Hi spec, I put together on a bread board the opto section of the schematic and used LEDs on the Opto's output emitter along with the 5k6 resistor to test it out and it passed with flying colors. they worked just as they are illustrated on the schematic.
Good news kal.
1-Mosfets: The absolute maximums of gate voltage is just that, maximums but the designed voltage should be determined by Rds On voltage? So for my N-Channel NTE2395 of +-20V maximum has an Rds-On of .028 Ohms @ Vgs of 10V+ and Id of 31Amps (is that amperage correct?). So how did we determine that 12V is best for this circuit?
Yes, the absolutes VGS of +- 20V is, as you say, the maximum voltage that can be applied without the danger of damaging the NMOSFET. Incidentally, if you did exceed this voltage you would probably break down (punch through) the extremely thin silicon oxide insulating layer between the gate and the D/S channel.
Indeed, the RDS-on is important, especially at high currents- the lower the RDS-on the better. The RDS-on of the NTE2395 would typically be twice the value at 25 deg C because the MOSFET will normally be heated by the power dissipation resulting from, IDSS * VDS, so that will give you a working RDS of around 56m Ohms. Suppose you wanted to conduct 36A, that would give a voltage drop of 2V. This would give a dissipation of 2V*36A = 72W. From this you can see that at this high current a low RDS-on is critical to the application.
On the other hand, take the case of your application. The maximum motor current is 3A. 3A*56m Ohms= 168m V. The NMOSFET power dissipation would be 168mV * 3A= 505m W.
So a low RDS-on is no longer that critical.
To get an RDs-on of 28m Ohms, as you say, requires a VGS of 10V or more. So, in theory, a VGS of 10V to the limit of 20V would do the job.
I just chose 12V, because it was a bit more than 10V and it meant that the resistors in the circuit would be the same value, ie 2K2. Also as the two 2K2 resistors divide the 24V of the supply rail in half, there is no danger of exceeding the max VGS of the MOSFET, so protection Zeners are not required.
In summary, a high power MOSFET, like the NTE2395, means that there are few critical design constraints for a relatively low current application like you motor drive H bridge. In fact the MOSFETs will think they are on holiday!
The data sheet for the NTE 2395 is not comprehensive, that is why you are having trouble deciphering what is going on. Also NTE products would not be a first choice, like Fairchild, International Rectifier, and Vishay, for example.
Does the MOSFET conduct current between drain and source at different voltage levels?
Because of the deficiencies of the NTE2395 data sheet, it would be better to refer to a proper MOSFET data sheet to answer this question, so take a look at the International Rectifier IRF540N. NMOSFET:
http://www.irf.com/product-info/datasheets/data/irf540n.pdf
Yes, just like a valve (tube), JFET, and BJT, a MOSFET conducts current proportional to an input control signal. With a MOSFET the control signal is a voltage between its gate and source (VGS), and the controlled current (IDS) flows from the drain to source The gate is insulated from the rest of the MOSFET so, for practical purposes, this means that no gate current flows. From this it follows that the power gain from the gate to the drain is infinite- a valve is the same, but as a BJT requires base current to make collector current flow, the power gain is not infinite, even at DC (note that as the input signal frequency increases parasitic capacitors, especially, have an effect which lowers the input resistance, which would now be called input impedance, of all devices. Once again this is a big subject, which you need not worry about at the moment).
If you look at the graph of Fig 1 on the IRF540N data sheet you can see ID plotted against VGS against drain source (VDS) with a junction temperature (TJ) of 25 deg C (junction is a misnomer here because there is no junction in A MOSFET, but the term is a convention)
Take the plot with a VGS of 4.5V. Although not shown on the graph with DS=0V there will be essentially no ID. As VDS is increased so will ID in proportion, so at this stage the D/S is acting like a resistor. But when VDS reaches around 1V, and IDS has reached 10A, IDS no longer increases but stays essentially constant right up to 50V, as shown on the graph. This second phase is called the drain current saturation region and is where the MOSFET would normally be operated as a linear amplifier.
You will notice that there is a whole family of plots with VGS ranging from 4.5V to 15V.
There's a "Note 5" beside the Rds-On rating that states : "Pulse width <= 300<=s; duty cycle 2%.".
If the manufacturer did high current/high voltage measurements at DC the device, a MOSFET in this case, would be destroyed by over dissipation. So instead they only apply the high current/voltage tests for a short period and the leave plenty of time for the device to cool down before applying the test again. In this case the on to off ratio is 1:50. This has nothing whatsoever to do with the switching characteristics of the device.
I intend to supply pulse in the range 20-60 Khz with 50% duty cycle. What's that going to do to the MOSFET?
That should be fine but why as high as 60Khz? I may have to add some speed up components here and there and adjust some values to optimize speed, but probably not.
2-Resistors: I understand the resistors in the schematic to be of two functions, voltage dividers and current limiters.
That is correct.
There are different resistors at the bases of same transistor type with the same specification. Why would one have a 10K and the other [2K2]
All the transistors in this circuit are either off (no collector current) or voltage saturated. Voltage saturated is where the collector current causes the whole supply voltage to be dropped across the collector load resistor, so that the VCE is essentially 0V.
To get this saturating collector current to flow you need to determine what current is necessary to saturate the collector by Vsupply/Rcollector. Having found the saturating collector current you then need to find out how much base current you will have to inject into the base to generate the collector current by Ib=ICsat/hFE. You normally push more base current than the bare minimum typically twice upwards. Data sheets, especially when specifying VCEsat, often say that IB should be VI sat/10. For most transistors Ib= IVsat/20 is a good figure. For a high hFE transistor, like a BC546 IVsat/100, would probably be good. One problem of having too much excess base current is that the transistor switching speed is slowed. In fact saturating a transistor slows it in general., but that will not be a problem for the sort of switching speeds you are considering.
The BC546 has an absolute maximum of 100mA and Hfe of 110-800. I think you used and number of 300 for a typical Hfe
Yes: for non critical applications, I would tend to assume an hFE of 300 on an initial design, but that high figure is only when a BC546/BC556 is in the linear region and not saturated.
Q33 and Q31 will be turned on at the same time with 24V+ signal and they have a 10K and 5K6 resistors respectively. The moment both transistors are turned on the resistors will have a current path to ground through Base-Emitter which will form a parallel circuit for the resistors with effective resistance being 3590Ohms @ 24V+ there will be a 6.6mA the base. 6.6x300= 2Amps. BUT the 270Ohm resistor will limit that collector current to a reasonable 18mA. How much current will be flowing through the Base-Emitter of the same transistor, the same 6.6mA, 2A or 18mA?
I think you mean: '
Q32 and Q31 will be turned on at the same time when ANTICLOCK goes to 5V. Q32 has a 5K6 resistor in series with its base, while Q31 has a 10K resistor in series with its base. Why are the base resistors different?'
Q32 collector is required to pass 10mA through the optocoupler LED. At this current Q32 collector, will be saturated by design. Assume that the hFE of Q32 when saturated is 100, them Q32 base current must be at least 10mA/100= 100uA. When ANTICLOCK is at 5V the base of Q32 will be 600mV as a result of Q32 base current. Thus, the base resistor (5K6) will have 5V- 600mV= 4.4 volts across it. This means that Q32 base current will be 4.4V/5K6 Ohms= 786uA. This is more than enough to ensure that Q32 IC supplies the 10mA required by the optocoupler LED.
By the same token, if CLOCK is also at 5V, 786uA would be flowing itno Q30 base thru R37. But in this situation, Q31 needs to divert the 786uA down to ground to protect the MOSFETs from all turning on at the same time. The hFE of Q31 is again 100 so Q31 needs a minimum base current of 786uA/100= 7.86 uA. But Q31 base resistor is 10K, so Q31 base current would be 4.4V/10K= 440uA. So Q31 will be highly saturated. This illustrates why the same transistor can have different base resistors. Q31 base resistor could have been as high as 4.4V/7.86u A=560K Ohms and still perform its task of diverting Q30 base current.
You may ask, as Q31 only requires 7.86 uA why hit it with 440uA. The reason is that it is unwise to have to higher resistance values or switching speed will be reduced and the circuit will be more susceptible to pick up and interference. Don't concern yourself with this aspect at the moment; it is a big subject, but not that difficult to fathom.
Is any of that correct or am I just hallucinating
Not at all- you ask some searching questions, which indicate that you are on the right lines and are keen to get an understanding of the detailed working of a circuit. If you carry on with this approach it will not be long before the operation of circuits like this will be second nature to you. Then you can advise others on ETO who ask questions about their circuits.
will measure again in the morning.
