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I'm assembling a multi channel audio amp using prebuilt channel boards and a donor amp case. The donor amp case has a small board mated to a glass panel on the front with one LED light per channel. The donor amp is basically providing some mains power on/off functionality which feeds mains to the first audio amp board which has power supplies and audio circuit. The main amp board then feeds +80VDC, +12VDC, -12VDC to two secondary amps which only have audio and no onboard power suppply.

I'd like to have an indication for each channel as to whether that channels amp is on. The amp boards do not have a pin which indicates "on", but they do have a "standby/protect" pin which is pulled up high when the amp is running normally. That pin will be pulled low internally when the amp protection circuit is running or can be pulled low externally if you want to put the amp in low power standby mode. In my case, the donor case contains power switching for standby mode so that function is handled.

What I'd like to do is figure out how to use the 12V standby/protect pin to drive the front LEDs. The voltage on the pin is very sensitive and I tried a test circuit like in the pic (green box) and figured out that getting the protect pin under about about 11V creates some weird behavior in the amp (drops out, pops on the output). running a large enough divider to keep the protect pin at the right voltage drops the current through the transistor too low for effective saturation.

One other note is that its not possible to pull 12V from the amp boards individually and the donor doesn't have a suitable DC supply so was planning on using a separate supply (I had 5V on hand which is why its shown).

So advice on what I can do to maintain the highest voltage I can on the protect pin while still operating the LEDs? Note I'm knowledgeable enough to be dangerous, but not very experienced with electronics outside basic V=IR type applications.

sample circuit

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  • \$\begingroup\$ Welcome, good description of the problem \$\endgroup\$
    – Jens
    Commented Jul 4, 2022 at 15:37
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    \$\begingroup\$ An emitter follower setup? Are you not able to disconnect the led ground instead? \$\endgroup\$
    – Passerby
    Commented Jul 4, 2022 at 15:49
  • \$\begingroup\$ @Passerby Well if I was clear what an emitter follower setup was I could probably give better response, but are you asking whether I can isolate the two grounds for the 12V and 5V? I could do so phyiscally, but still need a way to reference the 12V side at some point. \$\endgroup\$
    – Mike-d-s
    Commented Jul 4, 2022 at 22:31
  • \$\begingroup\$ Emitter follower is the way you connected a NPN transistor at the top, between the power and the load. It's an uncommon way of doing what you want. And no I mean can you disconnect the leds cathode from the 12v ground in the blue box. \$\endgroup\$
    – Passerby
    Commented Jul 4, 2022 at 22:41
  • \$\begingroup\$ Ok, so reading that what I hear is that its uncommon to put the load downstream of the transistor and it would be more typical to run the transistor behind the LED instead. \$\endgroup\$
    – Mike-d-s
    Commented Jul 4, 2022 at 22:48

2 Answers 2

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If you use an N-MOSFET in the place of the bipolar transistor, there will be no voltage drop on the Standby signal because the base current is removed.

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  • \$\begingroup\$ Appreciate the input. I'll do some homework on them. \$\endgroup\$
    – Mike-d-s
    Commented Jul 4, 2022 at 22:31
  • \$\begingroup\$ Ok, so N-MOSFET base has capacitive type behavior and therefore once charged there is minimal (zero or some leakage current level?) current flow through the gate. So this would seem to prevent the voltage sag. \$\endgroup\$
    – Mike-d-s
    Commented Jul 4, 2022 at 23:03
  • \$\begingroup\$ Switching frequency is not really an issue since it just needs to come on once and then stay on until the amp is turned off (or a channel goes into protect) which is essentially a very long time from the mosfet viewpoint. From the little bit of reading I did, there is some charge time based on the +- 1mA current available, but given I'm not really switching on and off for practical purposes the charge time may not be an issue unless its going to be much more than a few seconds. \$\endgroup\$
    – Mike-d-s
    Commented Jul 4, 2022 at 23:03
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    \$\begingroup\$ @Mike-d-s Yes, there is a gate charge current, but if you use a small signal FET like BSN20, the gate capacitance Ciss is below 40 pF. The charge time will be below 1 us. \$\endgroup\$
    – Jens
    Commented Jul 5, 2022 at 0:42
  • \$\begingroup\$ Thanks, for the help. I'll probably look for a through hole solution since my soldering skills are so so (BS170-D26Z looks close specs). Given that I really don't know what the internal circuit is inside the board connected to the 12V protect pin other than its primarily pulled up with a 10k resistor. would it be good practice to add a resistor (1 Mohm?) gate to ground to bleed any voltage remaining on the protect line? Since the pin can be left open I'm assuming the board won't be bother by residual voltage on that pin, but to be safe I could add one outboard. \$\endgroup\$
    – Mike-d-s
    Commented Jul 5, 2022 at 15:53
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You wrote a 25k divider but to minimize the loading on the 10k detection point, this needs to be at least 10x bigger. I chose to use a Darlington and use the 12V, 5V available to maximize hFE on Q1 and minimize Pd on Q2.

Use a twisted pair for the 1M path or keep the leads short to reduce noise.

I defined your LED pair as 2.4V @ 20 mA with the diode curve. The hFE was chosen at 100 but the difference in brightness is likely insignificant for the +/- 50% range in hFE. The main thing is it solves your loading the 10k test point. Removing the LED does not change the base voltage more than 10 mV which means the current with the LED in stays constant because of the 33k load. This also ensures the LED current is off.

SIM

enter image description here

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  • \$\begingroup\$ My phrasing was probably off. The 12V line is actually based on a 10k pullup resistor internal to the board. I was using 25k outboard to try and maintain a higher voltage level on the board pin. FYI, the LED is actually a green LED built into a dual LED assembly and the 2.4Vf at 20 mA is from the datasheet for that part. I should have also mentioned that the LEDs bar is actually two LEDs and they are wired in parallel on the daughter board, so the 100 ohm was to give 26 mA to both so should be ok. The brightness when testing without the transistor was fine at than current level. \$\endgroup\$
    – Mike-d-s
    Commented Jul 4, 2022 at 22:35
  • \$\begingroup\$ Forgot to add, but based on looking at what you posted, the 12V is based on the pullup being open, but adding the 25k + 25k to ground brings the actual protect pin voltage down to below 12v? unless I'm misunderstanding? \$\endgroup\$
    – Mike-d-s
    Commented Jul 4, 2022 at 22:44
  • \$\begingroup\$ OK let me revise a better solution \$\endgroup\$
    – D.A.S.
    Commented Jul 4, 2022 at 23:02
  • \$\begingroup\$ FETs make crude comparators but can work even with 2:1 range in Vgs(th). BJT's are better in this respect with < 10% variation in Vbe for a given base current. \$\endgroup\$
    – D.A.S.
    Commented Jul 4, 2022 at 23:27

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