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24

Rawbrawb's answer doesn't explain the actual mechanism by which saturation occurs, which is a fairly easy to understand: It helps to first understand how materials generate magnetic fields. A simple way to think of this is as each atom being a small loop of current which generates a magnetic field. A magnetic material has huge numbers of these loops. ...


21

Saturation is always an issue in both transformer and inductor design. If we're going to spend money on a heavy and expensive iron core, then we want to work it as near to saturation as we can. The reason inductors are gapped, and transformers are not, is that they are trying to do different things. The purpose of an inductor is to store energy. This ...


17

Best is to use manufacturers data. Test in an oscillator. (See below) Apply variable DC + AC and monitor effect on AC as DC increased. Oscillator method - 2. Above. Given: A flyback converter/oscillator (eg a typical smps boost converter) operating in 'discontinuous mode' An oscilloscope Variable load. Iin is a triangle wave plus an off period. As ...


15

To understand this you have to first understand the role of permeability in magnetic fields. When you have a material in a magnetic field that has higher permeability it intensifies the field. So a device that has a high permeability material will have higher inductance than the same device but without the material. This is a good property because it ...


13

The transistor going into saturation isn't a property of the transistor itself, but instead a property of the circuit surrounding the transistor and the transistor, as part of it. The simplest case to imagine is an NPN switch. I'll present two different such switch circuits to make the above point concretely clear: simulate this circuit – Schematic ...


12

There is a precise definition and a sloppy one for saturation. I'll start with the precise one. That's pretty much it. The saturation region is precisely defined here. The sloppy one comes about because the practical behavior of different parameters of the BJT don't all neatly fall so perfectly on those lines. Besides, those voltages aren't the only thing ...


11

The ADG708 can only handle negative inputs if it's in the -2.5 to 2.5V dual supply configuration. If you've got it hooked up to 0 and 5V, then that makes complete sense. There will be protection diodes in the ADG708 that will clamp to -0.7 or so below it's lower rail (VSS).


10

Saturation of the core depends on the core material but usually for ferrite type materials it is round about 0.4 teslas (an example is 3C90 material from ferroxcube): - Notice that despite an increasing H field (basically amps), the flux density is leveling out at about 400 mT. So what is the H field all about? The H field is the thing that makes magnetism ...


10

Why do we want gap in the core material while designing inductor? And... The only reason to leave gap that makes sense to is to increase the number of design parameters to obtain a closer resulting inductance value at the end. I can't find any other reason to leave gap. There is a major reason and it's clear from the formulas you quote: - What ...


8

You're over-saturating the transistor. In this context, saturation means \$V_{BE}>V_{CE} \$. If you imagine your NPN transistor as two diodes back-to-back as the image below shows, you can see that if you drive your base hard enough, \$V_{CE}\$ will drop to a very low voltage (e.g. 0.1 V). Since we are driving our base very hard, \$V_{BE}\$ might be about ...


8

Even for Vin=0 the output voltage will ramp up to the maximum (supply rails) due to the finite output offset voltage (input offset times open-loop gain). This cannot be avoided for stand-alone integrators. In many cases, a resistor in parallel to the feedback capacitor can limit this unwanted DC output. The corresponding DC output voltage is Vin*Acl (closed-...


8

Why is the real saturation voltage way higher than stated in the datasheet? Because you have configured your transistors as a darlington pair and no matter how hard Q4 is turned on it can only realistically shorts Q3's collector to Q3's base and, given that you need 0.7 volts on Q3's base to turn on Q3, you are left with Q3's collector at about 0.7 volts. ...


7

BJT transistor will be saturated the moment the Ic will not follow the linear relation of: \$I_c = HFE * I_b\$. Thus all we have to do is to limit the Ic from reaching this value. Since \$I_b\$ is determined by the value of the resistor connected to the base and the driving voltage on its other end, it is easy to force \$I_b\$ to any value. When ...


7

A zener diode connected from op-amp output to inverting input (possibly with a series std diode) and NOT switched by S1 plus a resistor from Vsense to inverting input will limit Vout+ excursion. If this is dual supply then back to back zeners will do the same thing symmetrically. When Vout approaches Vzener negative feedback is provided. The resistor from ...


7

A fast calculation: ON-state Ib is about 4.3mA. (= (5V-0.7V)/1kOhm). If the transistor has current gain =50, then the Ic should be 50 * 4.3mA = 215mA. R2 and +5VDC supply limits the Ic under 5V/100Ohm = 50mA. So there's at least 4x exessive Ib, the transistor is heavily saturated. Due led's voltage drop (often about 1.5V) the theoretical max current is ...


6

How can I calculate saturation current of a Toroid inductor, with a core? The manufacturer of the cores provides BH curves like this theoretical one: - The magnetizing force is easily calculated - it is how much current you are feeding into the device multiplied by the number of turns and divided by the length around the toroid: - Magnetizing force = \$...


6

I'm offering this answer as an alternative saturation detector. Pass a direct current thru the coil that can be varied from a milli amp to possibly several amps. This is easily achieved with an opamp and BJT as a constant current generator. Next, generate a low amplitude sinwave and capacitive couple it to the inductor. Measure or view the sine amplitude ...


5

A few basic calculations: The current through Rled, assuming the transistor is fully saturated: $$ I_R = \frac{3.7V - 2.4V - 0.3V}{39 \Omega} = 25.6 mA $$ Looking at a 2N3904 Datasheet, they define saturation as the point hfe=10. Thus: $$ I_b = 2.56 mA $$ This means your micro controller needs a control signal of: $$ V_m = 0.65V + I_b \cdot 3.9 k\Omega = ...


5

Under reverse polarization, the externally applied voltage V pulls the holes in the the p side and the electrons in the n side away from the junction. The width of the depletion Layer and the height of the barrier increase accordingly. The increase in the barrier energy is measured in eV. This rise of the barrier height reduces the current to a negligible ...


5

Here are the voltages as they are described: Now, if you want to calculate the resistor current, you can see that it is 9.4V / 1000-Ohm = 9.4 mA. Since the base is in series with the resistor, the resistor current = the base current. Does this help?


5

The easiest way to find reasonable numbers for the Ebers-Moll parameters is to look at existing SPICE models. For example, the BC856 ISC = 1.633E-14 (NXP model). ISC is base-collector saturation current. There's also IS transport saturation current and ISE base-emitter leakage saturation current. Deriving a reasonably accurate model from a BJT ...


5

Vout= Vin (1 +r2/r1) is the equation for a non-inverting op-amp with negative feedback. simulate this circuit – Schematic created using CircuitLab Figure 1. Non-inverting amplifier configuration. Your configuration is non-inverting but has positive feedback. This will give a Schmitt trigger effect. Remember that the op-amp output will be \$ (V_+ - ...


5

Operating at a high frequency means: - you transfer a smaller energy more times per second = same power handling as a conventional transformer. you avoid core saturation associated with running at a lower frequency Core saturation is related to the peak flux density of the core material. For a typical ferrite this is round about 0.35 teslas: - For ...


5

Using the saturated state means the output voltage doesn't depend on small variations of the input voltage. Using the saturation region (or triode region for MOSFETs) can result in very low power consumption when the gate is kept stable in the 1 or 0 state. However, there are logic families that use forward active mode for the output transistors in both 1 ...


5

LF356 is not a single-supply op-amp - ie. the input common mode range (and output swing with a load to the negative rail) do not include the negative rail. The CA3140 and LM358 are single-supply op-amps. The op-amp in your circuit can only be in balance when both inputs are at the negative rail, hence it cannot work with the LF356. As you can see from the ...


5

the negative output voltage saturates at -2.95V. Yup. If you look at Figure 2 of the data sheet, it shows a nominal maximum for +/- 5 volt supplies to be 6 volts pk-pk. In other words, about +/- 3 volts, and you're getting -3. Just about what you'd expect. The positive side is doing better than expected, but that's just an unexpected bonus. TL;DR - You ...


4

Check the datasheet of an LDO regulator. Things won't get as simple as one or two transistors though. The stadard trick is to do the comparing at a lower voltage (using a long tailed pair), and to drive a PNP series transistor. This design can be found at http://www.discovercircuits.com/Andy/Discrete_LDO_Voltage_Regulator.htm :


4

There are other things to consider. Here's the equivalent circuit of the transformer: - The extra components that need to be considered are: - Xp and Xs - these are leakage inductances and no-matter how hard you try to make a perfect transformer there will be some flux in the primary that just won't couple to the wires in the secondary - this is portrayed ...


4

Zeroeth order: You said: I tested it in the real world, on my breadboard, and it didn't work, BJTs saturated (\$V_C\$ is almost equal to \$+V_{CC}\$) . . ." No. Quite the opposite. BJTs are in cutoff. They want to be in FA (forward active) mode, i.e., the Vbe of both transistors (Q1 & Q2) should approximately equal 0.65V (approx.), just as your ...


4

To answer your second (sort-of) question first, you would normally operate a MOSFET in the ohmic (aka linear or triode) mode when it is used as a switch. The voltage drop across the MOSFET is more-or-less proportional to the drain-source current. In the saturation region, the current through the MOSFET is more-or-less independent of the drain-to-source ...


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