Gate driver, will it work ?

[tom911]

Hi,

Just wondering why in circuit below, the voltage keeps flowing constantly if its not open at all times ?
(cant simulate it properly and the designer doesn't know what's going on that simulation)

View attachment 62799

Trying to deliver at least 9V at to the output.
Clock+bjt represent the operation of optocoupler, (this simulator) doesn't have optocouplers and thus I've improvised.

We have actually build a circuit but the lab will not be open during the weekend and thus I can't wait to see how it failed

Thanks for feedback and directions.

 

[duffy]

Voltage doesn't "flow", current flows. You built a voltage divider with three resistors from +12V to ground, current will flow through that regardless of the state of the transistor.

 

[tom911]

You

I'm not sure If I've understood fully, but the voltage doesn't flow when switch is off, the current does. Because the output is "on" only when the voltage will be flowing therefore the circuit should work.

Thanks for the answer it has cleared a few points I will still try to make my own conclusion with oscilloscope but its good to know what I should expect.

 

[duffy]

No, voltage doesn't "flow", ever. Current flows. The yellow dots show current. The current will flow regardless of whether the transistor is conducting or not, because you built a voltage divider. Remove the 4.7K resistor. Now you don't have a divider. Watch what happens on the negative cycle.

 

[tom911]

So I've re-looked at the circuit in the form of resistors only and applied a simple Voltage calculations using:
Vout = Vin * (R2/(R1+R2))

Old circuit:
This gives me a 11.45V and 67mA out in the first Vout point (between supply and npn).
The second divider gives and output voltage of: 10.7V and current 2.3mA which is too small.

New circuit:
If I would remove the 4k7 Resistor I would have a 35mA with a 330Ω resistor between output.

If I would replace 330Ω with 120ΩResistor I would have 11.4Vout and 95mA which Is strongly recommended, therefore I have re-wired circuit to the following form:

Falstad simulation

The simulation shows 10.8Vout and 90mA.

I don't see the reason now why the circuit would not work. I believe that I've correctly understood the voltage division now, but with my limited knowledge I think there is more to it

Thanks for replay.

 

[tom911]

Hi again,

View attachment 62924

I've tested circuit and after 3h of trying to find a fault I've gave up.
Vin = 12V

Point A (Ground at that point) = Green dot showed no result on oscilloscope but Supply generator has changed to from 0.00A to 0.03A (30mA was also a calculated value for output current)
Point B (ground at that point) = Red dot showed no result on oscilloscope but supply generator has changed from 12V to 1.4V and 0.17A which suggest a short circuit, however there was no short circuit anywhere between those points.

I've tried to put a temporary link between Npn, testing with ground point at whole length of the cct and yet, after the junction point it also showed a SC result.

This therefore leads to conclusion that whole circuit doesn't work or I've done something wrong with my measurements.

6 of such optocouplers are grounded in parallel but I still don't believe that it could cause any problem as there is no SC anywhere between them.

Thanks for any kind of feedback.

 

[alec_t]

Your circuit in post #6 has nothing driving the base of the (presumably NPN) transistor.
If you short point A (green dot) to ground the current from the 12V supply will be ~ 36mA if the itself supply has negligible resistance but 30mA if the supply has an internal resistance of 62Ω.
I don't understand "Point B (ground at that point) = Red dot showed no result on oscilloscope but supply generator has changed from 12V to 1.4V and 0.17A". I would expect a current of ~0.17A only if you short the collector to ground and the supply has an internal resistance of 62Ω.
If you're trying to make a gate driver why do you have a resistor in the emitter lead? Try removing that, use the 180Ω for a collector resistor instead of the 8.2Ω, and use a 3k3 resistor in series between your pulse source and the transistor base.

 

[tom911]

Thank your for replay alec but today I've re-changed the design, making it a voltage follower from Emitter and It worked on a control cct (a single opto) but it didn't work with parallel connection of optocouplers.

My task for today is to determine what could be cause of no emitter voltage in paraller connection.

This post is more like a 'dairy' but it keeps me on track at all times and I've a chance to receive some feedback =)

Control circuit:
View attachment 62995

Actual 3-phase circuit (should be 6 optos but 3 should be fine to visualize):
View attachment 62991

Output waveform on 1 opto circuit:
View attachment 62994

An output clearly followed the output with new design without any problems, the voltage levels on oscilloscope however didn't reflect the acutal voltage values, but I assume that Vout calculation is correct.

Possible reason for unsuccessful operation:
- A parallel connection of Opto-couplers does not provide sufficient current drive to LED in one of the optos which therefore will not switch the gate on, leaving Vmax on the collector side.
- Control driver had a floating Base connection and thus in acutal 3-phase it might have caused some malfunction of voltage division.

Possible solution:
- A current divider analysis.
- remove Base resistor

 

[alec_t]

Why do you want a voltage follower? Since the base voltage is 5V max the emitter voltage (Vout) should be 5-0.7 = 4.3V max.

 

[duffy]

Why do you want a voltage follower? Since the base voltage is 5V max the emitter voltage (Vout) should be 5-0.7 = 4.3V max.

Probably even worse if he's driving it with an opto.

 

[alec_t]

@duffy
As far as I can gather the OP is trying, ultimately, to simulate the use of six optos in parallel to drive a FET gate. But I may be wrong

 

[tom911]

Today was the last day to demonstrate the working cct but I will keep working on it to make it fully work.


@Post 9 and 10:
Yes I've realized that this is a voltage follower photo-BJT (its an optocoupler) which is very limited however I don't understand why the low-side switching (post1) would work in this cct. On falstad it supplies 'high' to the output at all times the same was with the actual physical circuit measured with digital oscilloscope and thus I don't want the output to be on at all times it should be also noted that Opto is PWM driven and we had no time to re-design the low-side switching so we moved to a high-side design.

Why voltage follower:
- non-inverted signal
- I've found post yesterday that not the same rules apply to photo-bjt as for base BJT but I couldn't find exactly how to calculate what voltage will be on the output so I assumed a that its gonna behave like a voltage divider in ON condition.

Today's results:
In cct in post 8 I've replaced 180Ω with 100Ω which 'theoretically' should give me 110mA and 11.07V, this would satisfy the gate of mosfet incredibly well I think, but the result was different.

Testing cct:

View attachment 63024
A voltage drop across LED was 1.2V which was correct however the optos were not ran at full supply current i.e. The LED should be driven at least with 10mA but pulse generator supplied only 8mA that's why we concluded very low output on BJT side of the mosfet.

The output on BJT emitter-side was around 0.8-0.9V and we assumed that its because of the not sufficient current going to opto but this will be further analysed.
The other reason might be the fact as you mentioned above, common collector phototransitor theory might be wrong which need to researched in detail.

View attachment 63021
Output measured at point before the PIN1 but after resistor on optocoupler.
Output measured at the Gate of the mosfet S1

The actual value to the LED was 1.12-1.32V but it just keep giving different values at all times.
The value on Gate leaves many question and was oscillating between 0.1-1.2V which leaves even more questions.

Overall I'm happy that the Gate was supplied in respect with the input impulses, its not the matter of analyzing the circuit and where mistakes has been done.

 

[alec_t]

The LED should be driven at least with 10mA but pulse generator supplied only 8mA

What is the current transfer ratio of the opto? If it were only 1, for example, then you'd only get 8mA flowing in the output transistor.

The value on Gate leaves many question and was oscillating between 0.1-1.2V which leaves even more questions.

If the gate is not driven properly there may be ringing and the FET could oscillate at high frequency.

 

[tom911]

What is the current transfer ratio of the opto? If it were only 1, for example, then you'd only get 8mA flowing in the output transistor.

Its 100%.

But the fact that there is only 1V going out from emitter makes it totally useless I'm pretty sure that this topology will not work and I will back to this thread after my exams probably.

 

[tom911]

So I've submitted my final report and I came up with 2 conclusions why the circuit didn't fully worked:

1. common-emitter topology of phototransistor simply doesn't allow flow of voltage from Collector to Emitter no matter if its ON or OFF. 8 mA have flew from LED to phototransistor (assuming 100% CTR of opto) and thus voltage of 800-1000mV has been shown across gate of the mosfet (desired value was 10V).

2. 4N35 requires minimum of 10mA to operate properly where is our usb output could supply maximum of 8.5mA which might suggest a limited operation of phototransistor between collector and emitter.

I will work on those two possible issues after exam, but I'm pretty sure that the only fault was a common-collector connection of our phototransistor.