My half wave rectifier

The formulas I normally use don't have R1 on them and I don't know what to do.

  • 3
    \$\begingroup\$ It's just another voltage drop based on current draw. And a total current reduction based on increase resistance. \$\endgroup\$ – Aaron Jul 13 at 19:30
  • \$\begingroup\$ Start with specs for all the variables \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 13 at 20:33

Often a fast RC product ( time constant) is used to SLOW DOWN the turnon speed.

Slower turnon will reduce the magnetic interference, which depends on

  • Vinduce = L * dI/dT

and such slow_down_the_turnon is important in the 2,000 volt rectifiers in MicroWave Oven power supplies.

I've chatted with custom_home builders who accept responsibility for success with high_end homes having both regular power wiring AND computer_network wiring AND home entertainment (music, TV) wiring. The biggest problem, for which these builders use special installation guidelines, is appliances (MicroWave Ovens) inducing horrid trash into the computer_network wiring or into the music/TV wiring. If all else fails, they will use special PowerLine filters between the MicroWave Oven and the 110/220 power, to prevent entry of 100 nanosecond Trise spikes from coupling into Computer and music/TV cables.

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  • \$\begingroup\$ Perhaps it is just an example, but the 1N4007 diode has a 1 KV 1 AMP rating, so for AC input 330 VAC is safe. Microwave ovens supply -1500 VDC 1 AMP to the cathode of the magnetron tube. Now they use SMPS to get -1500 VDC, so maybe RF filters are needed. But the supply shown is profoundly smaller. \$\endgroup\$ – user105652 Jul 14 at 2:04

Because you already have some answers to your question. I will focus on a theoretical answer assuming ideal components.

Well, let's solve this mathematically. We have the following circuit:


simulate this circuit – Schematic created using CircuitLab

Where \$\text{V}_\text{i}\left(t\right)\$ is given by:

$$\text{V}_\text{i}\left(t\right)=\max\left(0,\hat{\text{u}}\sin\left(\omega t\right)\right)\tag1$$

When we use and apply KCL, we can write the following set of equations:


When we use and apply Ohm's law, we can write the following set of equations:

$$ \begin{cases} \text{I}_1=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_1}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_1}{\text{R}_3} \end{cases}\tag3 $$

Substitute \$(3)\$ into \$(2)\$, in order to get:


Now, we can solve for \$\text{V}_1\$:


Now, applying this to your circuit we need to use (from now on I use the lower case letters for the function in the 'complex' s-domain where I used Laplace transform):


So, we get:


In order to find the time-domain represeantation we need to use the convolution property of the Laplace transform, so we can write:


Using the table of selected Laplace transforms, we can see that:




Now, using \$\text{R}_1=100\space\Omega\$, \$\text{R}_3=100\space\Omega\$, \$\text{C}=100\cdot10^{-9}\space\text{F}\$, \$\hat{\text{u}}=1\space\text{V}\$, and \$\omega=2\pi\text{f}\$ where \$\text{f}=10^5\space\text{Hz}\$, we get the following:


Plotting the solution gives:

enter image description here

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R1 combined with Rout creates a voltage divider. It is assumed the resistors have enough wattage rating to be comfy with the heat build-up.

R1 combined with Cout creates a simple RC filter. 1/(2*pi)RC = -3 db at 15.9 KHZ. In this case C is very small for a supply filter, often at least 100 uF, which rolls off at 15.9 HZ. Larger 1,000 uF capacitors would roll off at 1.6 HZ, substantially lower than 60 HZ ripple.

This circuit could be a simple wave-shaper and not a power supply, so then the - 3 db at 15.9 KHZ has another meaning not defined here. The output is a DC version of an AC input, minus 1 diode drop/2. It is not a RMS peak value as the 100 ohm resistors are a substantial load and divide the diode output by 2.

It shows the frequency source as being a 5 volt 100 KHZ square wave, so a filter with a 15.9 KHZ cut-off means the output is mostly DC with little ripple, so it is 5 volts - one diode drop, or 4.3 volts/2 = 2.15 VDC output.

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