voltage drop question (pc speakers auto on / stand by mod) - Page 1
Forum section: electronics / beginners

author: eldon
author_rank: Newbie
timestamp: May 01, 2015, 05:58:44 pm
content: HI,i'm trying to modify a pair of nice little pc speakers Genius SP-HF1250B (http://www.geniusnet.com/Genius/wSite/ct?xItem=55433&ctNode=148).The amplifier is based on a TDA7379 :http://www.100y.com.tw/pdf_file/33-ST-TDA7379.PDFMy goal is to leave it always on and have it switch on automatically when there's some signal on the inputs..I'm planning on using this circuithttp://sound.westhost.com/project38.htmfor the sensor/switch.My question is not exactly about that, i've been studying the amplifier circuit and found out that the TDA7379 has a "stand by" pin.Of course as it is the circuit turns the stand by pin on (>3.5v) but i'm a bit confused as to how the voltage drop circuit works exactly (especially the D1//R1 part) and how i could get it to go to ground without damaging the parts in place..here's the circuit :Pin 7 stand by input is supposed to draw a max of 100nA.1 -I'm simply wondering why they added a diode in parallel to the resistor.2 -Then if i wanted to use the standby mode, possibly leaving all the parts in place, can i bring Pin7 to ground, directly / by mean of a resistor to ground ?Pin7 should be CMOS compatible so below 1.5v should put the amp in stand by..I'm not quite sure i'll do that in the end, even in stand by, the amplifier part would still be always powered on.And there's a nice coil that supplies VCC, i could put my "switch" there and apply power only when signal is detected which should be much better..This amp uses a switched mode power supply to produce the vcc rail (13.8v), i was thinking of using it to power the sensor circuit.There's also a 9v regulator powering the control board (2 op amps, volume and tone pots)humm there was a led there too but she died after a few hours..thx for your comments..eldon

author: flynwill
author_rank: Regular Contributor
timestamp: May 01, 2015, 07:08:53 pm
content: The diode is probably there to insure that the standby pin is pulled low as soon as power goes away. Otherwise there would be a delay as the two capacitors discharge through R1. I would speculate that was added to prevent nasty pops from a brief power interruption.If you don't want to remove any parts, you can pull pin7 to ground with either a discrete transistor (NPN BJT or MOSFET), or an "open collector" logic output, as long as that open collector output is ok with 7 Volts.You'll need to invert the logic of the output in the project 38 circuit you found. Could be a simple as replacing the relay & D6 with a 10-100k resistor and then use a second mosfet with the gate attached to the drain of the first, source connected to ground and drain connected to the Pin Standby input on the amp.

author: madires
author_rank: Super Contributor
timestamp: May 01, 2015, 07:50:25 pm
content: Quote from: flynwill on May 01, 2015, 07:08:53 pm... and add a low value resistor to limit the cap's discharging current.

author: eldon
author_rank: Newbie
timestamp: May 02, 2015, 09:13:02 am
content: thx for your comments.Quote from: flynwill on May 01, 2015, 07:08:53 pmright i see, i didn't think of the current flow during power off.I'm still not sure i understand why they added a zener to drop Vcc voltage on stand by pin7, there's no max voltage specified in the (very thin) datasheet.Then again maybe it is to speed up going down to stand by during power off, although i must say the voltage goes down quite slowly on that pin compared to Vcc. It stays above 1.5v for quite a long time, much longer than for Vcc to drop below min supply 8V..I've manually tested the stand by mode, and the current draw is still quite high (~100mA) in my opinion.So i think i'll switch Vcc on/off entirely.That leads me to some follow up questions.I can easily insert my switch by desoldering a coil bridging both ground and Vcc from the power supply circuit to the amplifier circuit. The coil is a small ferrite ring with very thick wire.There are two large capacitors 2200+1000uF before the coil (power supply side) and one 1000uF after the coil (amp side).I will probably use a relay to switch Vcc on/off.My questions are :- should i put my switch before or after the coil ?- is it okay to only switch Vcc and leave ground always connected ?- now that Vcc will be applied on/off "instantly" should i also add some "protective" measures on the amplifier side ?maybe replacing the 1000uF, amp side, with a larger value ?I may also try to use the stand by pin to make sure it goes on a short while after Vcc is switched on, i'll first see what happens "as is"..thx again.

author: madires
author_rank: Super Contributor
timestamp: May 02, 2015, 01:36:20 pm
content: Quote from: eldon on May 02, 2015, 09:13:02 amHave you seen that the stand-by input is the base of a BJT?Quote from: eldon on May 02, 2015, 09:13:02 amPlease have a look at the block diagram in the datasheet and find out how the stand-by function works. And switching Vcc based on the input signal won't make you happy.

author: eldon
author_rank: Newbie
timestamp: May 02, 2015, 04:12:44 pm
content: Quote from: madires on May 02, 2015, 01:36:20 pmokay i have been reading about op amps and mosfets so i've probably already forgotten about NPN.And i'm certainly not understanding the VSB Threshold voltages correctly.Anyways i'm more interested by your Vcc switching comment, could you elaborate ?And please remember this is the beginners forum, probably already too advanced for my current electronics knowledge, sorry about that.Let me have a shot at the Stand by circuit though.My guess would be that, when high (>3.5v), Pin7 enables a positive bias (from vcc pin13) to be applied to the non inverting inputs of the op amps, and when low (<1.5v), Pin13 gets "disconnected" and the non inverting inputs go to ground.I'm not understanding how the op amp amplifying part works, but i'm guessing that putting the non inverting input to ground doesn't help amplifying.But i don't see how switching Vcc on/off all together would make me unhappy, may i be as bold as to say that's what happens when i turn the power switch on/off ?By the way if you're wondering, that amplifier uses the tda7379 in double bridge mode.thx again.

author: madires
author_rank: Super Contributor
timestamp: May 02, 2015, 07:39:51 pm
content: Quote from: eldon on May 02, 2015, 04:12:44 pmThe circuit you've posted includes a delay, i.e. after powering up stand-by will be low for a short moment. This will prevent the famous power-on "blob" sound. If you switch Vcc based on the input signal, each time the amp needs some time to be operational and the delay of the stand-by part increases the time further while preventing the blob sound. That way you'll miss the beginning of the audio signal. By switching stand-by you won't have that delay. BTW, to get a fast switching you could place a transistor between the output of the circuit in post #1 and the stand-by pin of the TDA with a resistor between the stand-by pin and ground. The resistor gounds stand-by when the transistor isn't switched on. If you do it the other way around, like mentioned before, each time cap C2 needs to be charged to enable the TDA, i.e. causing a delay.

author: Zero999
author_rank: Super Contributor
timestamp: May 02, 2015, 08:08:19 pm
content: Quote from: eldon on May 01, 2015, 05:58:44 pmWhere did you get that circuit from?Quote from: eldon on May 02, 2015, 09:13:02 amThat's far too high for it to be in stantby mode. It sounds like it's either not going into standby mode or there's some other circuitary responsible for the current draw.You can probably remove the MOSFET and relay from the LM358-based circuit and connect the standby pin in place of the MOSFET's gate.

author: eldon
author_rank: Newbie
timestamp: May 03, 2015, 09:44:37 am
content: @madiresthx for explaining your concern about the Vcc switch and startup delay but it actually isn't a problem at all..I was worried you spotted some circuit problem that would blow off the tda :)I haven't explained the context of this project, but the goal is to improve some classrooms audio setups, so you don't have to manage power on/off..When the teacher logs into his computer the windows music will play and switch the speakers on, then i will make sure the switch off delay will be very long (~30min). That way if some audio has to be played the speakers should be already on, and if the speakers shut down, it won't really matter if the first 3-5 seconds of audio won't play.Audio volume and other windows sounds are initialized at logon, no system sounds and volume at 10%, that way the logon sound should not be too loud and the speakers being (full) on should not disturb the usual computer activity.Quote from: Hero999 on May 02, 2015, 08:08:19 pmit's in stand by alright, i can play something on the input and i won't hear a thing.I suspect the control board, led, and possibly some leakage current produce that current draw (more like 60-80mA actually), although i do agree this is high..I will try to measure the switching power supply power draw on its own but at that point i won't be able to do much about it, budget wise i can't add a separate power supply for the sensor board..2 pairs of those speakers have already been installed a year ago, and they're switched on all day long almost every day, sometimes days at a time when we forget to switch them off, and they're working fine so i'm not very worried about leaving the power supply always on.And i must say they are quite cheap, nicely build with a massive heat sink though, so it won't be too bad if they have to be replaced, or i can always replace the board with another amp kit..QuoteI will switch Vcc on/off through the relay, i'd like to preserve the tda as much as possible, i suspect that's the part that will fail first if left always on even in stand by..Not sure i'll have enough time or resources, but i will try to put two different delays on the sensor circuit, so i can manage Vcc and stand by switches separately.Well now all i have to do is make a pcb and order my boards from china, that'll be a first for me, wish me luck :)- One last thing though :i did read that it was better to always have a load on a switching power supply.When i switch Vcc off, the power supply will have no load, should i add a resistor to ground before my switch, so some mA always flow ? If so how many mA ?There are 2200+1000uF capacitors on the supply side, should their leakage current be considered enough ?thx again for your explanations.