I was looking at the d/s for the SN754410 h-bridge chip. Anybody got any idea what the inverted diode on the base of the high-side transistor of the output stage on page 2 does? Maybe temperature-compensation, ala leakage?
http://www.jameco.com/Jameco/Products/ProdDS/1054684.pdf
The rest of the ckt shown appears to be standard TTL totem-pole output stage stuff.
Protects the base emitter junction against inductive kick.
Hmm, could be. I was originally going to mention, and obviously forgot, there are the usual back-emf protection diodes on the output pins, but this is possible added protection.
By far the easiest way to destroy a transistor is pull the base more than a few volts below the
emitter (for NPN). The main protection diode is high current and thus prob. will not switch fast enough
for the base-emitter protection role.
The explanation that the diode protects the reverse voltage base-emitter breakdown of the pull-up NPN doesn't seem reasonable. There will be no sustained breakdown current since there is nothing actively holding the base voltage below the emitter. B-E junctions are sometimes reliably used as zener diodes on ICs though using them that way may degrade the beta of the device in normal operation. The breakdown is not destructive on silicon ICs if current is kept reasonable.
Anyway, the base-emitter resistor should be enough to prevent breakdown of the B-E junction. The base-emitter resistor must be sufficient to turn off the pull-up NPN quickly in normal operation so it should certainly be enough to prevent B-E breakdown. In the silicon processes I am familiar with, the B-E breakdown happens around 5.8V.
The breakdown is not destructive on silicon ICs if current is kept reasonable.
C6, I'm siding with the other guys. First off, inductive spikes can easily be 400-600V which is not quite "kept reasonable". As Mark indicated, the back-emf diodes may not turn on fast enough to adequately squelch very fast transient edges, so the extra diode will provide additional protection.
B-E junctions are sometimes reliably used as zener diodes
This would only be reliable if there is some current limiting, which doesn't exist here.
Anyway, the base-emitter resistor should be enough to prevent breakdown of the B-E junction. The base-emitter resistor must be sufficient to turn off the pull-up NPN quickly in normal operation so it should certainly be enough to prevent B-E breakdown.
I don't think so. In this case, the B-C junction of the output NPN would provide a path for the transient current, and all you would need is for the transient current I(t) to be large enough to produce a drop of I(t) * Rbe > Vbe(breakdown) to blow the base.
If not for inductive spike protection, do you have an alternate hypothesis for what you think the diode might be doing instead. ??
oric_dan:
If not for inductive spike protection, do you have an alternate hypothesis for what you think the diode might be doing instead. ??
I know exactly what the diode is doing. In a design review for the IC someone raised the same objection as MarkT and the designer went away with the task of including the diode. No proof. No need. Only intuition of someone in the back of the room. Zero real estate added to the IC. I've seen it happen hundreds of times. Not every component in a schematic is useful.
Well ok, but I don't quite see that this nullifies the arguments I made last time.
Your arguments hold no water. "inductive spikes can easily be 400-600V". Really? If you have that much sustained voltage on the output you will blow more that the B-E junction. You can't think in terms of forced voltage on the output because you don't have a forced voltage. The output will actually have to withstand ESD which can be several kV (don't believe me?...look here). This will not destroy the chip because it is transient and doesn't hold the energy to blow the ESD protection or the output circuitry. You have a similar situation with the back EMF.
And where did you get the notion that there are fast diodes and slow diodes on an IC? I am guessing that the diffusions used for the clamping diodes is the same as the NPN B-C junction. The only thing affecting turn on speed is the series resistance and this can be made as low as desired using layout techniques.
Your arguments hold no water. "inductive spikes can easily be 400-600V". Really? If you have that much sustained voltage on the output you will blow more that the B-E junction. You can't think in terms of forced voltage on the output because you don't have a forced voltage.
This discussion is already going in circles, so I'm going to drop it. Eg, I don't see how you can mix arguments like this in the same sentence:
Your arguments hold no water. "inductive spikes can easily be 400-600V". Really? If you have that much sustained voltage on the output you will blow more that the B-E junction. You can't think in terms of forced voltage on the output because you don't have a forced voltage.
Secondly, in regards "inductive spikes can easily be 400-600V", I've measured this myself, but here's one from online:
http://www.yoctopuce.com/EN/article/relay-and-inductive-loads
Thirdly, again, discharging a 100pF cap through a 1500 ohm R is not hardly the same as a motor. We're obviously arguing very different things.
the charged human body is modeled by a 100 pF capacitor and a 1500 ohm discharging resistance. During testing, the capacitor is fully charged to several kilovolts (2kV, 4kV, 6kV and 8kV are typical standard levels) and then discharged through the resistor connected in series to the device under test.
The "inductive spike" is a certain current value - the voltages rises as necessary to keep
that current flowing. The depletion zone in a reverse biased base-emitter junction is
narrow, and all the power(*) of the reverse breakdown current is dumped in it causing
it to heat extremely rapidly.
(*) ~5V times the load current - more power than the transistor as a whole dissipates
when on (when the heat is generated over a much larger portion of the device)
I'm inclined to agree with Charlie that it serves no useful purpose. If the voltage on the output rose above the supply rail by enough to damage the output transistor (around 5 or 6 V) that would mean that the protection diode on the output would be blown and the IC would be already US.
Russell.
MarkT:
The "inductive spike" is a certain current value - the voltages rises as necessary to keep
that current flowing. The depletion zone in a reverse biased base-emitter junction is
narrow, and all the power(*) of the reverse breakdown current is dumped in it causing
it to heat extremely rapidly.(*) ~5V times the load current - more power than the transistor as a whole dissipates
when on (when the heat is generated over a much larger portion of the device)
You do realize that you just described what happens normally in a reverse biased B-C junction of an NPN with no harm.
russellz:
I'm inclined to agree with Charlie that it serves no useful purpose. If the voltage on the output rose above the supply rail by enough to damage the output transistor (around 5 or 6 V) that would mean that the protection diode on the output would be blown and the IC would be already US.Russell.
As I see it, diodes are used in these apps because they can handle forward "surge" currents that are much larger than their continuous current ratings. Eg, for 1N4004, 30A vs 1A, and 160A pulse-current for the MOSFET. So, it'll take much more to blow the back-emf protection-diodes than the NPN B-E junction.
http://www.diodes.com/datasheets/ds28002.pdf
http://www.jameco.com/Jameco/Products/ProdDS/669951IR.pdf
oric_dan:
As I see it, diodes are used in these apps because they can handle forward "surge" currents that are much larger than their continuous current ratings. Eg, for 1N4004, 30A vs 1A, and 160A pulse-current for the MOSFET. So, it'll take much more to blow the protection-diodes than the NPN B-E junction.
http://www.diodes.com/datasheets/ds28002.pdf
http://www.jameco.com/Jameco/Products/ProdDS/669951IR.pdf
Quite, so the voltage cannot rise to the 5V+ above the rail to cause damage to the BE junction. So the small low power diode across that junction does nothing.
Russell.
charliesixpack:
You do realize that you just described what happens normally in a reverse biased B-C junction of an NPN with no harm.
Read again, this is reverse breakdown of the BE junction, which is much more heavily doped
(so narrower depletion zone), not just reverse biasing. You only get amplification if there is
a massive ratio in doping levels between the emitter, base, collector. Its the power per
unit volume that is the big issue here.
russelz:
I'm inclined to agree with Charlie that it serves no useful purpose. If the voltage on the output rose above the supply rail by enough to damage the output transistor (around 5 or 6 V) that would mean that the protection diode on the output would be blown and the IC would be already US.
You assume the diode can respond as fast as is necessary - and that the wiring has no stray inductance.
MarkT:
Read again, this is reverse breakdown of the BE junction, which is much more heavily doped
(so narrower depletion zone), not just reverse biasing. You only get amplification if there is
a massive ratio in doping levels between the emitter, base, collector. Its the power per
unit volume that is the big issue here.
I forgot to mention that you also proved zener diodes impossible.
charliesixpack:
I forgot to mention that you also proved zener diodes impossible.
As mentioned already, this is just going in circles. In the previous loop, this was mentioned re zeners, and ignored.
This would only be reliable if there is some current limiting, which doesn't exist here.
This shows how to make a bad zener --> don't use any current-limiting.
http://cr4.globalspec.com/thread/72501
Do you know what it means - [almost] vertical slope?
MarkT:
You assume the diode can respond as fast as is necessary
Well, the built in output protection diodes should be as fast as the output transistors and, as I said, they will be dead before allowing the output voltage to rise by 5 V or so.
- and that the wiring has no stray inductance
The stray inductance in series with the built in output protection diodes should be insignificant.
Russell.
Well, the built in output protection diodes should be as fast as the output transistors and, as I said, they will be dead before allowing the output voltage to rise by 5 V or so.
Once again, you're not thinking hard enough about dynamic characteristics here.
Took me a while to run this stuff down - as most of the online info involves diode turn-"off" characteristics, rather than turn-on. In any case, you can watch the video [pay special attention around time 45-sec plus, and also at the very end of the video], and/or read the appnote [figures 2,12,etc] - and note, he is only talking about 1-Amp currents here, nothing like what you might get with an inductive motor transient. Note the bit where it says "exceeding IC breakdown limit".
http://www.digikey.com/videos/en/v/Diode-Turn-On-Time/1653078115001