Basic problem with transistors

[Hero999]

That won't work at all.

For a start, there's no base resistors on PA1.

The high side transistors are now PNP so are turned on when their bases are connected to 0V.

When PA1 is high and PA0 is low no transistors will turn on because the collectors will be higher than the bases.

 

[giftiger_wunsch]

I pointed out that the transistors that you removed from Hero999's offering were too important to leave out.

I didn't leave out any transistors, though. Hero's updated schematic removed two of the transistors and simply made the top two transistors high-gain. This is the schematic on which I based mine.

Connecting PA0 and PA1 (8mA) to the motor (1A) shouldn't be done.

I was concerned this was the main drawback with the simplified ground scheme used in my second schematic, where I ditched all of the schottky diodes. I was also concerned about this at the very start when I was formulating these schematics. But what's the big problem with an extra 3.3mA flowing through the motor while it's on anyway?

And, early in the thread, if you are avoiding protection diodes, 0.1 uf across the motors to limit the inductive kickback and noise.

The motors already have capacitors soldered across them; perhaps I should have included this detail in the schematic.

 

[giftiger_wunsch]

When PA1 is high and PA0 is low no transistors will turn on because the collectors will be higher than the bases.

I see your point: I had overlooked that. Perhaps a couple of diodes will still be required to make my system work the way I wanted.

Essentially all I'm trying to do is put the base-emitter of Tr1a across PA0-PA1, and the base-emitter of Tr1b across PA1-PA0, and eliminate any alternative routes by which the current might flow. In my haste to amend my over-schottkied circuit though, I may have overlooked a few things.

If anyone can suggest the best way of wiring it to produce this effect, it will be very much appreciated.

 

[Hero999]

I think the best way is to adopt my circuit.

Here's a picture that illustrates the conventional current flow in the circuit. It shows you how the Schottky diode protects the h-bridge from both inputs going high.

The 0V of this circuit must be joined to the 0V rail of microcontroller or else it won't work.

Do you understand that the problem with your phantom ground idea is that won't work with more than one h-bridge in the circuit?

Do you now understand how the Schottky diode works in my circuit?

Do you understand that if both the inputs are 0 on a neighbouring h-bridge the other h-bridge won't be protected from both inputs being high because the lows on its neighbour will provide a ground?

 

[marcbarker]

The top transistors are configured as emitter followers which will loose 0.7V plus the saturation voltage. .

Because of the resistivity of the silicon or something like that it can be more like 0.9 V.

In any case, the top transistor isn't strictly an emitter follower in the usual way because it's 'bootstrapped' in a sense. This is because the base drive voltage (up to 3.3 V) is higher than its collector voltage (2.4 V). This means that the loss is not ~0.9 V, but a lot lower. The simulations I did confirm this.

One problem with the 4 BJT circuit is that the lower transistors B-E junction shunts the drive signal. I got a round this problem by replacing them with trenchfets.

 

[giftiger_wunsch]

Do you understand that the problem with your phantom ground idea is that won't work with more than one h-bridge in the circuit?

Do you understand that if both the inputs are 0 on a neighbouring h-bridge the other h-bridge won't be protected from both inputs being high because the lows on its neighbour will provide a ground?

I got around these problems before by adding extra schottky diodes to further isolate the groundwork, as you could see in my earlier schematics. But if your way does provide this protection, then I apologise as I didn't realise that this was the case; I thought the only H-H protection your circuit offered was to limit the current to a low enough value that it won't cause damage. I'll take a look at the diagram you posted now and maybe it'll shed some light. It was the H-H protection that all of my 'phantom ground' (in your words) circuitry was designed for, but if your schematic does the job in a simpler way that I'll adopt the schematic.

Thanks.

 

[mneary]

It's not that grounding through the 'low' outputs "won't work". It's about how well it will work, and what problem does it solve. What are the trade offs?

Your logic levels are degraded. The 3.3V "high" (which was really less than 3.3V) is robbed of a diode drop or a Vbe, or both.

If you do have two outputs such as PA0, PA1 high, then their 8mA+8mA will both want to flow somewhere else such as PA3 (which is already carrying 8mA from PA2).

If somebody walks across a carpet on a dry day and generates a 1kV spark (well below what you would notice) on the robot, it will flow through PA0..PA7 towards the earth on the microcontroller chassis and the wall wart. Microcontroller inputs are likely to get fried.

Incoming electromagnetic noise such as from nearby (1km) radio and television broadcasters (or very nearby cell phones) may cause unexplained upsets. Similarly, an Earth accompanying signals can reduce the likelihood of causing interference to others.

Advantage: If all of the PA0..PA7 lines are driven high (or floating), the motors will shut off. This protection is defeated if any line is driven low, in which case the sole low PA signal is overloaded.

[edit] As usual, I'm the slower typist. [/edit\]

 

[swallow_159]

Because the Power Ground and Control Groud are only connected by a single point only, means there cannot be a current flow between them.

The H-B control circuit should also include isolation of these currents from oneanother. Series-resistors between the uC output and H-B is an extra level of protection to protect the uC if the H-B failed, say it tried to send a current to the uC.

A fundemantal law of electricity is current passes through the path of least resistance. The art of design is programming this current flow.

Think about currents more, and less about voltages.

how can I get help ?

 

[marcbarker]

how can I get help ?

Sorry, I don't understand the intended context of how I've been quoted.

 

[giftiger_wunsch]

how can I get help ?

Try either reading the information in this thread and learning from it, or starting your own rather than hijacking my already overcomplicated thread.

 

[Hero999]

Are you talking about this schematic?
https://www.electro-tech-online.com/attachments/sam-mp-complete1-jpg.32312/

I don't see how it could work for the same reasons I mentioned here https://www.electro-tech-online.com/threads/basic-problem-with-transistors.95681/#post776592

Then there's the issue of the extra voltage loss caused by the Schottky which wouldn't solve the problem of other ports grounding neighbouring h-bridges.

 

[giftiger_wunsch]

Do you now understand how the Schottky diode works in my circuit?

I'm not quite sure - is it simply that when Tr4 is activated, it draws away enough of the current from the CW terminal through Tr4 and to ground, that the current through Tr3's base isn't enough to activate it?

If that's all it is, then I shall slap my forehead in disgust that all of my careful and unfortunately inefficient designs could have been avoided by something that simple all this time...

Of course, if someone had explained this at the start the schematic war could have been avoided

 

[mneary]

If Tr6 is already turned on Tr4 will not be able to turn Tr3 off to protect the circuit.

In this case you might need a transistor instead of the Schottky. I am looking bacck for the post where I showed this.

 

[giftiger_wunsch]

Then there's the issue of the extra voltage loss caused by the Schottky which wouldn't solve the problem of other ports grounding neighbouring h-bridges.

I went over the schematic very carefully and I'm fairly sure that the schottky diodes would have solved the problem of other ports grounding neighbouring H-Bridges, that was the reason for using a Schottky to connect each H-Bridge to the battery ground. But as you have shown, it can be done in a simpler way so that works for me.

 

[mneary]

If the saturation voltage of Tr4 (with the motor or Tr6 as load) is a problem in trying to shut off Tr3, the concept that I posted in #215 can be adapted.

 

[Hero999]

I went over the schematic very carefully and I'm fairly sure that the schottky diodes would have solved the problem of other ports grounding neighbouring H-Bridges, that was the reason for using a Schottky to connect each H-Bridge to the battery ground. But as you have shown, it can be done in a simpler way so that works for me.

Yes you're right, the Schottky diodes will solve the problem of neighbouring h-bridges grounding it.

The problem is the extra voltage drop and that your phantom grounding method does not work with PNP high side transistors.

I'm not quite sure - is it simply that when Tr4 is activated, it draws away enough of the current from the CW terminal through Tr4 and to ground, that the current through Tr3's base isn't enough to activate it?

Yes, you're right.

It takes about 0.7V to turn on a BJT but the Schottky diode will drop just 0.3V.

If that's all it is, then I shall slap my forehead in disgust that all of my careful and unfortunately inefficient designs could have been avoided by something that simple all this time...

Don't get me wrong, your original design idea would work, just not very well.

Of course, if someone had explained this at the start the schematic war could have been avoided

I tried to explain it to you on several occasions, obviously a verbal description wasn't good enough and needed illustrating.

In any case, the top transistor isn't strictly an emitter follower in the usual way because it's 'bootstrapped' in a sense. This is because the base drive voltage (up to 3.3 V) is higher than its collector voltage (2.4 V). This means that the loss is not ~0.9 V, but a lot lower. The simulations I did confirm this.

I get it now, because of the phantom ground, the MCU voltage is partly in series with the motor's power supply.

 

[giftiger_wunsch]

Sorry Hero, when you mentioned protection before, it seemed you were talking about a current limiter to simply limit the current which can flow through when both Tr3 and Tr4 were activated; now I see how the schottky does more or less the same job that my rather less efficient design did...

Anyway, it's certainly been a learning experience... I'll start ordering some of the components I'll need finally ^_^

 

[giftiger_wunsch]

Looks like rapid electronics are extremely sneaky with their prices... all prices are without VAT, delivery costs £5, and only becomes free when the value of the goods is >£30, *without* the VAT. I'm starting to think i'd be better off just sticking with maplin. Except I couldn't find any high-gain transistors on there.

I guess I'll have to wait until I need other stuff totalling £30...

 

[Hero999]

I did warn you about that.
https://www.electro-tech-online.com/threads/basic-problem-with-transistors.95681/#post775467

Even if you don't spend the full £30 and have to pay £5 for delivery, it still might work out cheaper to buy all the components from Rapid than from Maplin.

Have you checked out RS Components and Farnell? Cheaper than Maplin a bit more expensive than Rapid but do free delivery.

 

[giftiger_wunsch]

I did warn you about that.
https://www.electro-tech-online.com/threads/basic-problem-with-transistors.95681/#post775467

Even if you don't spend the full £30 and have to pay £5 for delivery, it still might work out cheaper to buy all the components from Rapid than from Maplin.

Have you checked out RS Components and Farnell? Cheaper than Maplin a bit more expensive than Rapid but do free delivery.

Thanks to the cheaper p&p, the lack of pre-VAT prices, and the lower amounts required for bulk prices, maplin seem to be cheaper. But as I said, I didn't see any high-gain transistors on maplin. Couldn't find them on RS Components either; haven't tried farnell yet.