Racking my brain on a simple circuit - Page 1
Forum section: electronics / repair

author: ajcrm125
author_rank: Contributor
timestamp: June 17, 2017, 03:23:00 pm
content: It's been years since I've done analog design (I do digital design full-time) but for some reason I can't pinpoint what's going on with this circuit.I'm troubleshooting an old Nintendo arcade game (Space Firebird) and there's a problem with some of the analog sounds.Attached is a snippet of the schematics.What I've deduced is that the shooting sounds and enemy explosion sounds are in need of the pre-amp that involves IC26. So this looks to me like a simple OR type structure that enables IC26 when either of these events occur. IS23 (74ls05) receives these 2 inputs. Pin 13 goes low when you shoot and pin 11 goes low when an enemy explodes. Lets start with the shooting....When you shoot, the inverter turns on and since it's open collector, lets the 10k pullup at R28 take over. Looking at the base of TR6 it appears to start at -5V due to the 200k pulldown at R5 and then when you shoot, I would expect it to ramps up to around 4.5V because of the resistor divider between R28 and R5. But it doesn't... it climbs to 0v and stays there.Another thing I noticed was that the emitter starts at -5V and then climbs only a few hundred millivolts and stays there. That tells me T6 is at least turning on. Shouldn't it be reaching close to 5V when T6 turns on because the collector is at 5V?Let me know what I'm missing.....Thanks!

author: alm
author_rank: Super Contributor
timestamp: June 17, 2017, 03:45:24 pm
content: I have not done a full analysis, but for starters C19/C20 form a high-pass filter, so on a rising edge the base of TR6 would only be high for maybe 10 ns. What does the base of TR6 look like on a scope when you shoot?Are D5/D6 zeners? It certainly looks that way because of their orientation. Can't really see how they would be forward biased.Not sure why the base would float to 0 V as long as IC23-12 is not switching, is C19 leaky (for DC current)?

author: Armadillo
author_rank: Super Contributor
timestamp: June 17, 2017, 05:32:28 pm
content: Quote from: ajcrm125 on June 17, 2017, 03:23:00 pmNo, it depends on the pulse width, the time, the window period that permits the cap to charge up and it will be transient.Quote from: ajcrm125 on June 17, 2017, 03:23:00 pmTry replacing capacitor C7 and C8.

author: SeanB
author_rank: Super Contributor
timestamp: June 17, 2017, 06:08:46 pm
content: IC 26 is a voltage controlled amplifier, used to give an envelope to the constant sound on the input as required. The capacitors C19,20 are there as level shifters, so that a 5V shift on the input is transferred to a 5V shift on the base of the transistor, so it will give a -5V to 0V swing on the emitter on each pulse, charging the 22uF capacitor up to 0V and thus giving, via the diodes D11,12 and the resistors around the VCA a fast attack volume up, that then drops down to muted as the capacitors discharge at the end of the trigger pulse.If it only has one sound or the other, then the associated capacitors with that sound are dry. If both do not work then the VCA side or the diodes are faulty, or the sound generator side is faulty. To test if the VCA is faulty simply use a 10k resistor between pin 2 of IC26 and the junction of R20 and R21, which should give a continuous sound. If the VCA is working then check the emitters go to 0V from -5V, and decay appropriately.What is the fault you get, no sound or distorted sound or only one working.Edit to attach datasheet link, good old Intersil part there.http://www.intersil.com/content/dam/intersil/documents/ca30/ca3080-a.pdf

author: ajcrm125
author_rank: Contributor
timestamp: June 18, 2017, 01:52:50 am
content: Quote from: SeanB on June 17, 2017, 06:08:46 pmOk so I guess my first question is why does the negative side of C19 (and therefore the base of TR6) only charge up to 0V? What's causing it to charge up to that level and no further?Same question for the charging of C7.

author: Armadillo
author_rank: Super Contributor
timestamp: June 18, 2017, 07:04:52 am
content: At transient max at 4.4v, 1 diode drop at the base of the transistor.The voltage at the emitter should rise up in tandem, otherwise the transistor is Kapood!

author: SeanB
author_rank: Super Contributor
timestamp: June 18, 2017, 08:40:20 am
content: C19 at power up charges from the 5V rail via the 10k resistor, and the low level of the OC buffer in IC23 keeps the positive side from rising more than 0.2V above ground, the negative side being pulled down to -5V by R5 and a minor charging by the transistor base emitter junction, but the BE junction current will be small.In use C7 is discharged via R29, D5, R25, R2 back to -5V, and is charged up by 5V when the gate puts charge into the transistor base by R68, so it normally sits at -5V on the positive side when in standby, but is charged up to 0V when triggerred.

author: Armadillo
author_rank: Super Contributor
timestamp: June 18, 2017, 09:51:51 am
content: Bread boarded;Its less than 1 diode drop to positive rail [in parallel with 200K ohm].As measured from the negative rail, its 9.8v.Diode used is 1N4148If you cant get the voltage up, replace the capacitors.The voltage remains valid even frequency adjusted up to 1MHz.

author: ajcrm125
author_rank: Contributor
timestamp: June 18, 2017, 11:50:25 am
content: Well when the OC buffer is enabled then yes, I would agree that the positive side of that cap would be ground but once that buffer disables, the fact that it's OC means it's like it's not even there. It should climb to approx 5V and that's what I'm seeing via the scope as well.But I think you guys made me realize something....If if D5 and D6 are zeners then that makes sense why the base wouldn't climb any higher. Whatever the reverse voltage drop is for that zener... that's it.. that's all I'm gonna see. And as far as the emitter goes, it's only going to climb so long as the transistor is active and that's only gonna happen so long as the Vbe (base-emitter voltage drop) is greater than the transistor's threshold voltage Vt. Once the emitter gets to be 0-Vt, that's it for that guy as well and the transistor will start shutting off.Let me see what the output side of D11 and D12 are doing. I want to say they didn't even move but let me confirm.