 1k.

If all is well and there is no DC at the amp output, no signal will be coming from the previous stages. This means that under these conditions, the + input of the OPamp needs to be slightly higher than the - input to insure that the relay is energized. In order to do that, you need to make the - input lower than 0.2V, after an initial delay period. You can do that by pulling - input down with a large (100k) resistor to gnd or -Vcc of the OP-amp. The diodes from the previous stage will prevent it going any lower than some -0.7V, which guarantees that the relay will be energized if all is well on power-up.

In order to introduce an initial delay, you can add another diode with the cathode on the common point of the ones coming from the DC detect circuit. Insert R2 between R13 and C2, and make it about 1/4 of R13. The anode of the aforementioned diode to the midpoint between R13 and R2. The initial 0V at C1 as it starts to charge will put the anode of the diode at about 9.2V, way over the comparator threshold and insures that the relay is turned off on power up. You only need to put a reverse-connected diode across the R2 and R13 series so that when power is switched off, C1 discharges into the falling rail to insure proper delay when power is applied again even if it is very quickly after switching it off. The point of having R2 is to prevent the initial voltage on the anode of the diode from pulling the - input of the OP-amp all the way to the + rail, because the OPamp can sometimes latch-up under these conditions.

Finally, you need to forget the resistor divider on the input or at least set up the first OP-amp to have gain about the same or higher as the divider attenuation. The corrected circuit will work whenever the DC value on the output of the first OPamp is above about 0.9V. You can insure that the input to it never sees more than the power supply of the OPamp minus a few V by inserting a 1:3 divider (say 2k2 on top, 1k on the bottom), but it would also be prudent to amplify this signal back up so that your detector sensitivity does not drop by the same amount (detect threshold increases by a factor of 3 if input is divided by 3). I'd go with 0.5V on input to trip the protection, giving the first OPamp more gain than the input atenuator attenuation. Keep in mind that given a proper time constant, (about 100ms - 300ms) the average of the input signal will remain zero even with very low frequency content in the audio signal, but serious subsonics will trip the DC protect.

Now a word on how to make this work with only 2 OPamps.
Imagine you set the - input of the comparator connected OPamp at 0V by connecting it to GND via a resistor. Now, in order to have the comparator energize the relay initially, you need to set the + input at something higher. Lets assume 1/2 of a diode voltage drop, i.e. ~~0.3V. You can do this by appropriately dimensioning the resistor ratios from output of the comparator to the + input, and the + input to ground.
What you want to accomplish is, if your input DC goes positive, the - input must go over the threshold voltage, 0.3V. You can accomplish that by pulling the - input over 0.3V via a diode with anode on the output of your 1st OPamp, and cathode on - input of the comparator. This will trip the comparator and turn the relay off. The input OPamp needs to generate 0.3V + 1 diode drop = ~~1V at it's output to trip the comparator.
If your input DC goes negative, you again want the - input to be higher than the + input of the comparator. Instead of pulling the - input higher, you can now pull the + input lower, using a diode with anode on the + input and cathode to the output of your first OPamp. Since the - input is normally zero, you need to pull the + input 0.3V down from it's initial 0.3V treshold. In order to do this, the output of the first OP-amp needs to go 1 diode drop minus 0.3V negative, i.e. some 0.3V negative. In order to make this the same as for the positive DC detect which is 1V positive to trip the comparator, instead of using one diode for the pulling down of the + input, use two in series. Now the output of the first OPamp needs to go 2 diode drops minus 0.3V negative, which is roughly 1V negative. So now you have symetrical DC detect thresholds and one OPamp less.
Also, the initial power-on delay circuit remains the same as in the above description of the 'fixed' circuit.
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Old 2nd June 2005, 06:49 AM   #4
palesha is offline palesha  India
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"You could also make a similar one that also includes power fail detect and delayed switch-on with only 3 transistors "

How about the Schematics?
Better put it here.
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Old 2nd June 2005, 07:04 AM   #5
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And whatabout this circuit : it's delayed output switch, combinated with simply DC protection.
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Old 2nd June 2005, 07:23 AM   #6
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Something is broken, I'm sorry
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Old 2nd June 2005, 07:51 AM   #7
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Ilimzn, thanks for your very thoughtful reply.

I will fettle the comparator threshold as described so it trips at a volt (having removed the input attenuator - I'll put a pair of back-to-back zeners across C2 to protect the opamp from high volts) and adjust RC to 100-300mS.

>Put a reverse biassed diode (1N4148 or similar small diode) in parallel with the B-E junction of the transtor, so that it clamps the negative voltage to -0.5..-0.7V.

check

>your DC detect stage must also charge C1 in order to change the voltage on the - input of the OPamp, which will take time and introduce a further delay when the speakers need to be turned off.

no - C1 actually introduces a switch-on delay as it lifts the - input of the comparator til it's charged through R13. DC detect stage doesn't need to charge it.

>You could also make a similar one that also includes power fail detect and delayed switch-on

Yes, I have plans to connect it up to another cct that detects power-fail as part of the soft-start. I'll do that by feeding a TTL high to the junction of R2/R13 via a diode.

>You could also make a similar one that also includes power fail detect and delayed switch-on with only 3 transistors

That sounds cunning - can we see it?

>Now a word on how to make this work with only 2 OPamps....

I'll read this again after a coffee...thanks again!
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Old 2nd June 2005, 03:49 PM   #8
ilimzn is offline ilimzn  Croatia
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Quote:
Originally posted by bremen nacht
Quote:
your DC detect stage must also charge C1 in order to change the voltage on the - input of the OPamp, which will take time and introduce a further delay when the speakers need to be turned off.
no - C1 actually introduces a switch-on delay as it lifts the - input of the comparator til it's charged through R13. DC detect stage doesn't need to charge it.
Yes it does, when a DC fault occurs. When the diodes become forward biassed, they have to pass current into the power rail through C1, discharging it to a lower voltage.

Quote:
Quote:
You could also make a similar one that also includes power fail detect and delayed switch-on
Yes, I have plans to connect it up to another cct that detects power-fail as part of the soft-start. I'll do that by feeding a TTL high to the junction of R2/R13 via a diode.
It is actually simpler to do by powering the relay through a separate diode off of the transformer secondary. You use it to make a simple half-wave rectifier, and deliberately use a small smoothing cap so that with the mains on, the lowest voltage in the sawtooth resulting from minimal smoothing by the small cap can reliably hold the relay on (you can then also reduce the series resistor). The cap has to hold that voltage only for a fraction of a second, say 10 cycles of AC power. Other, cleverer solutions that offer a better relay trip point in case of AC brownout (temporary drop in voltage just to the point of the relay disengaging) are possible, but this, as simple as it is, is enough in 99.99% of all cases.

Quote:
Quote:
You could also make a similar one that also includes power fail detect and delayed switch-on with only 3 transistors
That sounds cunning - can we see it?
It is even possible to make it with only 2 transistors but the thresholds are quite high, see attached image - it is from a Kenwood amp from the 70s. The 3 transistor version is better as it has better defined thresholds.
Performance is not as good as in these simple circuits as for the OPamp version - the thresholds are not so easily customized, and there is no hysteresis. But, it will work good enough to protect the speaker.

A little explanation of the points in the circuit:
A is the input from the amp output.
B and B' are the relay contact, to be put in series with the speaker connection.
C is the ground line.
D is the power fail detect supply, derived from a deliberately 'bad' half-wave rectifier, like the one I described above.
E is the +V power rail for the amp, 35-40V.

The design will also be improved if a diode is placed between Qe27 collector and point D (cathode to D).
This will discharge Ce28, the power on delay cap, into the relay in the event of a power fail.
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