5
Wider traces exhibit less resistance, but more significantly less inductance. The lower resistance and inductance helps a little bit with efficiency, and a little more with ripple, but the larger gains are in EMC. A lower inductance trace radiates less, particularly in conjunction with a solid ground plane.
4
A regulator is not a good option. Too slow and possibly not stable without capacitors.
You could use a resistive divider, but another option is to use a purpose-designed level translator chip such as the 74LVC1T45, which can work from 1.65V to 5.5V in either direction. When converting 5V to 3.3 it has a propagation delay of less than 5.4ns over the whole ...
3
If you want to solve the question manually, you need to do iterative calculations:
You start with assuming the zener isn't there, calculate the voltage drop across R2.
If the voltage drop across R2 is smaller than the rated clamp voltage (5.1V in your case) of the zener diode, you can stop and neglect the current through the (leakage) zener diode
If the ...
2
It is a lot. Even large packages have a listed thermal resistance from junction to ambient of about 65K/W. Small packages are more than double that. Try not to use a linear regultaor step something down by more than a few volts.
Your LDO is listed as 63K/W or 235K/W depending on the package you choose. That's really hot.
2
The resistive divider solution so the simplest one. Go ahead with those. If you see loading effects (sagging of signal transition) you can also opt for active clamping using zener diode or BJTs (consider the inverted logic, if it going to MCU it can. be handled, else invert it again..or go with Zener clamping or dedicated level shifter IC) depending on what ...
2
An LDO would probably not turn on in 100 ns (note standard SI units). Instead you can probably use a voltage divider.
simulate this circuit – Schematic created using CircuitLab
Figure 1. A voltage divider to convert a 5 V signal to 3.6 V.
The resistor values were chosen for easy calculation. You can scale up or down but watch out for loading effect ...
2
In this case, can I use LDO to drop the voltage?
No because virtually all (LDO or non LDO) regulators require input and output capacitors and, all voltage regulators are far slower in operation than that needed to main the signal's shape and edge integrity.
Use a resistive potential divider.
2
Trace at point A carries high di/dt current so it should have low inductance to the decoupling capacitor to minimize voltage spikes.
\$ e = L \frac{di}{dt} \$ and in dc-dc converters di/dt is pretty high so you want to minimize L. A few amps switched in a few ns with a few nH inductance = VIN drops a lot when it switches.
I've had a case where this trace ...
2
You can consider the drop out voltage parameter to be a design constraint. For a output voltage of 5 V, 6.5 V should be the minimum input voltage.
Please note that, the 1.5 V is the drop considering load current of 1.5A. for a lower load current, the drop will be lesser.
2
Is it feasible? It really depends on your definition of feasible.
First, let's talk about the heat sink. Any suitable heat sink you might buy would cost much much more than simply buying a commercial, off-the-shelf high current bench power supply, so you'll need to already have a heat sink roughly this size on hand:
If you do, great! You have passed the ...
2
Let's see what size Cap you would need for 100Hz.
Ic=CdV/dt for dt=10ms and dV~10% input ripple of 34V ~ 4V
C=Ic*dt/dV = 40A * 10ms/4Vpp = 100 mF = 0.1 farad @50V
I don't think you planned on this.
Above is the discharge equation.
The charge equation depends on the duty cycle which happens to nearly equal the ripple assumed to be near 10~15% of the DC ...
answered Feb 9 at 0:55
Tony Stewart Sunnyskyguy EE75
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2
The simplified calculation for \$V_{out}\$:
\$V_{out} = 1.25 (1+\frac{R2}{R1} )\$
Is only accurate when the current through \$R_1\$ and \$R_2\$ are the same.
When that's not the case (when \$I_{Adj}\$ is not much smaller than the current through \$R_1\$) then the calculation will be slightly wrong, we would need to use the formula form the datasheet:
\$...
1
But I'm not so sure if it's ok to directly connect buck converter and the LDO to a single 20V,5A power supply rail.
It depends on the behavior of your supplies and of the rail, but in general that's not a problem.
What could be a problem even with just the one converter to 18V is that if that rail is coming from a switching regulator that's does ...
1
The open circle is an unlabeled terminal for the input voltage. The input should be at least the output rating (5V) plus (per the HT1050 data sheet) the 300mv dropout voltage. Call the input range 6v to 12v to be safe.
The symbol (mis)labeled 9V is a polarized capacitor. Again, per the HT1050 data sheet, 10uf @16v should be fine.
The ground symbol you ...
1
If you look at the thermal resistance figures in the datasheet you can calculate the resulting die temperature rise.
It’s not remotely practical.
Use a switching regulator. You can buy modules from Murata and others if you don’t want to worry about choosing the inductor etc.
1
Choosing a higher value for R2 does not make Iadj any smaller.
Iadj is the regulator's quiescent operating current (and could vary from one part to the next) - it is not determined by the value of R2, but by the regulator IC.
As you can see in the formula for the output voltage, you need to account for Iadj in your calculations, but you need to remember that ...
1
Look at the front page of the data sheet: -
The internal switch that causes the inductor to "charge up" is limited to 250 mA for the TPS61041. This means that going "balls-out", the maximum power that can be taken from the 3.3 volt supply is 0.825 watts. Inefficiencies might mean only 0.75 watts is delivered to the output load.
You want to drive 40 mA at ...
1
Your situation needs caps for three purposes.
Hold up. Since you are using rectified AC, you need to hold up the voltage until the next sine wave. An approximation with margin is to assume that you need to hold up until the next peak. This is also referred to as bulk capacitance to keep the ripple to an acceptable level (as Tony described in his comments).
...
1
You have fairly unreasonable demands for your power supply. As pointed out in the comments, your specs would require a linear supply to dissipate 1.4KW under the worst case condition of output voltage and current. This, as well as the schematic included, indicates that power supply design is not really your area of expertise.
This module appears to meet ...
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