67

Probably you already have it, and just didn't know it. If you are driving a motor with a half-bridge or H-bridge and PWM or similar, you have regenerative braking. Let's consider a half-bridge, since for this analysis we will run the motor in only one direction: First, let's consider non-regenerative braking. If the bridge output is high (S1 closed, S2 open)...


34

In short: You have linear control of the 'speed' by applying a pwm signal, now the frequency of that signal has to be high enough so that your DC Motor only passes the DC component of the PWM signal, which is just the average. Think of the motor as a low pass filter. If you look the transfer function or relationship angular speed to voltage, this is what ...


19

Not a ground loop. But a short circuit: Always use fuses with batteries. Instead, you can make it like this, so that there is only one common ground: Drawback is unbalanced load on the battery. Meaning the left battery depletes faster, possible damaging it when the set is deep discharged. I recommend a battery balancer, or a 24V-12V power supply instead.


18

For low voltages, it seems like the DRV8837 is pretty good: - With an 800mA load, the volt drop is: - \$I_O\cdot R_{OS(ON)}\$ = 800mA x 0.33 ohms = 0.264 volts. At this current, the power dissipation will be 0.8 x 0.8 x 0.33 watts = 211 mW. Compare this with the L293 power dissipation at about 800mA - maybe about 3V is lost giving rise to a power ...


16

The PWM frequency supplied to a (presumably) brushed DC motor needs to be high enough that the combination of mechanical inertia and inductance of the coils is sufficient to smooth out the mechanical impulses of each pulse. This minimum would differ from motor to motor. Too low a frequency, and the motor motion will be perceived as a series of jerks, or a ...


15

The triangle is a symbol for a buffer, driver or amplifier, whatever you call it. The transistors drawn inside the triangle are there to let you know what is the type of your output. Thanks to this drawing you can see that the outputs are switched via BJTs between the power supply rails using some kind of Push-Pull topology. Knowledge about the type of ...


15

Flogging the FREDs Voltage fed converters with transformer isolation will exhibit ringing in the secondary. Ringing is caused by parasitic inductances and capacitances in the circuit, with the dominant elements will being the transformer leakage inductance (\$ L_ {\text {Lk}}\$) and junction capacitance ( \$ C_j\$)of the bridge diodes. The diode data ...


15

Recently, I bought the H-Bridge VNH7070BAS from ST because it would drive up to 15A It will not drive anything like this level of current for anything more than a few micro seconds before the automatic current limit circuit operated. The internal transistors together contribute circa 0.1 ohm impedance in the current path and, at (say) 10 amps continuous, ...


14

Each pin on an Arduino can handle 40 milli amperes, not 40 Amperes. That too in ideal conditions (temperature, Vcc), actual allowable current can be a fair bit lower. Besides current limitation of the Arduino pins, a key failure condition is the LDO voltage regulator on the Arduino board, which will overheat and can get destroyed if it is made to source high ...


12

Try probing on the power supply rail. I bet you see those spikes on there. It'll be due to the lead length between your bench supply and the MOSFETs. Clearly you won't see it on the lower FET side because your scope is referenced to that rail but, if you probed back at the power supply I bet you would. Try a 1uF or 10uF ceramic across the power rails close ...


11

I think you will encounter two challenges in designing this. The first is getting the high side to turn on. The source of the high-side moves up and down relative to ground, so a voltage source referenced to ground won't be able to drive it without exceeding the maximum gate-to-source voltage \$V_{GS}\$. For RFG50N06, this is 20V. The second is that you ...


11

Your first concern in selecting a gate driver is to find one that can drive enough current to switch your selected MOSFETs fast enough for your application. As a rough estimation, you can divide the total gate charge of your MOSFET by the current the driver can sink/supply. $$ t_{on}=\frac{Q_g}{I_g}$$ Using the worst-case values for IRF1405 and the slower ...


11

At minimum, you need to use a frequency so that the motor "sees" the average and doesn't react to individual pulses. That is usually a few 100 Hz. However, there are other effects that the motor doesn't care about but you might. Individual sections of wire in the windings may vibrate slightly with the PWM frequency, which causes audible whine. This is ...


11

How are you decoupling the 12V supply? One possible failure mode is that inductive spikes from switching off the motor current (i.e. at the PWM rate) are dumped into the 12V supply via the flyback diodes. Yes, that's supposed to happen, but... If the 12V supply is not decoupled, and is sourced from a PSU not a rechargeable battery, or is sourced via a long ...


11

The diodes serve two distinct purposes. Under regenerative braking, they return the generated voltage to the power supply (where with suitable electronics, it can be used to recharge the battery). Note that unless the motor is being run above its normal speed, the generated voltage will be no more than the supply voltage, so it is within the voltage rating ...


10

The transfer function makes sense only for linear systems (in summary, systems for which \$f(a+b)=f(a)+f(b)\$ and \$a\cdot f(x) = f(a \cdot x)\$). Yours is not a linear system, because it contains nonlinear elements (the diodes, due to the exponential relationship between voltage and current). You can linearize the model at the neighborhood of a certain ...


10

The datasheet shows this for a reason. If you are not completely sure what it's purpose is and know that you don't need it, then you have to include it. In this case, the FET driver is intended for N channel FETs both top and bottom. When the top FET turns on, it's drain and source voltage will be close. After all the purpose is for the FET to act like a ...


10

Q1 and Q3 turn ON when you connect +5V to their resistors. Q2 and Q4 turn OFF when you connect +12V to their resistors. Connecting 5V to R2 or R4 will result in the transistor turning on. So, if you connect 5V to R3 and less than 12V (or whatever the power supply voltage is) to R4, both transistors turn on creating a short circuit. If you want to turn ...


10

I first saw the following circuit in a TI app note many years ago. It works well and is robust. I have used it or variations of it in several projects (and products). simulate this circuit – Schematic created using CircuitLab There are a couple of things to consider: 1) This circuit inverts the drive signal. 2) R3 is added to ensure the FET drive ...


10

To use this circuit with a 12V supply R5 & R6 must be rated for at least 0.25W power and the 100 ohm resistors must be rated for at least 2W power. Depending on the transistors used these 4 resistors (100 ohm) may need recalculation to avoid excessive base current and power dissipation. However this circuit will not work. One problem is that it ...


10

If the motor is producing power, the net power into the motor must be positive, so the net current out of the batteries must be in the direction that drains them, so you are fine. If the motor is being regeneratively braked, then power can flow out of the motor and can push the supply voltage up and charge the batteries (this is used to advantage in ...


9

This is a classic snubbering problem. A diode can't instantaneously go from conduction to blocking; the charge in the PN junction needs to get swept out, and an RC snubber across each diode should help this. I used to design industrial soft starters and on the medium-voltage units we had a lot of design work around this particular aspect. It's been a long ...


9

The reasons are different depending on if it is a BJT or a MOSFET. The effect, though, is the same - it's there to reduce current and protect the IO pins on the controller that are driving the transistors. On the BJT the base -> emitter junction is essentially like a diode, and without a resistor would be like a near short circuit. The resistor stops the ...


9

A TEC is polarized in the sense that how it is connected matters. If you want to be able to heat and cool an element then a full Hbridge will work. This will allow you to pump current both directions across a TEC's terminals. If you apply a positive voltage to a TEC in one polarization then side A will get warm and side B will cool. If you then reverse ...


9

P-channel MOSFETs tend to have a higher Rds(on), making them less efficient (for the same price). For a small H-bridge, the simplicity of using them makes them practical However, for high power applications, the extra complexity of driving N channel FETs can be justified by improved efficiency from lower Rds(on).


9

Your motor is likely geared down, because 150 rpm is only 2.5 revolutions per second. At 50 rpm, your motor will require more than a second to perform one revolution. That having been said, the switches in your h-bridge don't dissipate much power when they are on (essentially zero volts) or when they are off (zero current). They only have both voltage and ...


9

Your P channel MOSFETs are connected upside down: - Notice the little diode symbols inside the MOSFETs? They will permanently conduct current and the MOSFETs will not switch off.


8

There seems to be some confusion as to the power requirements. This device does not need to dissipate the energy produced by a 1kV drop at 50A; that power is dissipated by the load. This device will operates in two states: 1kV blocking at milliamps (or less) of leakage current, and 50A of current at milliohms (or less) of resistance. That results in ...


8

I'm not sure why you think BJTs are significantly slower than power MOSFETs; that's certainly not an inherent characteristic. But there's nothing wrong with using FETs if that's what you prefer. And MOSFET gates do indeed need significant amounts of current, especially if you want to switch them quickly, to charge and discharge the gate capacitance — ...


7

Yes, you can definitely parallel the two outputs of an L293D. I made a few stepper drivers based on L293D with parallel outputs and had no problems. As this application note from ST Microelectronics (APPLICATIONS OF MONOLITHIC BRIDGE DRIVERS) states: Higher output currents can be obtained by paralleling the outputs of both bridges. For example, the ...


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