unfamiliar symbol

The two collector transistors are current mirrors. The area of the two collectors is in a precise ratio, usually one to one. Therefore the current through the transistor divides per that ratio. If it's one to one, the current divides 50%- 50%.

In a current mirror, one collector current is defined by the appropriate voltage drops and the other collector follows or mirrors that current. This kind of structure is available within integrated circuits but not available in discrete devices. You can make discrete current mirrors but not with divided collector areas.

Apparently the sync input must pulse high to restart the ramp. If you do not use it, the device will self oscillate asynchronously.

The common mode range of the error amp is 1.5 volts to 5.2 volts to stay within the specified common mode rejection values. It's in the spec. on page

Hello,

If you look on page 4 of the datasheet, the minimum voltage to operate the sync line + 1.2 volts. That is because you have to turn on the transistor (0.7 volts) called bias, and there is a voltage divider in front of it too. This pin "sync" is for an external oscillator, if you don't want to use the built in one plus Ct and Rt. You would have to supply it with a signal that is greater than + 1.2 volts on the peak or add a positive offset voltage to the signal.

The dual collector trasistor is common for transistors in an IC, don't ask me why, I don't know what the advantage is.

Page 4 also gives information for the error amplifier.

Shaun

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Okay, I know about current mirrors like Q12 and Q13, now I know that symbol means something similar.

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Then a high signal on the sync would have to turn on Q14 in order to discharge the timing capacitor; but when I try following the current mirrors around the circuit it looks like just the opposite happens. I can settle this by sticking a chip in the breadboard.

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Thanks!

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Okay, I know about current mirrors like Q12 and Q13, now I know that symbol means something similar.

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Then a high signal on the sync would have to turn on Q14 in order to discharge the timing capacitor; but when I try following the current mirrors around the circuit it looks like just the opposite happens. I can settle this by sticking a chip in the breadboard.

Yes I agree it does look weird. I checked the Texas instruments version of this part, UC3525 and they do NOT show a connection between the collector of Q13 and the base of Q14. I have not studied it enough to figure out the actual circuit or who is right. Obviously there is at least one error in at least one of the schematics. So who knows?

Tell us how the breadboard reacts and then maybe we can make sense of the schematics.

Keep in mind that the schematics are for reference and are abbreviated and stylized for simplicity. They are NOT the actual circuitry used with complete detail.

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Thanks!

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The TI schematic is all wet. Q14 is floating there in space, doing nothing... the base doesn't connect to anything that can bias it on. Likewise with the whole discharge circuit: it has no connection to the components in the timing section. So the TI diagram is missing connections at the base and emitter of Q14 that the OnSemi circuit diagram shows. The OnSemi diagram starts to make sense as I look at it now. I had the mirrors turned around or something when I looked at it before. Without going too deep into it-- The 7.4k and 14k resistors set a trip point for the timing ramp, not far from one timing constant. At that point the mirrors turn on Q14 (I had it backwards the first time Iooked). That discharges the timing cap. Same thing happens if you pull up the sync pin. You were right, Bob.