# Overvoltage protection circuit transistor draws 2x load current then suddenly stablizes

I've built an over voltage protection circuit according to this schematic by Maxim:

If the load is low power consumption like MCU or other ICs, it works just fine.

If the load is higher for example, 350mA (maybe DCDC circuit,) the circuit starts to draw 2x current of the load (would be up to 700mA) in a constant speed upon power up. At the same time Q1 is generating heat (I guess this is where the extra current goes.) Once it reaches around 700mA it suddenly drops back to around 440mA then it is stable.

I initially thought it was the load circuit causing this, but if I power the load circuit without OVP it doesn't have this behavior.

Question:

1. Why does the current ramp up?
2. How can I avoid this?
3. In the stable condition is the 90mA difference consumed by Q1?
• Can you attach an oscillogram of your measurement where it does? Have you tried to simulate it? Does the simulation match reality? Mar 29, 2023 at 7:24

The circuit you are using is intended for low current use.

From the Maxim application note:

The measurements for Figure 3 were conducted using a sensor signal conditioning IC, the MAX1455, as the load circuit with a 1uF decoupling capacitor across the IC power-supply pins. The nominal current draw of the IC circuit during the test was 3mA.

At low current there is a voltage drop across Q1 of only a few millivolts.

At the far higher current you want to use in your DC-DC converter, the voltage drop will be much higher.

If the DC-DC converter needs 5V as an input voltage, then it will not get it while using this circuit. The voltage to the DC-DC converter will be closer to 4V (possibly lower) at start up. The converter will have to draw more current to try to get its output to the correct level - that will drop the input voltage even more. As the converter reaches steady state (its output voltage reaches its set level,) it will draw less current and the voltage drop across Q1 will be smaller. Because less voltage is dropped across Q1, the converter will be able to operate better - it will draw less current.

This circuit isn't made for what you are trying to do with it. It cannot work with high current, where "high current" means more than a few milliamperes.

It is possible that Q1 will start to oscillate in combination with the heavy load.

Q1 is essentially the same as the pass transistor of a low drop out linear regulator. Such regulators are notorious for oscillating when the capacitance on the output is not right (too much, too little, not enough series resistance, etc.) You might have similar problems with this circuit.

Wow, It's horrible looking at supply on the right, load on the left! Let me redraw it the way around I'm used to:

simulate this circuit – Schematic created using CircuitLab

This is never going to give you more than 200mA out, without a seriously diminished output voltage. That's because base current for Q1 is limited by R3. If the input voltage is +5V, then the voltage across R3 is 4.3V, for a current of around 600μA.

That particular model of transistor has very high gain β. Assuming β=300, maximum collector current will be:

$$I_C=\beta \times I_B = 300 \times 600\mu A = 180mA$$

A simulation will give you some idea of what the output voltage will do as we vary load current (I did not try to model the FMMT718 accurately, so this is approximate):

The point at which output voltage plummets is heavily dependent on Q1's current gain, β. Lower gain transistors will begin to switch off much earlier.

Don't expect this circuit to work well for load currents above this, which will cause Q1 to un-saturate, in the absence of sufficient base current to keep it saturated. Consequently, emitter-collector voltage $$\V_{CE}\$$ starts to rise, along with $$\P=IV\$$ power dissipation, causing it to get hot.

You could reduce R3, to get more base current. If, for instance, you require 500mA output, you must first calculate what minimum base current is needed:

$$I_B = \frac{I_C}{\beta} = \frac{0.5A}{300} = 1.7mA$$

We can never be sure of β, so to be safe, we'll double that, to about 3mA. Now calculate what value of R3 will pass 3mA with 4.3V across it:

$$R_3 = \frac{V}{I} = \frac{4.3V}{3mA} = 1.4k$$

With R3 = 1400Ω we do indeed increase the load current at which Q1 leaves saturation, but we find another problem:

$$\V_{CE}\$$ can rise significantly as collector current rises, and this is manifest as a drop in output voltage under heavier loads.

All these problems can be mitigated or avoided altogether by replacing Q1 with a P-channel MOSFET, and alter some resistances to suit:

simulate this circuit

Choose a MOSFET with low $$\R_{DS(ON)}\$$. For operation at 5V, you also need $$\V_{GS(TH)} < 2.5V\$$, to ensure that it's properly switched on. Above I set the MOSFET model's $$\V_{GS(TH)} = 2.0V \$$, to demonstrate a little quirk of behavior. The MOSFET can only begin to switch on when input voltage rises above $$\V_{GS(TH)}\$$. This is a plot of output voltage vs. input voltage:

As you can see, the cut-off behaviour, when input exceeds 6V, is still present, but the output is also suppressed for inputs less than the MOSFET's gate threshold.

• by switching to mosfet it would still consume a bit of power when connecting to a heavy load is it? Apr 12, 2023 at 6:51
• The right MOSFET will dissipate less power than any BJT. A BJT will always have base current, and will never have less than 0.1V across it. A MOSFET has no appreciable gate current, and could have significantly less than 0.1V across it, for the same load current. Apr 12, 2023 at 8:30
• If I may ask, could you go further and replace Q2 with a small PMOS as well? Wouldn't that also help with power consumption? Jan 8 at 14:42
• @BestQualityVacuum Yes, a P-channel MOSFET would work in the place of Q2, but there is consistency of $V_{BE}=0.7V$ across all models of BJT, and no such consistency in $V_{GS(TH)}$ between FETs. You'd find the circuit "tripping" at different voltages for different FETs, even from the same batch. Also, most power wasted here is in D1 and R1; Replacing Q2 only, with a MOSFET, wouldn't have much impact from an efficiency perspective, unless you also increased R1. Jan 8 at 14:51