Originally Posted by audioguru

I think we have all seen manufactured products that are machine-made with such tiny little surface-mount parts. But can we hand-etch a pcb and hand-solder these tiny parts?
Can we buy these little parts in small quantities?

that's a good question... I just noticed the physical dimensions of these devices, and I don't like my chances of being able to solder something 2.9mm in length with pins 0.95mm apart

Since I haven't yet got myself an etching tray, etc. which I would need to make my own PCBs, I was planning on building these circuits on stripboard. Clearly even if I could solder such a small device (and I have little experience in soldering) I would need to be able to make a suitable PCB as well.

The devices suggested by marcbarker seemed to be appropriate, though perhaps I should check the dimensions in the datasheet for those items as well...

I could solder that fairly easily.

I woudln't know where to get hold of it though, I found the model on LTspice and thought it looked like a nice little MOSFET.

Originally Posted by Hero999

I could solder that fairly easily.

Please bear in mind that I have done very little soldering and I'm likely to have a lot of trouble soldering something so small, especially components which may be destroyed by the heat if the soldering isn't completed quickly. Thanks for the suggestion though.

Originally Posted by marcbarker

I chose MOSFETs for the lower transistors

I assume that what was meant here is that only the two lower transistors on the circuits need to be MOSFETs - could anyone explain why only these two would need to be replaced? And was the implication then, that the upper two transistors should be BJTs as originally discussed?


Thanks.

Giftiger, the lower fets are there because the top BJTs are activated by the (motor current /hFE) and the lower FETs are activated by the 'reflected' motor voltage coming back. If they were BJTs too, they would shunt the reflected voltage. I didn't know if it was a good or bad thing, so I changed your BJTs for fets instead.

Originally Posted by audioguru

I think we have all seen manufactured products that are machine-made with such tiny little surface-mount parts. But can we hand-etch a pcb and hand-solder these tiny parts?
Can we buy these little parts in small quantities?

I used to have amazing people working for me , they could work under a microscope and assemble prototype surface mount circuits, using bits of copper-clad blank PCB and these funny self-adhesive surface mount IC footprint bases, all wired up with almost hair-thin enamelled copper wires.

RS and Farnell here in UK sell SMD parts in small quantities, if it's capacitors you buy a strip of 50 off for about £0.50, individual transistors if the price is not less than that amount.

The good thing about the AO6407/8 is that the four of the pins all go to the drain so it doesn't matter if there's some bridging between them.

The gate and source pins are at the bottom corners of the IC whcih makes it much easier to solder them without bridging.

Originally Posted by marcbarker

Giftiger, the lower fets are there because the top BJTs are activated by the (motor current /hFE) and the lower FETs are activated by the 'reflected' motor voltage coming back. If they were BJTs too, they would shunt the reflected voltage. I didn't know if it was a good or bad thing, so I changed your BJTs for fets instead.

Sorry maybe I'm being dumb again but I don't understand what you mean I thought the transistors are activated by the current from base to emitter, or the voltage between gate and source in the case of MOS-FETs... how can they be activated by the motor current?

I ought point out that I've not 100 % analysed the circuit, so there has been some assumptions that I've made. In true engineering style I've actually made an 'engineering trade-off' in my understanding of the circuit, it's not perfect. I will tend to of built a circuit by around this time, because experience has taught me that analysis and discussion can only take us so far, and real hardware is needed as a 'sanity check' every so often.

I believe that the top BJTs will have a current passing base/emitter that will switch on the transistor (and connect the battery + to motor), I expect something like 2 mA. I think the transistor will behave as a 'common collector' or 'emitter follower'. (I wonder if I'm right!). The same voltage potential (originating from the logic H) somehow, finds it's way to the lower FET gate and turns it on. The "reflected voltage from the motor", what I mean is the + voltage on the motor.

Your schematic looks roughly like what I am expecting, but I'm still not really sure I understand why the MOS-FETs are required at the bottom, if they're not required at the top. I thought this was mainly because the microprocessor output would not have enough current to drive the BJTs fully on...

In your schematic, I'm not sure how you arrived at the conclusion that the current on the microprocessor circuit will be ~2mA either. You have inserted 100ohm resistors, and if fully on, the transistors should have more or less negligible impedance values between emitter/collector and drain/source, the motor having a resistance of approximately 5ohms, so also pretty much negligible. So the current should be approximately 3.3 / 100 = 33mA. This is why I planned to use a higher-value resistor.

I believe you're right about the BJTs, they should behave as a 'common collector'.

I just had another look at the datasheet for the MOS-FETs marcbarker suggested, and they have similar dimensions to those suggested by Hero999. It looks like I'm going to have to buy the equipment to make my own PCBs, and I'll most likely have difficulty in soldering these components on even then. Perhaps I can find some IC-holders of the correct dimensions so that I don't have to worry about heat sensitivity while trying to do this fiddly soldering.

Originally Posted by giftiger_wunsch

1.why the MOS-FETs are required at the bottom, if they're not required at the top.

2.[why] current on the microprocessor circuit will be ~2mA either.

3.You have inserted 100ohm resistors,

4.motor having a resistance of approximately 5ohms, so also pretty much negligible. So the current should be approximately 3.3 / 100 = 33mA.

5.This is why I planned to use a higher-value resistor.

1. A mosfet won't operate as an emitter follower

2. emitter follower base current, a function of Ic(200mA) / hFE (100 guess)

3. No I didn't add them, they're already there, 100 ohm res. are the inherent output source impedance of the uC

4. I thought the motor took 0.2 A? I suppose the motor stalled will be 5 ohm.

5. Sorry you've lost me! what higher resistor?

Originally Posted by giftiger_wunsch

It looks like I'm going to have to buy the equipment to make my own PCBs, and I'll most likely have difficulty in soldering these components on even then. Perhaps I can find some IC-holders of the correct dimensions so that I don't have to worry about heat sensitivity while trying to do this fiddly soldering.

I've posted this link to Analogue Breadboarding from AD
scroll down to fig 20. They're called "Wainwright Mini-Mount". If you can still get them. Fig 21 shows something made with it.
Analog Devices : Rarely Asked Questions (RAQs) : Analog Breadboarding

Where did you find out that 100ohm is the uC's output source impedance?

4 - sorry, I confused my figures. My friend measured the resistance of the motor at an average of about 5ohm, then found the current was about 200mA while running, so yes it was about 5ohm while stalled. 200mA from 3V = it only reaches about 15ohm while running anyway though.

5 - I discussed in previous posts that to limit the output of the microprocessor to its max recommended current of 8mA, the resistance of the circuit should be around 3.3 / 0.008 = 412ohm. Since most of the components have much lower resistance than this anyway, I will use resistor(s) adding up to close to this value. If you are right about the uC's input impedance of 100 ohms, I'll subtract this from the resistor value.

Best place for the resistor would probably be the ground portion (green on my original schematics), since all of the uC current will pass through this.

Originally Posted by giftiger_wunsch

Where did you find out that 100ohm is the uC's output source impedance?
<..> Best place for the [400 ohm] resistor would probably be the ground portion (green on my original schematics), since all of the uC current will pass through this.

Using data from Table 28.1 in datasheet:

"Vol= 0.4 V @ 8 mA" = equivalent source impedance '50 Ohms' (not 100 as drawn)
"Voh=2.4 V @ 8 mA" = e. s. i. '110 Ohms'

If you add "8 mA" limiting resistors, I don't think the circuit will work any more.

Why does the uC output need to be limited to 8 mA? You don't have to operate the outputs at 'data-sheet specification', because you're not driving another IC. It won't blow it up if there is a momentary short either. The output current into a 'short circuit' (stalled motor) will be 20-25 mA.

If I don't limit the current to 'datasheet specification', then I'll have to be careful about using the other parallel ports for other purposes at the same time, since it was established earlier in the thread that the average current through all the ports shouldn't exceed 8mA. If 8mA is sufficient to run the BJTs and MOS-FETs, there's no reason to exceed the recommended maximum current through the port.

If my understanding is correct, the base current only exceeds 2 mA (guess) if the motor is stalled. A stalled motor for a significant time will flatten its battery, or damage the motor. And I think that'll happen before the uC get hurt.

Talking of other IO ports... Are you able to 'tri-state out' the uC ports that drive the other motors? And I don't know if you can drive more than one motor simulataneously when they all share the same motor battery. I might be wrong though.