Schematic help

[Decrash]

I attached a schematic of a ESC...can someone please tell me what diodes D1 and D2 are for? and why are they facing the same direction?

I got the schematics from here
http://home.t-online.de/home/042073629-0001/

thanks

[vintage1]

They appear to be a cride form of flyback suppression. One is a zener diode, wich breaks down and conducts when the voltage across teh motor rises somewhat beyond the rail voltage, and switches on teh mosfet to clamp any positive going spikes.

The second is to stop the combination drawing current off the gate when the mosfet is driven by the chip itself.

The combination together will ONLY conduct in one direction (the ordinary diode takes care of that) ad then only when the voltage exceeeds the zener voltage.

It looks crude and not particularly effective to me, frankly. I'd certainl;y want to take care of any leakage curent by putting a siurce to gate restsitor on the MOSFET, and put the 47 ohm resistor right up by the FET gate.

I also am dubious about poytentiall driving teh chip output higher than the rail voltage. I'd put another resistor in between it and the diodes where the 47 ohm is now, and a clamp diode up to Vcc...

Looks like a digital designer did some analogue electronics, and got one to work, once :)

[jeffs555]

I agree with everything Vintage1 said. I also don't know why they would be necessary. If the schottky diode is properly sized, it will suppress the flyback pulses and D1/D2 will never really do anything.

Jeff

[steve lewin]

To some extent it depends what MOSFETs you use. If you need protection diodes they're best close to the MOSFETs rather than at the end of possibly long (inductive) wires on the motor. That circuit is straight out of International Rectifier's old application notes though the latest ones say external zeners should be avoided as "Unfortunately they also contribute to oscillations and have been known to cause device failures."

Tricky call.

Steve

[Decrash]

Ok thanks, that makes sense.
I'll do what vintage1 said.

[Chippie]

Originally posted by vintage1
I'd certainl;y want to take care of any leakage curent by putting a siurce to gate restsitor on the MOSFET, and put the 47 ohm resistor right up by the FET gate.



most esc designs I've seen include a resistor between the pic output, feeding the fet, and ground........so that on power up the pic output and hence the fet gate is grounded....basically to stop the fet conducting before the pic has had chance to initialiaze....


I fail to see the value of the 2 diode combination,perhaps the circuit designer could enlighten us of their purpose. I can see the purpose of the diode across the motor....back emf from an inductive load, but not the others......curious ;)

[Decrash]

I fail to see the value of the 2 diode combination

I believe it is basically supposed to do the same thing as the schotty diode across the motor. Like jefss555 said I don't see why you would need both if there is a schotty already across the motor.

D2 is a Zener diode; I didn't realize that the first time I looked at the schematic.

Steve lewin, I heard that having the diode closer to the motor could elminate some of the RF noise from having the spike travel across the long wire. But yeah, I don't think that would matter with a low amp ESC like this one.
anyways thanks again

[whanderson]

Originally posted by Chippie

I fail to see the value of the 2 diode combination
There is, I believe, a different explanation for the 2 diodes. The reverse diode connection is reminiscent of a charge pump (Vishay app note). The combination of diodes, gate capacitance, and motor inductance form a resonant circuit. This bootstraps the microcontroller's pulse to reduce the MOSFET's on resistance.

[Chippie]

Originally posted by whanderson
There is, I believe, a different explanation for the 2 diodes. The combination of diodes, gate capacitance, and motor inductance form a resonant circuit. This bootstraps the microcontroller's pulse to reduce the MOSFET's on resistance.

Bill, with my somewhat limited knowledge of electronics, could explain the term "bootstrap" in the context you describe please?

[whanderson]

Originally posted by Chippie
Bill, with my somewhat limited knowledge of electronics, could explain the term "bootstrap" in the context you describe please?
The microcontroller's pulse swings between 0 and +5V. The charge pump "bootstraps" the 5V to something higher like 12V or more. As in "pulled up by the bootstraps."

[jeffs555]

I don't see how that circuit could possibly act as a charge pump, and if it could, it would be applying your "12V or more" directly to the output of the pic processor thru the 47 ohm resistor which would be extremely detrimental to the pic. I think its purpose is just like Vintage1 originally suggested, flyback suppression.

Jeff

[vintage1]

Nah. he's right. Its not a bootstrap. Typically you would use a bootstrap when dring a FET in source follower mode, not grounded source like this is. It works backwards when the drain is the output!

Brushless controllre use em when they want to use e.g. all N channel FETS rather than complenetary pairs. of N and P channel.

We used to use them in cheaper audio power designs as well. Until peple realised that teh better answer was to have a little supply at a slightly higher voltage than the main rails. Bootstrapping goes back to valve days!

[whanderson]

Originally posted by jeffs555
I don't see how that circuit could possibly act as a charge pump, and if it could, it would be applying your "12V or more" directly to the output of the pic processor thru the 47 ohm resistor which would be extremely detrimental to the pic. I think its purpose is just like Vintage1 originally suggested, flyback suppression.

Jeff
Where in the Microchip datasheets does it say 12V thru a 47 Ohm resistor is detrimental to the PIC? I can't find such a reference.

[whanderson]

Originally posted by vintage1
Nah. he's right. Its not a bootstrap. Typically you would use a bootstrap when dring a FET in source follower mode, not grounded source like this is. It works backwards when the drain is the output!

Brushless controllre use em when they want to use e.g. all N channel FETS rather than complenetary pairs. of N and P channel.

We used to use them in cheaper audio power designs as well. Until peple realised that teh better answer was to have a little supply at a slightly higher voltage than the main rails. Bootstrapping goes back to valve days!

The following picture is an excerpt form an STMicroelectronics datasheet.

If there is a "better answer" to charge pumps why does Digikey list 300 types of these devices from 9 different manufacturers? Why do modern battery power supplies use then all over the place? Bootstrapping does go back to "valve days" as does just about all other electronic circuitry. As you point out, brushless controllers use a charge pump of some kind for high side N channel drivers. Why are you so quick to discount my comments?

[jeffs555]

11.0 ELECTRICAL CHARACTERISTICS - PIC12C508/PIC12C509
Absolute Maximum Ratings†
Voltage on VDD with respect to VSS .................................................. .................................................. .............0 to +7.5 V
Voltage on MCLR with respect to VSS............................................... .................................................. ..............0 to +14 V
Voltage on all other pins with respect to VSS .................................................. ............................. –0.6 V to (VDD + 0.6 V)


Max voltage on VDD is 7.5V, max voltage on any pin other than MCLR is 0.6V higher than VDD. Therefore, max voltage on any pin other than MCLR is 8.1 volts.

Jeff

[jeffs555]

I know that charge pumps are used all the time, but there is no series inductor in the dirve circuit of the ESC circuit we are talking about. Without the inductor, it won't boost the gate voltage.

Jeff

[whanderson]

Originally posted by jeffs555
11.0 ELECTRICAL CHARACTERISTICS - PIC12C508/PIC12C509
Absolute Maximum Ratings†
Voltage on VDD with respect to VSS .................................................. .................................................. .............0 to +7.5 V
Voltage on MCLR with respect to VSS............................................... .................................................. ..............0 to +14 V
Voltage on all other pins with respect to VSS .................................................. ............................. –0.6 V to (VDD + 0.6 V)


Max voltage on VDD is 7.5V, max voltage on any pin other than MCLR is 0.6V higher than VDD. Therefore, max voltage on any pin other than MCLR is 8.1 volts.

Jeff
Still doesn't answer my question. What is the max volts on GPIO pins? PIC's with a Port B have an external pull up option. It doesn't make much sense if this external pull up is limitied to Vdd. I doubt the GPIO transistors are different than those used in Port B.

[whanderson]

Originally posted by jeffs555
I know that charge pumps are used all the time, but there is no series inductor in the dirve circuit of the ESC circuit we are talking about. Without the inductor, it won't boost the gate voltage.

Jeff
The motor winding is the inductor.

[jeffs555]

The Microchip specs say "Absolute Maximum Ratings". The maximum voltage on any pin is 0.6 volts above VDD. This is not due to the GPIO transistors, it is due to the ESD protection diodes which are on each pin and connected to VSS and VDD. If you apply a voltage 0.6 volts higher than VSS, or 0.6 volts lower than VSS, the diodes will become forward biased, and draw excessive current.

The reason for allowing open drain drive is to allow bidirectional busses, and wired or functions. The pullup should not be connected to voltages higher tha VDD.

Jeff

[jeffs555]

The motor inductance is not in series with the driver.

Jeff

[Mr.RC-CAM]

What is the max volts on GPIO pins?There is substantial ESD protection on the I/O pins. So, in an indirect way, it is the max current applied to the I/O pin that is most important. With some external series resistance, you can even apply household 115VAC power to a PIC's input (I/O only, not power pins).

Just use a little math to ensure that the input currents are limited to within the chip's max allowed. On some of the parts a few mA's are possible, but generally you would want to limit the currents (with a resistor) to the uA region. There is a Microchip App Note (#AN521) on this trickery.


RC-CAM

[whanderson]

Originally posted by jeffs555
The Microchip specs say "Absolute Maximum Ratings". The maximum voltage on any pin is 0.6 volts above VDD. This is not due to the GPIO transistors, it is due to the ESD protection diodes which are on each pin and connected to VSS and VDD. If you apply a voltage 0.6 volts higher than VSS, or 0.6 volts lower than VSS, the diodes will become forward biased, and draw excessive current.

The reason for allowing open drain drive is to allow bidirectional busses, and wired or functions. The pullup should not be connected to voltages higher tha VDD.

Jeff
The PIC I/O ports can be tristated. Setting Tris bits to a 1 sets the corresponding I/O pin to hi Z allowing wired or with or without pull ups. Mr.RC-Cam nailed the voltage on I/O pin issue, thank you.

The "flyback suppression" argument has the same flaw you say my analysis has. Since the zener is 24V, the PIC GPIO will see volts well above Vdd. Should be smoke city based on what you say. In addition, flyback supression is done accross the motor not the MOSFET. The MOSFET has an internal Schottky supression diode. The gate has ESD protection. The flyback suppression theroy doesn't work.

[jeffs555]

Whanderson,
I believe Mr RC-CAM understands the app note he posted. The voltage on any pin other than MCLR cannot possibly rise more than 0.6 volts above VDD without blowing a protection diode on the PIC. All the app note he posted says is that by limiting the current by using a very large resistor(5 megOhms), you can apply 115 VAC to one end of the resistor and the protection diodes will prevent the other end of the resistor from rising more than 0.6v above VDD or 0.6vbelow VSS. When Microchip says the "Absolute Maximum" voltage on any pin is 0.6v higher than VDD they mean it. Ignore their specs at your own risk.

Do you understand how a zener diode works? It does not supply voltage. The 24volt zener won't conduct until the drain of the mosfet and the side of the motor connected to it rises above 24 volts. When the output of the PIC goes to ground, the mosfet turns off, and the flyback from the motor tries to raise the drain of the mosfet way above 12volts. If the schottky and the R-C snubber, and the internal zener(the internal drain protection is a zener not a schottky) were not there, the drain voltage would rise high enough to turn on the ZD24 zener, which would pull the gate high, causing the mosfet to turn on and pull the drain voltage back down thereby suppressing the flyback pulse, and hopefully protect the mosfet and PIC. Like everyone said before, because of the schottky, the feedback zener is redundant and probably not necessary. It looks like whoever designed that esc put on every form of flyback supression he could find.

Jeff

[JMP_blackfoot]

Originally posted by whanderson
The flyback suppression theroy doesn't work.
When the PIC outpout goes "0", the IRF7413 switches off. The motor inductance causes the drain voltage to rise. If the drain voltage rises above 24V, ZD24 conducts and rises the gate voltage of IRF7413 until the threshold voltage is reached (1 to 3V in the datasheet). IRF7413 conducts and clamps the drain voltage at 25 to 28V, which is within its specifications (30V max drain-source voltage) and prevents damage to itself. At the same time, the PIC output pin only sees a max voltage of about 3V (the IRF7413 gate voltage), which is not harmful, especially since it lasts a very short time. The combination of Mosfet + Zener diode + diode acts as a power zener to suppress dangerous flyback voltage.

[vintage1]

Originally posted by whanderson
The following picture is an excerpt form an STMicroelectronics datasheet.

If there is a "better answer" to charge pumps why does Digikey list 300 types of these devices from 9 different manufacturers? Why do modern battery power supplies use then all over the place? Bootstrapping does go back to "valve days" as does just about all other electronic circuitry. As you point out, brushless controllers use a charge pump of some kind for high side N channel drivers. Why are you so quick to discount my comments?

Becuae if you look, you will se there the combo is between gate and source, which is positive feedback.
In teh scehmatic posted, teh diodes are between gate and drain, negaticve feedback.

Look up 'Miller effect'; to see what caps between invertiung outputs and inputs do to circuits.

Plus, no inductor,.

[vintage1]

Originally posted by whanderson

The "flyback suppression" argument has the same flaw you say my analysis has. Since the zener is 24V, the PIC GPIO will see volts well above Vdd. S

I agree, with the schottky, its total *****s. I.e. its a useless piece of design by someone who copied somethibg without uinderstanding it, it worked, so he left it in. It would not be the first time I have seen such in my career.

BTW a car ignition coil, when switched off in a hurry, achieves up to 400v across its primary, and some 25kv across its secondary. Don't underestimate how a fast switch can gertenret HUGE voltages in conjunction with an indictor.

[JMP_blackfoot]

Originally posted by whanderson

Since the zener is 24V, the PIC GPIO will see volts well above Vdd.

ABSOLUTELY WRONG !
See my post #24 for a correct explanation of the operation of this circuit.

Also read : http://www.semikron.com/applica_help/e/3_6_3_2.pdf
go to the 3rd page : "Active clamping". All is revealed.:rolleyes:

Or : http://www.ep.liu.se/exjobb/isy/2002/233/exjobb.pdf

[Bruce Abbott]

Since the PIC's output is low when the MOSFET is turned off, it can handle quite a bit of current - and it needs to. Typical impedance of a 12C508 output is about 30 Ohms. Assuming the MOSFET's gate has a threshhold of 3V, the PIC would sink 39mA, and the voltage at its output pin would be 1.2V (only for a short time until the MOSFET turns on). If for some reason the MOSFET was slow to turn on, the voltage could rise a lot higher, possibly causing the PIC to latch up, or even damaging its output.

Even though this circuit probably works well, I would be wary of allowing potentially damaging voltages to get to the PIC. When Zener diodes fail, they usually become a short. In this case it would probably be fatal for the PIC, but I guess you have to decide which is more in need of protection - the PIC or the MOSFET. I am operating several home-built ESC's that don't use clamping circuits, and have not experienced any MOSFET failures.

One other point:- flyback voltage from the motor is suppressed by the Schottky diode, but there is still inductance in the wiring. Adding a low-esr electrolytic capacitor across the battery input reduces spikes caused by the battery wires. To suppress motor wiring spikes a Schottky diode should be installed on the PCB, either in addition to or in place of the one on the motor.