How to Convert an Inverter to an UPS

An inverter is an equipment which will convert a battery voltage or any DC (normally a high current) into a higher mains equivalent voltage (120V, or 220V), however unlike an UPS inverters may lack one feature, that is these may not be able to switch from mains battery charging mode to inverter mode and vice versa during grid power failure and restoration situations.

Converting an Inverter to UPS

An inverter can be easily converted to an UPS with a few simple modifications or rather additions with their existing circuitry.

The lacking or missing changeover feature in an inverter can be upgraded by including a few number of relay stages within its circuit, as explained in the following sections:

Referring to the figure below, we see that the above requirement is implemented by using 4 SPDT relays whose coils are wired up in parallel and joined with a mains operated DC source, which could very well be the battery charger DC output.

It means during the presence of mains input the relays would be energized such that their N/O contacts get connected with the individual relay poles and the respective electrical gadgets which could be seen connected with the poles..

The left two relays can be seen with their N/O contacts connected with the mains AC input, while the N/Cs are terminated with the inverter mains output.

The relays at the right side have their N/O contacts rigged with battery charger (+)/(-) inputs, and the N/Cs are integrated to the inverter DC input.

The above data ensures the following actions during mains presence and failure situations:

When mains AC is present, the appliances get connected to the available mains power via the left pair of relay poles, while the battery is able to get the required charging voltage through the right hand relay poles. This also ensures that the inverter is cut-off via the N/C points from the battery and is no longer able to operate.

In a situation when mains AC fails, the relay contacts revert to their N/C contacts, giving rise to the following actions:

The battery instantly gets connected with the inverter DC input via the right hand side relay N/C contacts, such that the inverter becomes operative and its output starts producing the required mains back up voltage.

At the same instant the above inverter mains voltage now gets switched to the appliances via the left hand side relay N/C contacts ensuring that the appliances do not experience an interruption while the positions revert in the course of the above actions.

Selecting the Relays

The relays must selected with low coil resistance type so that they operate under higher switching currents, and therefore are able to "hammer" the contacts much harder and quicker compared to the lower resistance coil relays.

This will ensure the changeover time to be rapid within milliseconds which happens to be the most crucial factor with UPSs and inverters needing to be converted into UPS systems.

In the above diagram if an automatic battery charger is used, the supply would be cut off once the battery is fully charged, which would also cut off the supply to the relays forcing the inverter to switch ON even while the mains is present.

To avoid this issue, the relays must be powered through a separate power supply as shown in the following diagram. A capacitive type of power supply circuit could be seen here, which makes the design much compact.

Note: Please connect a 1K resistor across the filter capacitor associated with the bridge rectifier, this is to ensure its quick discharging during a mains failure, and an instant switching of the relevant relays.

Main Components and Their Functions:

Bridge Rectifier (1N4007 x 4)

So we have this bridge rectifier that consists of four 1N4007 diodes and its main job is to take the alternating current (AC) from the mains and convert it into direct current (DC). This conversion is really important because we need that DC power to make the relays work properly.

Filter Capacitor (100uF/100V)

Next up we are looking at the filter capacitor 100uF rated for 100 volts. These capacitors are essential because they help smooth out the rectified DC voltage making it much more stable and reliable for our circuit.

Relay Coils

Now let us talk about the relay coils. These coils function like switches that allow us to toggle between using mains power and inverter power. They play a crucial role in controlling how our entire system operates.

Formulas and Calculations:

DC Voltage Across the Relay Coil

We can figure out the DC voltage across the relay coil using this formula:

V_dc = (V_rms * √2) - V_f

In this equation the V_rms represents the input mains AC voltage (for example we might say it is 220 volts) and V_f is the total forward voltage drop across the bridge diodes. For our 1N4007 diodes we can calculate that as

V_f = 2 * 0.7V = 1.4V.

If we do a quick calculation for a 220V mains input, it would look like this:

V_dc = (220 * √2) - 1.4

V_dc = 310.6 - 1.4

V_dc = 309.2V

Capacitor Selection (Ripple Voltage)

When it comes to selecting capacitors we need to think about ripple voltage which we can calculate using this formula:

Δ_V = I_load / (f * C)

In this formula I_load is the load current drawn by the relay, f is the rectified AC frequency (which would be around 100 Hz if we are using a full-wave rectifier with a 50 Hz mains supply) and C represents the capacitance measured in Farads.

For example if we are working with a 100uF capacitor and our relay coil is drawing about 50mA of current then we can plug those values into our formula like this:

Δ_V = 0.05 / (100 * 0.0001)

Δ_V = 5V

Relay Coil Voltage

It is really important for us to ensure that the voltage rating of the relay coil matches up with the rectified DC voltage we calculated earlier. So if our rectified voltage is sitting at 309.2V then we need to make sure to use a relay that can handle that specific voltage.

Relay Switching Time

When we want to find out how quickly our relay can switch then we can use this formula:

t_s = L / R

Here L stands for the inductance of the relay coil and R represents the coil resistance.

For instance if we have L = 50mH and R = 500 ohms then we can calculate it like this:

t_s = 0.05 / 500

t_s = 0.1ms

Current Limiting by a Capacitor in AC Circuits

The current limited by the 2uF/400V capacitor is given by the formula:

I = V_rms / X_c

Where:

• I = Current through the capacitor (in amperes)
• V_rms = Mains AC RMS voltage (e.g., 220V)
• X_c = Capacitive reactance of the capacitor (in ohms)

The capacitive reactance is calculated as:

X_c = 1 / (2 * π * f * C)

Where:

f = Frequency of the mains AC (e.g., 50 Hz)

C = Capacitance (in farads)

Step-by-Step Calculation

Calculate X_c (Capacitive Reactance):

Given:

C = 2uF = 2 × 10^-6 F

f = 50 Hz

Formula: X_c = 1 / (2 * π * f * C)

Substitution: X_c = 1 / (2 * 3.1416 * 50 * 2 × 10^-6)

X_c = 1 / (0.00062832)

X_c ≈ 1591 ohms

Calculate Current (I):

Given:

V_rms = 220V

X_c ≈ 1591 ohms

Formula: I = V_rms / X_c

Substitution: I = 220 / 1591 I ≈ 0.138 amperes (or 138mA)

Design Considerations:

As we design our circuit we need to make sure that the diodes are capable of handling both peak inverse voltage (which will be equal to V_piv = V_dc) and also the current (which will be I_peak = I_load).

We should also select a relay that has a sufficient contact rating so it can safely manage whatever load we connect to it.

Lastly do not forget about that 1M resistor, its job is to discharge the filter capacitors when we disconnect power to prevent any accidental shocks.